College Trigonometry Version π Corrected Edition. July 4, 2013

College Trigonometr Version Corrected Edition b Carl Stitz, Ph.D. Lakeland Communit College Jeff Zeager, Ph.D. Lorain Count Communit College Jul, 0

ii Acknowledgements While the cover of this tetbook lists onl two names, the book as it stands toda would simpl not eist if not for the tireless work and dedication of several people. First and foremost, we wish to thank our families for their patience and support during the creative process. We would also like to thank our students - the sole inspiration for the work. Among our colleagues, we wish to thank Rich Basich, Bill Previts, and Irina Lomonosov, who not onl were earl adopters of the tetbook, but also contributed materials to the project. Special thanks go to Katie Cimperman, Terr Dkstra, Frank LeMa, and Rich Hagen who provided valuable feedback from the classroom. Thanks also to David Stumpf, Ivana Gorgievska, Jorge Gerszonowicz, Kathrn Arocho, Heather Bubnick, and Florin Muscutariu for their unwaivering support and sometimes defense of the book. From outside the classroom, we wish to thank Don Anthan and Ken White, who designed the electric circuit applications used in the tet, as well as Drs. Wend Marle and Marcia Ballinger for the Lorain CCC enrollment data used in the tet. The authors are also indebted to the good folks at our schools bookstores, Gwen Sevtis Lakeland CC and Chris Callahan Lorain CCC, for working with us to get printed copies to the students as inepensivel as possible. We would also like to thank Lakeland folks Jeri Dickinson, Mar Ann Blakele, Jessica Novak, and Corrie Bergeron for their enthusiasm and promotion of the project. The administrations at both schools have also been ver supportive of the project, so from Lakeland, we wish to thank Dr. Morris W. Beverage, Jr., President, Dr. Fred Law, Provost, Deans Don Anthan and Dr. Steve Oluic, and the Board of Trustees. From Lorain Count Communit College, we wish to thank Dr. Ro A. Church, Dr. Karen Wells, and the Board of Trustees. From the Ohio Board of Regents, we wish to thank former Chancellor Eric Fingerhut, Darlene McCo, Associate Vice Chancellor of Affordabilit and Efficienc, and Kell Bernard. From OhioLINK, we wish to thank Steve Acker, John Magill, and Stac Brannan. We also wish to thank the good folks at WebAssign, most notabl Chris Hall, COO, and Joel Hollenbeck former VP of Sales. Last, but certainl not least, we wish to thank all the folks who have contacted us over the interwebs, most notabl Dimitri Moonen and Joel Wordsworth, who gave us great feedback, and Antonio Olivares who helped debug the source code.

Table of Contents vii 0 Foundations of Trigonometr 69 0. Angles and their Measure................................ 69 0.. Applications of Radian Measure: Circular Motion.............. 706 0.. Eercises..................................... 709 0.. Answers...................................... 7 0. The Unit Circle: Cosine and Sine............................ 77 0.. Beond the Unit Circle............................. 70 0.. Eercises..................................... 76 0.. Answers...................................... 70 0. The Si Circular Functions and Fundamental Identities................ 7 0.. Beond the Unit Circle............................. 75 0.. Eercises..................................... 759 0.. Answers...................................... 766 0. Trigonometric Identities................................. 770 0.. Eercises..................................... 78 0.. Answers...................................... 787 0.5 Graphs of the Trigonometric Functions......................... 790 0.5. Graphs of the Cosine and Sine Functions................... 790 0.5. Graphs of the Secant and Cosecant Functions................ 800 0.5. Graphs of the Tangent and Cotangent Functions............... 80 0.5. Eercises..................................... 809 0.5.5 Answers...................................... 8 0.6 The Inverse Trigonometric Functions.......................... 89 0.6. Inverses of Secant and Cosecant: Trigonometr Friendl Approach..... 87 0.6. Inverses of Secant and Cosecant: Calculus Friendl Approach........ 80 0.6. Calculators and the Inverse Circular Functions................. 8 0.6. Solving Equations Using the Inverse Trigonometric Functions........ 88 0.6.5 Eercises..................................... 8 0.6.6 Answers...................................... 89 0.7 Trigonometric Equations and Inequalities....................... 857 0.7. Eercises..................................... 87 0.7. Answers...................................... 877 Applications of Trigonometr 88. Applications of Sinusoids................................. 88.. Harmonic Motion................................ 885.. Eercises..................................... 89.. Answers...................................... 89. The Law of Sines..................................... 896.. Eercises..................................... 90.. Answers...................................... 908. The Law of Cosines.................................... 90

viii Table of Contents.. Eercises..................................... 96.. Answers...................................... 98. Polar Coordinates..................................... 99.. Eercises..................................... 90.. Answers...................................... 9.5 Graphs of Polar Equations................................ 98.5. Eercises..................................... 958.5. Answers...................................... 96.6 Hooked on Conics Again................................. 97.6. Rotation of Aes................................. 97.6. The Polar Form of Conics............................ 98.6. Eercises..................................... 986.6. Answers...................................... 987.7 Polar Form of Comple Numbers............................ 99.7. Eercises..................................... 00.7. Answers...................................... 007.8 Vectors........................................... 0.8. Eercises..................................... 07.8. Answers...................................... 0.9 The Dot Product and Projection............................ 0.9. Eercises..................................... 0.9. Answers...................................... 05.0 Parametric Equations.................................. 08.0. Eercises..................................... 059.0. Answers...................................... 06 Inde 069

Preface Thank ou for our interest in our book, but more importantl, thank ou for taking the time to read the Preface. I alwas read the Prefaces of the tetbooks which I use in m classes because I believe it is in the Preface where I begin to understand the authors - who the are, what their motivation for writing the book was, and what the hope the reader will get out of reading the tet. Pedagogical issues such as content organization and how professors and students should best use a book can usuall be gleaned out of its Table of Contents, but the reasons behind the choices authors make should be shared in the Preface. Also, I feel that the Preface of a tetbook should demonstrate the authors love of their discipline and passion for teaching, so that I come awa believing that the reall want to help students and not just make mone. Thus, I thank m fellow Preface-readers again for giving me the opportunit to share with ou the need and vision which guided the creation of this book and passion which both Carl and I hold for Mathematics and the teaching of it. Carl and I are natives of Northeast Ohio. We met in graduate school at Kent State Universit in 997. I finished m Ph.D in Pure Mathematics in August 998 and started teaching at Lorain Count Communit College in Elria, Ohio just two das after graduation. Carl earned his Ph.D in Pure Mathematics in August 000 and started teaching at Lakeland Communit College in Kirtland, Ohio that same month. Our schools are fairl similar in size and mission and each serves a similar population of students. The students range in age from about 6 Ohio has a Post-Secondar Enrollment Option program which allows high school students to take college courses for free while still in high school. to over 65. Man of the non-traditional students are returning to school in order to change careers. A majorit of the students at both schools receive some sort of financial aid, be it scholarships from the schools foundations, state-funded grants or federal financial aid like student loans, and man of them have lives busied b famil and job demands. Some will be taking their Associate degrees and entering or re-entering the workforce while others will be continuing on to a four-ear college or universit. Despite their man differences, our students share one common attribute: the do not want to spend $00 on a College Algebra book. The challenge of reducing the cost of tetbooks is one that man states, including Ohio, are taking quite seriousl. Indeed, state-level leaders have started to work with facult from several of the colleges and universities in Ohio and with the major publishers as well. That process will take considerable time so Carl and I came up with a plan of our own. We decided that the best wa to help our students right now was to write our own College Algebra book and give it awa electronicall for free. We were granted sabbaticals from our respective institutions for the Spring

Preface semester of 009 and actuall began writing the tetbook on December 6, 008. Using an opensource tet editor called TeNicCenter and an open-source distribution of LaTeX called MikTe.7, Carl and I wrote and edited all of the tet, eercises and answers and created all of the graphs using Metapost within LaTeX for Version 0.9 in about eight months. We choose to create a tet in onl black and white to keep printing costs to a minimum for those students who prefer a printed edition. This somewhat Spartan page laout stands in sharp relief to the eplosion of colors found in most other College Algebra tets, but neither Carl nor I believe the four-color print adds anthing of value. I used the book in three sections of College Algebra at Lorain Count Communit College in the Fall of 009 and Carl s colleague, Dr. Bill Previts, taught a section of College Algebra at Lakeland with the book that semester as well. Students had the option of downloading the book as a.pdf file from our website www.stitz-zeager.com or buing a low-cost printed version from our colleges respective bookstores. B giving this book awa for free electronicall, we end the ccle of new editions appearing ever 8 months to curtail the used book market. During Thanksgiving break in November 009, man additional eercises written b Dr. Previts were added and the tpographical errors found b our students and others were corrected. On December 0, 009, Version was released. The book remains free for download at our website and b using Lulu.com as an on-demand printing service, our bookstores are now able to provide a printed edition for just under $9. Neither Carl nor I have, or will ever, receive an roalties from the printed editions. As a contribution back to the open-source communit, all of the LaTeX files used to compile the book are available for free under a Creative Commons License on our website as well. That wa, anone who would like to rearrange or edit the content for their classes can do so as long as it remains free. The onl disadvantage to not working for a publisher is that we don t have a paid editorial staff. What we have instead, beond ourselves, is friends, colleagues and unknown people in the opensource communit who alert us to errors the find as the read the tetbook. What we gain in not having to report to a publisher so dramaticall outweighs the lack of the paid staff that we have turned down ever offer to publish our book. As of the writing of this Preface, we ve had three offers. B maintaining this book b ourselves, Carl and I retain all creative control and keep the book our own. We control the organization, depth and rigor of the content which means we can resist the pressure to diminish the rigor and homogenize the content so as to appeal to a mass market. A casual glance through the Table of Contents of most of the major publishers College Algebra books reveals nearl isomorphic content in both order and depth. Our Table of Contents shows a different approach, one that might be labeled Functions First. To trul use The Rule of Four, that is, in order to discuss each new concept algebraicall, graphicall, numericall and verball, it seems completel obvious to us that one would need to introduce functions first. Take a moment and compare our ordering to the classic equations first, then the Cartesian Plane and THEN functions approach seen in most of the major plaers. We then introduce a class of functions and discuss the equations, inequalities with a heav emphasis on sign diagrams and applications which involve functions in that class. The material is presented at a level that definitel prepares a student for Calculus while giving them relevant Mathematics which can be used in other classes as well. Graphing calculators are used sparingl and onl as a tool to enhance the Mathematics, not to replace it. The answers to nearl all of the computational homework eercises are given in the

i tet and we have gone to great lengths to write some ver thought provoking discussion questions whose answers are not given. One will notice that our eercise sets are much shorter than the traditional sets of nearl 00 drill and kill questions which build skill devoid of understanding. Our eperience has been that students can do about 5-0 homework eercises a night so we ver carefull chose smaller sets of questions which cover all of the necessar skills and get the students thinking more deepl about the Mathematics involved. Critics of the Open Educational Resource movement might quip that open-source is where bad content goes to die, to which I sa this: take a serious look at what we offer our students. Look through a few sections to see if what we ve written is bad content in our opinion. I see this opensource book not as something which is free and worth ever penn, but rather, as a high qualit alternative to the business as usual of the tetbook industr and I hope that ou agree. If ou have an comments, questions or concerns please feel free to contact me at jeff@stitz-zeager.com or Carl at carl@stitz-zeager.com. Jeff Zeager Lorain Count Communit College Januar 5, 00

ii Preface

Chapter 0 Foundations of Trigonometr 0. Angles and their Measure This section begins our stud of Trigonometr and to get started, we recall some basic definitions from Geometr. A ra is usuall described as a half-line and can be thought of as a line segment in which one of the two endpoints is pushed off infinitel distant from the other, as pictured below. The point from which the ra originates is called the initial point of the ra. P A ra with initial point P. When two ras share a common initial point the form an angle and the common initial point is called the verte of the angle. Two eamples of what are commonl thought of as angles are P An angle with verte P. Q An angle with verte Q. However, the two figures below also depict angles - albeit these are, in some sense, etreme cases. In the first case, the two ras are directl opposite each other forming what is known as a straight angle; in the second, the ras are identical so the angle is indistinguishable from the ra itself. P A straight angle. The measure of an angle is a number which indicates the amount of rotation that separates the ras of the angle. There is one immediate problem with this, as pictured below. Q

69 Foundations of Trigonometr Which amount of rotation are we attempting to quantif? What we have just discovered is that we have at least two angles described b this diagram. Clearl these two angles have different measures because one appears to represent a larger rotation than the other, so we must label them differentl. In this book, we use lower case Greek letters such as α alpha, β beta, γ gamma and θ theta to label angles. So, for instance, we have β α One commonl used sstem to measure angles is degree measure. Quantities measured in degrees are denoted b the familiar smbol. One complete revolution as shown below is 60, and parts of a revolution are measured proportionatel. Thus half of a revolution a straight angle measures 60 = 80, a quarter of a revolution a right angle measures 60 = 90 and so on. One revolution 60 80 90 Note that in the above figure, we have used the small square to denote a right angle, as is commonplace in Geometr. Recall that if an angle measures strictl between 0 and 90 it is called an acute angle and if it measures strictl between 90 and 80 it is called an obtuse angle. It is important to note that, theoreticall, we can know the measure of an angle as long as we The phrase at least will be justified in short order. The choice of 60 is most often attributed to the Bablonians.

0. Angles and their Measure 695 know the proportion it represents of entire revolution. For instance, the measure of an angle which represents a rotation of of a revolution would measure 60 = 0, the measure of an angle which constitutes onl of a revolution measures 60 = 0 and an angle which indicates no rotation at all is measured as 0. 0 0 0 Using our definition of degree measure, we have that represents the measure of an angle which constitutes 60 of a revolution. Even though it ma be hard to draw, it is nonetheless not difficult to imagine an angle with measure smaller than. There are two was to subdivide degrees. The first, and most familiar, is decimal degrees. For eample, an angle with a measure of 0.5 would represent a rotation halfwa between 0 and, or equivalentl, 0.5 60 = 6 70 of a full rotation. This can be taken to the limit using Calculus so that measures like make sense. The second wa to divide degrees is the Degree - Minute - Second DMS sstem. In this sstem, one degree is divided equall into sit minutes, and in turn, each minute is divided equall into sit seconds. 5 In smbols, we write = 60 and = 60, from which it follows that = 600. To convert a measure of.5 to the DMS sstem, we start b noting that.5 = + 0.5. Converting the partial amount of degrees to minutes, we find 0.5 60 = 7.5 = 7 + 0.5. Converting the partial amount of minutes to seconds gives 0.5 60 = 0. Putting it all together ields.5 = + 0.5 = + 7.5 = + 7 + 0.5 = + 7 + 0 = 7 0 On the other hand, to convert 7 5 5 to decimal degrees, we first compute 5 60 = and 5 600 = 80. Then we find This is how a protractor is graded. Awesome math pun aside, this is the same idea behind defining irrational eponents in Section 6.. 5 Does this kind of sstem seem familiar?

696 Foundations of Trigonometr 7 5 5 = 7 + 5 + 5 = 7 + + 80 = 98 80 = 7.65 Recall that two acute angles are called complementar angles if their measures add to 90. Two angles, either a pair of right angles or one acute angle and one obtuse angle, are called supplementar angles if their measures add to 80. In the diagram below, the angles α and β are supplementar angles while the pair γ and θ are complementar angles. β θ α γ Supplementar Angles Complementar Angles In practice, the distinction between the angle itself and its measure is blurred so that the sentence α is an angle measuring is often abbreviated as α =. It is now time for an eample. Eample 0... Let α =.7 and β = 7 8 7.. Convert α to the DMS sstem. Round our answer to the nearest second.. Convert β to decimal degrees. Round our answer to the nearest thousandth of a degree.. Sketch α and β.. Find a supplementar angle for α. 5. Find a complementar angle for β. Solution.. To convert α to the DMS sstem, we start with.7 = + 0.7. Net we convert 0.7 60 =.6. Writing.6 = + 0.6, we convert 0.6 60 = 5.6. Hence,.7 = + 0.7 = +.6 = + + 0.6 = + + 5.6 = 5.6 Rounding to seconds, we obtain α 6.

0. Angles and their Measure 697. To convert β to decimal degrees, we convert 8 60 = 7 it all together, we have 5 7 8 7 = 7 + 8 + 7 = 7 + 7 = 897 600 7.7 5 and 7 600 = 7 600. Putting + 7 600. To sketch α, we first note that 90 < α < 80. If we divide this range in half, we get 90 < α < 5, and once more, we have 90 < α <.5. This gives us a prett good estimate for α, as shown below. 6 Proceeding similarl for β, we find 0 < β < 90, then 0 < β < 5,.5 < β < 5, and lastl,.75 < β < 5. Angle α Angle β. To find a supplementar angle for α, we seek an angle θ so that α + θ = 80. We get θ = 80 α = 80.7 = 68.69. 5. To find a complementar angle for β, we seek an angle γ so that β + γ = 90. We get γ = 90 β = 90 7 8 7. While we could reach for the calculator to obtain an approimate answer, we choose instead to do a bit of seagesimal 7 arithmetic. We first rewrite 90 = 90 0 0 = 89 60 0 = 89 59 60. In essence, we are borrowing = 60 from the degree place, and then borrowing = 60 from the minutes place. 8 This ields, γ = 90 7 8 7 = 89 59 60 7 8 7 = 5. Up to this point, we have discussed onl angles which measure between 0 and 60, inclusive. Ultimatel, we want to use the arsenal of Algebra which we have stockpiled in Chapters through 9 to not onl solve geometric problems involving angles, but also to etend their applicabilit to other real-world phenomena. A first step in this direction is to etend our notion of angle from merel measuring an etent of rotation to quantities which can be associated with real numbers. To that end, we introduce the concept of an oriented angle. As its name suggests, in an oriented 6 If this process seems hauntingl familiar, it should. Compare this method to the Bisection Method introduced in Section.. 7 Like latus rectum, this is also a real math term. 8 This is the eact same kind of borrowing ou used to do in Elementar School when tring to find 00 5. Back then, ou were working in a base ten sstem; here, it is base sit.

698 Foundations of Trigonometr angle, the direction of the rotation is important. We imagine the angle being swept out starting from an initial side and ending at a terminal side, as shown below. When the rotation is counter-clockwise 9 from initial side to terminal side, we sa that the angle is positive; when the rotation is clockwise, we sa that the angle is negative. Terminal Side Initial Side Terminal Side Initial Side A positive angle, 5 A negative angle, 5 At this point, we also etend our allowable rotations to include angles which encompass more than one revolution. For eample, to sketch an angle with measure 50 we start with an initial side, rotate counter-clockwise one complete revolution to take care of the first 60 then continue with an additional 90 counter-clockwise rotation, as seen below. 50 To further connect angles with the Algebra which has come before, we shall often overla an angle diagram on the coordinate plane. An angle is said to be in standard position if its verte is the origin and its initial side coincides with the positive -ais. Angles in standard position are classified according to where their terminal side lies. For instance, an angle in standard position whose terminal side lies in Quadrant I is called a Quadrant I angle. If the terminal side of an angle lies on one of the coordinate aes, it is called a quadrantal angle. Two angles in standard position are called coterminal if the share the same terminal side. 0 In the figure below, α = 0 and β = 0 are two coterminal Quadrant II angles drawn in standard position. Note that α = β + 60, or equivalentl, β = α 60. We leave it as an eercise to the reader to verif that coterminal angles alwas differ b a multiple of 60. More precisel, if α and β are coterminal angles, then β = α + 60 k where k is an integer. 9 widdershins 0 Note that b being in standard position the automaticall share the same initial side which is the positive -ais. It is worth noting that all of the pathologies of Analtic Trigonometr result from this innocuous fact. Recall that this means k = 0, ±, ±,....

0. Angles and their Measure 699 α = 0 β = 0 Two coterminal angles, α = 0 and β = 0, in standard position. Eample 0... Graph each of the oriented angles below in standard position and classif them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative.. α = 60. β = 5. γ = 50. φ = 750 Solution.. To graph α = 60, we draw an angle with its initial side on the positive -ais and rotate counter-clockwise 60 60 = 6 of a revolution. We see that α is a Quadrant I angle. To find angles which are coterminal, we look for angles θ of the form θ = α + 60 k, for some integer k. When k =, we get θ = 60 +60 = 0. Substituting k = gives θ = 60 60 = 00. Finall, if we let k =, we get θ = 60 + 70 = 780.. Since β = 5 is negative, we start at the positive -ais and rotate clockwise 5 60 = 5 8 of a revolution. We see that β is a Quadrant II angle. To find coterminal angles, we proceed as before and compute θ = 5 + 60 k for integer values of k. We find 5, 585 and 95 are all coterminal with 5. α = 60 β = 5 α = 60 in standard position. β = 5 in standard position.

700 Foundations of Trigonometr. Since γ = 50 is positive, we rotate counter-clockwise from the positive -ais. One full revolution accounts for 60, with 80, or of a revolution remaining. Since the terminal side of γ lies on the negative -ais, γ is a quadrantal angle. All angles coterminal with γ are of the form θ = 50 + 60 k, where k is an integer. Working through the arithmetic, we find three such angles: 80, 80 and 900.. The Greek letter φ is pronounced fee or fie and since φ is negative, we begin our rotation clockwise from the positive -ais. Two full revolutions account for 70, with just 0 or of a revolution to go. We find that φ is a Quadrant IV angle. To find coterminal angles, we compute θ = 750 + 60 k for a few integers k and obtain 90, 0 and 0. γ = 50 φ = 750 γ = 50 in standard position. φ = 750 in standard position. Note that since there are infinitel man integers, an given angle has infinitel man coterminal angles, and the reader is encouraged to plot the few sets of coterminal angles found in Eample 0.. to see this. We are now just one step awa from completel marring angles with the real numbers and the rest of Algebra. To that end, we recall this definition from Geometr. Definition 0.. The real number is defined to be the ratio of a circle s circumference to its diameter. In smbols, given a circle of circumference C and diameter d, = C d While Definition 0. is quite possibl the standard definition of, the authors would be remiss if we didn t mention that buried in this definition is actuall a theorem. As the reader is probabl aware, the number is a mathematical constant - that is, it doesn t matter which circle is selected, the ratio of its circumference to its diameter will have the same value as an other circle. While this is indeed true, it is far from obvious and leads to a counterintuitive scenario which is eplored in the Eercises. Since the diameter of a circle is twice its radius, we can quickl rearrange the equation in Definition 0. to get a formula more useful for our purposes, namel: = C r

0. Angles and their Measure 70 This tells us that for an circle, the ratio of its circumference to its radius is also alwas constant; in this case the constant is. Suppose now we take a portion of the circle, so instead of comparing the entire circumference C to the radius, we compare some arc measuring s units in length to the radius, as depicted below. Let θ be the central angle subtended b this arc, that is, an angle whose verte is the center of the circle and whose determining ras pass through the endpoints of the arc. Using proportionalit arguments, it stands to reason that the ratio s should also be a r constant among all circles, and it is this ratio which defines the radian measure of an angle. s r θ r The radian measure of θ is s r. To get a better feel for radian measure, we note that an angle with radian measure means the corresponding arc length s equals the radius of the circle r, hence s = r. When the radian measure is, we have s = r; when the radian measure is, s = r, and so forth. Thus the radian measure of an angle θ tells us how man radius lengths we need to sweep out along the circle to subtend the angle θ. r r r α r r r β r r r α has radian measure β has radian measure Since one revolution sweeps out the entire circumference r, one revolution has radian measure r =. From this we can find the radian measure of other central angles using proportions, r

70 Foundations of Trigonometr just like we did with degrees. For instance, half of a revolution has radian measure =, a quarter revolution has radian measure =, and so forth. Note that, b definition, the radian measure of an angle is a length divided b another length so that these measurements are actuall dimensionless and are considered pure numbers. For this reason, we do not use an smbols to denote radian measure, but we use the word radians to denote these dimensionless units as needed. For instance, we sa one revolution measures radians, half of a revolution measures radians, and so forth. As with degree measure, the distinction between the angle itself and its measure is often blurred in practice, so when we write θ =, we mean θ is an angle which measures radians. We etend radian measure to oriented angles, just as we did with degrees beforehand, so that a positive measure indicates counter-clockwise rotation and a negative measure indicates clockwise rotation. Much like before, two positive angles α and β are supplementar if α + β = and complementar if α + β =. Finall, we leave it to the reader to show that when using radian measure, two angles α and β are coterminal if and onl if β = α + k for some integer k. Eample 0... Graph each of the oriented angles below in standard position and classif them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative.. α = 6. β =. γ = 9. φ = 5 Solution.. The angle α = 6 is positive, so we draw an angle with its initial side on the positive -ais and rotate counter-clockwise /6 = of a revolution. Thus α is a Quadrant I angle. Coterminal angles θ are of the form θ = α + k, for some integer k. To make the arithmetic a bit easier, we note that = 6, thus when k =, we get θ = 6 + 6 = 6. Substituting k = gives θ = 6 6 = 6 and when we let k =, we get θ = 6 + 6 = 5 6.. Since β = / is negative, we start at the positive -ais and rotate clockwise = of a revolution. We find β to be a Quadrant II angle. To find coterminal angles, we proceed as before using = 6, and compute θ = + 6 k for integer values of k. We obtain, 0 and 8 as coterminal angles. The authors are well aware that we are now identifing radians with real numbers. We will justif this shortl. This, in turn, endows the subtended arcs with an orientation as well. We address this in short order.

0. Angles and their Measure 70 α = 6 β = α = 6 in standard position. β = in standard position.. Since γ = 9 is positive, we rotate counter-clockwise from the positive -ais. One full revolution accounts for = 8 of the radian measure with or 8 of a revolution remaining. We have γ as a Quadrant I angle. All angles coterminal with γ are of the form θ = 9 + 8 k, where k is an integer. Working through the arithmetic, we find:, 7 7 and.. To graph φ = 5, we begin our rotation clockwise from the positive -ais. As =, after one full revolution clockwise, we have or of a revolution remaining. Since the terminal side of φ lies on the negative -ais, φ is a quadrantal angle. To find coterminal angles, we compute θ = 5 + k for a few integers k and obtain, and 7. φ = 5 γ = 9 γ = 9 in standard position. φ = 5 in standard position. It is worth mentioning that we could have plotted the angles in Eample 0.. b first converting them to degree measure and following the procedure set forth in Eample 0... While converting back and forth from degrees and radians is certainl a good skill to have, it is best that ou learn to think in radians as well as ou can think in degrees. The authors would, however, be

70 Foundations of Trigonometr derelict in our duties if we ignored the basic conversion between these sstems altogether. Since one revolution counter-clockwise measures 60 and the same angle measures radians, we can radians use the proportion 60, or its reduced equivalent, radians 80, as the conversion factor between the two sstems. For eample, to convert 60 to radians we find 60 radians 80 = radians, or simpl 80. To convert from radian measure back to degrees, we multipl b the ratio radian. For eample, 5 6 radians is equal to 5 6 radians 80 radians = 50. 5 Of particular interest is the fact that an angle which measures in radian measure is equal to 80 57.958. We summarize these conversions below. Equation 0.. Degree - Radian Conversion: ˆ ˆ To convert degree measure to radian measure, multipl b radians 80 To convert radian measure to degree measure, multipl b 80 radians In light of Eample 0.. and Equation 0., the reader ma well wonder what the allure of radian measure is. The numbers involved are, admittedl, much more complicated than degree measure. The answer lies in how easil angles in radian measure can be identified with real numbers. Consider the Unit Circle, + =, as drawn below, the angle θ in standard position and the corresponding arc measuring s units in length. B definition, and the fact that the Unit Circle has radius, the radian measure of θ is s r = s = s so that, once again blurring the distinction between an angle and its measure, we have θ = s. In order to identif real numbers with oriented angles, we make good use of this fact b essentiall wrapping the real number line around the Unit Circle and associating to each real number t an oriented arc on the Unit Circle with initial point, 0. Viewing the vertical line = as another real number line demarcated like the -ais, given a real number t > 0, we wrap the vertical interval [0, t] around the Unit Circle in a counter-clockwise fashion. The resulting arc has a length of t units and therefore the corresponding angle has radian measure equal to t. If t < 0, we wrap the interval [t, 0] clockwise around the Unit Circle. Since we have defined clockwise rotation as having negative radian measure, the angle determined b this arc has radian measure equal to t. If t = 0, we are at the point, 0 on the -ais which corresponds to an angle with radian measure 0. In this wa, we identif each real number t with the corresponding angle with radian measure t. 5 Note that the negative sign indicates clockwise rotation in both sstems, and so it is carried along accordingl.

0. Angles and their Measure 705 θ s t t t t On the Unit Circle, θ = s. Identifing t > 0 with an angle. Identifing t < 0 with an angle. Eample 0... Sketch the oriented arc on the Unit Circle corresponding to each of the following real numbers.. t =. t =. t =. t = 7 Solution.. The arc associated with t = is the arc on the Unit Circle which subtends the angle in radian measure. Since is 8 of a revolution, we have an arc which begins at the point, 0 proceeds counter-clockwise up to midwa through Quadrant II.. Since one revolution is radians, and t = is negative, we graph the arc which begins at, 0 and proceeds clockwise for one full revolution. t = t =. Like t =, t = is negative, so we begin our arc at, 0 and proceed clockwise around the unit circle. Since. and.57, we find that rotating radians clockwise from the point, 0 lands us in Quadrant III. To more accuratel place the endpoint, we proceed as we did in Eample 0.., successivel halving the angle measure until we find 5 8.96 which tells us our arc etends just a bit beond the quarter mark into Quadrant III.

706 Foundations of Trigonometr. Since 7 is positive, the arc corresponding to t = 7 begins at, 0 and proceeds counterclockwise. As 7 is much greater than, we wrap around the Unit Circle several times before finall reaching our endpoint. We approimate 7 as 8.6 which tells us we complete 8 revolutions counter-clockwise with 0.6, or just sh of 5 8 of a revolution to spare. In other words, the terminal side of the angle which measures 7 radians in standard position is just short of being midwa through Quadrant III. t = t = 7 0.. Applications of Radian Measure: Circular Motion Now that we have paired angles with real numbers via radian measure, a whole world of applications awaits us. Our first ecursion into this realm comes b wa of circular motion. Suppose an object is moving as pictured below along a circular path of radius r from the point P to the point Q in an amount of time t. Q r θ s P Here s represents a displacement so that s > 0 means the object is traveling in a counter-clockwise direction and s < 0 indicates movement in a clockwise direction. Note that with this convention the formula we used to define radian measure, namel θ = s, still holds since a negative value r of s incurred from a clockwise displacement matches the negative we assign to θ for a clockwise rotation. In Phsics, the average velocit of the object, denoted v and read as v-bar, is defined as the average rate of change of the position of the object with respect to time. 6 As a result, we 6 See Definition. in Section. for a review of this concept.

0. Angles and their Measure 707 have v = displacement time = s length. The quantit v has units of t time and conves two ideas: the direction in which the object is moving and how fast the position of the object is changing. The contribution of direction in the quantit v is either to make it positive in the case of counter-clockwise motion or negative in the case of clockwise motion, so that the quantit v quantifies how fast the object is moving - it is the speed of the object. Measuring θ in radians we have θ = s r v = s t = rθ t = r θ t thus s = rθ and The quantit θ is called the average angular velocit of the object. It is denoted b ω and is t read omega-bar. The quantit ω is the average rate of change of the angle θ with respect to time and thus has units radians time. If ω is constant throughout the duration of the motion, then it can be shown 7 that the average velocities involved, namel v and ω, are the same as their instantaneous counterparts, v and ω, respectivel. In this case, v is simpl called the velocit of the object and is the instantaneous rate of change of the position of the object with respect to time. 8 Similarl, ω is called the angular velocit and is the instantaneous rate of change of the angle with respect to time. If the path of the object were uncurled from a circle to form a line segment, then the velocit of the object on that line segment would be the same as the velocit on the circle. For this reason, the quantit v is often called the linear velocit of the object in order to distinguish it from the angular velocit, ω. Putting together the ideas of the previous paragraph, we get the following. Equation 0.. Velocit for Circular Motion: For an object moving on a circular path of radius r with constant angular velocit ω, the linear velocit of the object is given b v = rω. We need to talk about units here. The units of v are length time, the units of r are length onl, and the units of ω are radians length time. Thus the left hand side of the equation v = rω has units time, whereas the right hand side has units length radians time = length radians time. The supposed contradiction in units is resolved b remembering that radians are a dimensionless quantit and angles in radian measure are identified with real numbers so that the units length radians time reduce to the units length time. We are long overdue for an eample. Eample 0..5. Assuming that the surface of the Earth is a sphere, an point on the Earth can be thought of as an object traveling on a circle which completes one revolution in approimatel hours. The path traced out b the point during this hour period is the Latitude of that point. Lakeland Communit College is at.68 north latitude, and it can be shown 9 that the radius of the earth at this Latitude is approimatel 960 miles. Find the linear velocit, in miles per hour, of Lakeland Communit College as the world turns. Solution. To use the formula v = rω, we first need to compute the angular velocit ω. The earth radians makes one revolution in hours, and one revolution is radians, so ω = hours = hours, 7 You guessed it, using Calculus... 8 See the discussion on Page 6 for more details on the idea of an instantaneous rate of change. 9 We will discuss how we arrived at this approimation in Eample 0..6.

708 Foundations of Trigonometr where, once again, we are using the fact that radians are real numbers and are dimensionless. For simplicit s sake, we are also assuming that we are viewing the rotation of the earth as counterclockwise so ω > 0. Hence, the linear velocit is v = 960 miles miles 775 hours hour It is worth noting that the quantit revolution hours in Eample 0..5 is called the ordinar frequenc of the motion and is usuall denoted b the variable f. The ordinar frequenc is a measure of how often an object makes a complete ccle of the motion. The fact that ω = f suggests that ω is also a frequenc. Indeed, it is called the angular frequenc of the motion. On a related note, the quantit T = is called the period of the motion and is the amount of time it takes for the f object to complete one ccle of the motion. In the scenario of Eample 0..5, the period of the motion is hours, or one da. The concepts of frequenc and period help frame the equation v = rω in a new light. That is, if ω is fied, points which are farther from the center of rotation need to travel faster to maintain the same angular frequenc since the have farther to travel to make one revolution in one period s time. The distance of the object to the center of rotation is the radius of the circle, r, and is the magnification factor which relates ω and v. We will have more to sa about frequencies and periods in Section.. While we have ehaustivel discussed velocities associated with circular motion, we have et to discuss a more natural question: if an object is moving on a circular path of radius r with a fied angular velocit frequenc ω, what is the position of the object at time t? The answer to this question is the ver heart of Trigonometr and is answered in the net section.

0. Angles and their Measure 709 0.. Eercises In Eercises -, convert the angles into the DMS sstem. Round each of our answers to the nearest second.. 6.75. 00.5. 7.06. 79.999 In Eercises 5-8, convert the angles into decimal degrees. Round each of our answers to three decimal places. 5. 5 50 6. 0 7. 50 5 8. 7 58 In Eercises 9-8, graph the oriented angle in standard position. Classif each angle according to where its terminal side lies and then give two coterminal angles, one of which is positive and the other negative. 9. 0 0. 5. 0. 05. 70. 7.. 5 6 8.. 7 6 5. 6. 5. 9. 7. 5 7. 5 6. 0. 5. 8. 6 In Eercises 9-6, convert the angle from degree measure into radian measure, giving the eact value in terms of. 9. 0 0. 0. 5. 70. 5. 50 5. 5 6. 5 In Eercises 7 -, convert the angle from radian measure into degree measure. 7. 8. 9. 7 6 0. 6.. 5. 6.

70 Foundations of Trigonometr In Eercises 5-9, sketch the oriented arc on the Unit Circle which corresponds to the given real number. 5. t = 5 6 6. t = 7. t = 6 8. t = 9. t = 50. A o-o which is.5 inches in diameter spins at a rate of 500 revolutions per minute. How fast is the edge of the o-o spinning in miles per hour? Round our answer to two decimal places. 5. How man revolutions per minute would the o-o in eercise 50 have to complete if the edge of the o-o is to be spinning at a rate of miles per hour? Round our answer to two decimal places. 5. In the o-o trick Around the World, the performer throws the o-o so it sweeps out a vertical circle whose radius is the o-o string. If the o-o string is 8 inches long and the o-o takes seconds to complete one revolution of the circle, compute the speed of the o-o in miles per hour. Round our answer to two decimal places. 5. A computer hard drive contains a circular disk with diameter.5 inches and spins at a rate of 700 RPM revolutions per minute. Find the linear speed of a point on the edge of the disk in miles per hour. 5. A rock got stuck in the tread of m tire and when I was driving 70 miles per hour, the rock came loose and hit the inside of the wheel well of the car. How fast, in miles per hour, was the rock traveling when it came out of the tread? The tire has a diameter of inches. 55. The Giant Wheel at Cedar Point is a circle with diameter 8 feet which sits on an 8 foot tall platform making its overall height is 6 feet. Remember this from Eercise 7 in Section 7.? It completes two revolutions in minutes and 7 seconds. 0 Assuming the riders are at the edge of the circle, how fast are the traveling in miles per hour? 56. Consider the circle of radius r pictured below with central angle θ, measured in radians, and subtended arc of length s. Prove that the area of the shaded sector is A = r θ. Hint: Use the proportion A area of the circle = s circumference of the circle. r θ s r 0 Source: Cedar Point s webpage.

0. Angles and their Measure 7 In Eercises 57-6, use the result of Eercise 56 to compute the areas of the circular sectors with the given central angles and radii. 57. θ = 6, r = 58. θ = 5, r = 00 59. θ = 0, r = 9. 60. θ =, r = 6. θ = 0, r = 5 6. θ =, r = 7 6. Imagine a rope tied around the Earth at the equator. Show that ou need to add onl feet of length to the rope in order to lift it one foot above the ground around the entire equator. You do NOT need to know the radius of the Earth to show this. 6. With the help of our classmates, look for a proof that is indeed a constant.

7 Foundations of Trigonometr 0.. Answers. 6 5. 00 9 0. 7 6. 79 59 56 5. 5.8 6..7 7. 50.58 8. 7.979 9. 0 is a Quadrant IV angle coterminal with 690 and 0 0. 5 is a Quadrant III angle coterminal with 5 and 95. 0 is a Quadrant II angle coterminal with 80 and 0. 05 is a Quadrant I angle coterminal with 5 and 5. 70 lies on the positive -ais coterminal with 90 and 60. 5 is a Quadrant II angle 6 coterminal with 7 6 and 7 6

0. Angles and their Measure 7 5. is a Quadrant I angle coterminal with and 5 6. 5 is a Quadrant III angle coterminal with and 7. is a Quadrant II angle coterminal with and 5 8. is a Quadrant IV angle coterminal with 5 and 7 9. 7 lies on the negative -ais 0. is a Quadrant I angle coterminal with and coterminal with 9 and 7

7 Foundations of Trigonometr. lies on the negative -ais coterminal with and 5. 7 is a Quadrant III angle 6 coterminal with 9 6 and 5 6. 5 is a Quadrant I angle. lies on the negative -ais coterminal with and coterminal with and 5. lies on the positive -ais 6. is a Quadrant IV angle coterminal with and coterminal with 7 and 9

0. Angles and their Measure 75 7. 5 is a Quadrant IV angle coterminal with 7 and 8. is a Quadrant IV angle 6 coterminal with 6 and 6 9. 0 0.. 7. 5 6. 5.. 6. 5 7. 80 8. 0 9. 0 0. 0. 60. 00. 0. 90 5. t = 5 6 6. t = 7. t = 6 8. t =

76 Foundations of Trigonometr 9. t = between and revolutions 50. About 0. miles per hour 5. About 67.5 revolutions per minute 5. About. miles per hour 5. About 5.55 miles per hour 5. 70 miles per hour 55. About. miles per hour 57. square units 58. 650 square units 59. 79.85 9.07 square units 60. 6. 50 square units square units 6. 8.05 9.6 square units

0. The Unit Circle: Cosine and Sine 77 0. The Unit Circle: Cosine and Sine In Section 0.., we introduced circular motion and derived a formula which describes the linear velocit of an object moving on a circular path at a constant angular velocit. One of the goals of this section is describe the position of such an object. To that end, consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. B associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The -coordinate of P is called the cosine of θ, written cosθ, while the -coordinate of P is called the sine of θ, written sinθ. The reader is encouraged to verif that these rules used to match an angle with its cosine and sine do, in fact, satisf the definition of a function. That is, for each angle θ, there is onl one associated value of cosθ and onl one associated value of sinθ. P cosθ, sinθ θ θ Eample 0... Find the cosine and sine of the following angles.. θ = 70. θ =. θ = 5. θ = 6 5. θ = 60 Solution.. To find cos 70 and sin 70, we plot the angle θ = 70 in standard position and find the point on the terminal side of θ which lies on the Unit Circle. Since 70 represents of a counter-clockwise revolution, the terminal side of θ lies along the negative -ais. Hence, the point we seek is 0, so that cos 70 = 0 and sin 70 =.. The angle θ = represents one half of a clockwise revolution so its terminal side lies on the negative -ais. The point on the Unit Circle that lies on the negative -ais is, 0 which means cos = and sin = 0. The etmolog of the name sine is quite colorful, and the interested reader is invited to research it; the co in cosine is eplained in Section 0..

78 Foundations of Trigonometr θ = 70 P, 0 θ = P 0, Finding cos 70 and sin 70 Finding cos and sin. When we sketch θ = 5 in standard position, we see that its terminal does not lie along an of the coordinate aes which makes our job of finding the cosine and sine values a bit more difficult. Let P, denote the point on the terminal side of θ which lies on the Unit Circle. B definition, = cos 5 and = sin 5. If we drop a perpendicular line segment from P to the -ais, we obtain a 5 5 90 right triangle whose legs have lengths and units. From Geometr, we get =. Since P, lies on the Unit Circle, we have + =. Substituting = into this equation ields =, or = ± Since P, lies in the first quadrant, > 0, so = cos 5 = = sin 5 =. = ±. and with = we have P, P, θ = 5 5 θ = 5 Can ou show this?

0. The Unit Circle: Cosine and Sine 79. As before, the terminal side of θ = 6 does not lie on an of the coordinate aes, so we proceed using a triangle approach. Letting P, denote the point on the terminal side of θ which lies on the Unit Circle, we drop a perpendicular line segment from P to the -ais to form a 0 60 90 right triangle. After a bit of Geometr we find = so sin 6 =. Since P, lies on the Unit Circle, we substitute = into + = to get =, or = ±. Here, > 0 so = cos 6 =. P, P, θ = 6 60 θ = 6 = 0 5. Plotting θ = 60 in standard position, we find it is not a quadrantal angle and set about using a triangle approach. Once again, we get a 0 60 90 right triangle and, after the usual computations, find = cos 60 = and = sin 60 = P,. P, θ = 60 0 θ = 60 Again, can ou show this?

70 Foundations of Trigonometr In Eample 0.., it was quite eas to find the cosine and sine of the quadrantal angles, but for non-quadrantal angles, the task was much more involved. In these latter cases, we made good use of the fact that the point P, = cosθ, sinθ lies on the Unit Circle, + =. If we substitute = cosθ and = sinθ into + =, we get cosθ + sinθ =. An unfortunate convention, which the authors are compelled to perpetuate, is to write cosθ as cos θ and sinθ as sin θ. Rewriting the identit using this convention results in the following theorem, which is without a doubt one of the most important results in Trigonometr. Theorem 0.. The Pthagorean Identit: For an angle θ, cos θ + sin θ =. The moniker Pthagorean brings to mind the Pthagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimatel derived. 5 The word Identit reminds us that, regardless of the angle θ, the equation in Theorem 0. is alwas true. If one of cosθ or sinθ is known, Theorem 0. can be used to determine the other, up to a ± sign. If, in addition, we know where the terminal side of θ lies when in standard position, then we can remove the ambiguit of the ± and completel determine the missing value as the net eample illustrates. Eample 0... Using the given information about θ, find the indicated value.. If θ is a Quadrant II angle with sinθ = 5, find cosθ.. If < θ < with cosθ = 5 5, find sinθ.. If sinθ =, find cosθ. Solution.. When we substitute sinθ = 5 into The Pthagorean Identit, cos θ + sin θ =, we obtain cos θ + 9 5 =. Solving, we find cosθ = ± 5. Since θ is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the -coordinates are negative in Quadrant II, cosθ is too. Hence, cosθ = 5.. Substituting cosθ = 5 5 into cos θ + sin θ = gives sinθ = ± 5 = ± 5 5. Since we are given that < θ <, we know θ is a Quadrant III angle. Hence both its sine and cosine are negative and we conclude sinθ = 5 5.. When we substitute sinθ = into cos θ + sin θ =, we find cosθ = 0. Another tool which helps immensel in determining cosines and sines of angles is the smmetr inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of θ = 5 6. We plot θ in standard position below and, as usual, let P, denote the point on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies 6 half revolution. In Eample 0.., we determined that cos 6 = and sin 6 This is unfortunate from a function notation perspective. See Section 0.6. 5 See Sections. and 7. for details. radians short of one =. This means

0. The Unit Circle: Cosine and Sine 7 that the point on the terminal side of the angle 6, when plotted in standard position, is From the figure below, it is clear that the point P, we seek can be obtained b reflecting that point about the -ais. Hence, cos 5 6 = and sin 5 6 =.,. P, θ = 5 6 P, θ = 5 6, 6 6 6 In the above scenario, the angle 6 is called the reference angle for the angle 5 6. In general, for a non-quadrantal angle θ, the reference angle for θ usuall denoted α is the acute angle made between the terminal side of θ and the -ais. If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive -ais; if θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative -ais. If we let P denote the point cosθ, sinθ, then P lies on the Unit Circle. Since the Unit Circle possesses smmetr with respect to the -ais, -ais and origin, regardless of where the terminal side of θ lies, there is a point Q smmetric with P which determines θ s reference angle, α as seen below. P = Q P Q α α α Reference angle α for a Quadrant I angle Reference angle α for a Quadrant II angle

7 Foundations of Trigonometr Q Q α α α α P P Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle We have just outlined the proof of the following theorem. Theorem 0.. Reference Angle Theorem. Suppose α is the reference angle for θ. Then cosθ = ± cosα and sinθ = ± sinα, where the choice of the ± depends on the quadrant in which the terminal side of θ lies. In light of Theorem 0., it pas to know the cosine and sine values for certain common angles. In the table below, we summarize the values which we consider essential and must be memorized. Cosine and Sine Values of Common Angles θdegrees θradians cosθ sinθ 0 0 0 0 6 5 60 90 0 Eample 0... Find the cosine and sine of the following angles.. θ = 5. θ = 6. θ = 5. θ = 7 Solution.. We begin b plotting θ = 5 in standard position and find its terminal side overshoots the negative -ais to land in Quadrant III. Hence, we obtain θ s reference angle α b subtracting: α = θ 80 = 5 80 = 5. Since θ is a Quadrant III angle, both cosθ < 0 and

0. The Unit Circle: Cosine and Sine 7 sinθ < 0. The Reference Angle Theorem ields: cos 5 = cos 5 = sin 5 = sin 5 =. and. The terminal side of θ = 6, when plotted in standard position, lies in Quadrant IV, just sh of the positive -ais. To find θ s reference angle α, we subtract: α = θ = 6 = 6. Since θ is a Quadrant IV angle, cosθ > 0 and sinθ < 0, so the Reference Angle Theorem gives: cos 6 = cos 6 = and sin 6 = sin 6 =. θ = 5 5 θ = 6 6 Finding cos 5 and sin 5 Finding cos 6 and sin 6. To plot θ = 5, we rotate clockwise an angle of 5 from the positive -ais. The terminal side of θ, therefore, lies in Quadrant II making an angle of α = 5 = radians with respect to the negative -ais. Since θ is a Quadrant II angle, the Reference Angle Theorem gives: cos 5 = cos = and sin 5 = sin =.. Since the angle θ = 7 measures more than = 6, we find the terminal side of θ b rotating one full revolution followed b an additional α = 7 = radians. Since θ and α are coterminal, cos 7 = cos = and sin 7 = sin =. θ = 7 θ = 5 Finding cos 5 and sin 5 Finding cos 7 and sin 7

7 Foundations of Trigonometr The reader ma have noticed that when epressed in radian measure, the reference angle for a non-quadrantal angle is eas to spot. Reduced fraction multiples of with a denominator of 6 have 6 as a reference angle, those with a denominator of have as their reference angle, and those with a denominator of have as their reference angle.6 The Reference Angle Theorem in conjunction with the table of cosine and sine values on Page 7 can be used to generate the following figure, which the authors feel should be committed to memor.,, 5 6, 0,,, 6,, 0 0,, 0,, 7 6 5, 5 7 6,,, 0, Important Points on the Unit Circle 6 For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a natural wa to match oriented angles with real numbers!

0. The Unit Circle: Cosine and Sine 75 The net eample summarizes all of the important ideas discussed thus far in the section. Eample 0... Suppose α is an acute angle with cosα = 5.. Find sinα and use this to plot α in standard position.. Find the sine and cosine of the following angles: a θ = + α b θ = α c θ = α d θ = + α Solution.. Proceeding as in Eample 0.., we substitute cosα = 5 into cos α + sin α = and find sinα = ±. Since α is an acute and therefore Quadrant I angle, sinα is positive. Hence, sinα =. To plot α in standard position, we begin our rotation on the positive -ais to the ra which contains the point cosα, sinα = 5,. 5, α Sketching α. a To find the cosine and sine of θ = + α, we first plot θ in standard position. We can imagine the sum of the angles +α as a sequence of two rotations: a rotation of radians followed b a rotation of α radians. 7 We see that α is the reference angle for θ, so b The Reference Angle Theorem, cosθ = ± cosα = ± 5 and sinθ = ± sinα = ±. Since the terminal side of θ falls in Quadrant III, both cosθ and sinθ are negative, hence, cosθ = 5 and sinθ =. 7 Since + α = α +, θ ma be plotted b reversing the order of rotations given here. You should do this.

76 Foundations of Trigonometr θ θ α α Visualizing θ = + α θ has reference angle α b Rewriting θ = α as θ = + α, we can plot θ b visualizing one complete revolution counter-clockwise followed b a clockwise revolution, or backing up, of α radians. We see that α is θ s reference angle, and since θ is a Quadrant IV angle, the Reference Angle Theorem gives: cosθ = 5 and sinθ =. θ θ α α Visualizing θ = α θ has reference angle α c Taking a cue from the previous problem, we rewrite θ = α as θ = + α. The angle represents one and a half revolutions counter-clockwise, so that when we back up α radians, we end up in Quadrant II. Using the Reference Angle Theorem, we get cosθ = 5 and sinθ =.

0. The Unit Circle: Cosine and Sine 77 α α θ Visualizing α θ has reference angle α d To plot θ = + α, we first rotate radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediatel applicable. It s important that ou see wh this is the case. Take a moment to think about this before reading on. Let Q, be the point on the terminal side of θ which lies on the Unit Circle so that = cosθ and = sinθ. Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted lines in the figure below are equal. Hence, = cosθ = 5. Similarl, we find = sinθ =. θ P 5, α Q, α α Visualizing θ = + α Using smmetr to determine Q,

78 Foundations of Trigonometr Our net eample asks us to solve some ver basic trigonometric equations. 8 Eample 0..5. Find all of the angles which satisf the given equation.. cosθ =. sinθ =. cosθ = 0. Solution. Since there is no contet in the problem to indicate whether to use degrees or radians, we will default to using radian measure in our answers to each of these problems. This choice will be justified later in the tet when we stud what is known as Analtic Trigonometr. In those sections to come, radian measure will be the onl appropriate angle measure so it is worth the time to become fluent in radians now.. If cosθ =, then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at =. This means θ is a Quadrant I or IV angle with reference angle. One solution in Quadrant I is θ =, and since all other Quadrant I solutions must be coterminal with, we find θ = +k for integers k.9 Proceeding similarl for the Quadrant IV case, we find the solution to cosθ = here is 5, so our answer in this Quadrant is θ = 5 + k for integers k.. If sinθ =, then when θ is plotted in standard position, its terminal side intersects the Unit Circle at =. From this, we determine θ is a Quadrant III or Quadrant IV angle with reference angle 6. 8 We will stud trigonometric equations more formall in Section 0.7. Enjo these relativel straightforward eercises while the last! 9 Recall in Section 0., two angles in radian measure are coterminal if and onl if the differ b an integer multiple of. Hence to describe all angles coterminal with a given angle, we add k for integers k = 0, ±, ±,....

0. The Unit Circle: Cosine and Sine 79 6 6 In Quadrant III, one solution is 7 6, so we capture all Quadrant III solutions b adding integer multiples of : θ = 7 6 + k. In Quadrant IV, one solution is 6 so all the solutions here are of the form θ = 6 + k for integers k.. The angles with cosθ = 0 are quadrantal angles whose terminal sides, when plotted in standard position, lie along the -ais. While, technicall speaking, isn t a reference angle we can nonetheless use it to find our answers. If we follow the procedure set forth in the previous eamples, we find θ = + k and θ = + k for integers, k. While this solution is correct, it can be shortened to θ = + k for integers k. Can ou see wh this works from the diagram? One of the ke items to take from Eample 0..5 is that, in general, solutions to trigonometric equations consist of infinitel man answers. To get a feel for these answers, the reader is encouraged to follow our mantra from Chapter 9 - that is, When in doubt, write it out! This is especiall important when checking answers to the eercises. For eample, another Quadrant IV solution to sinθ = is θ = 6. Hence, the famil of Quadrant IV answers to number above could just have easil been written θ = 6 + k for integers k. While on the surface, this famil ma look

70 Foundations of Trigonometr different than the stated solution of θ = 6 + k for integers k, we leave it to the reader to show the represent the same list of angles. 0.. Beond the Unit Circle We began the section with a quest to describe the position of a particle eperiencing circular motion. In defining the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider for the moment the acute angle θ drawn below in standard position. Let Q, be the point on the terminal side of θ which lies on the circle + = r, and let P, be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, OP A and OQB. These triangles are similar, 0 thus it follows that = r = r, so = r and, similarl, we find = r. Since, b definition, = cosθ and = sinθ, we get the coordinates of Q to be = r cosθ and = r sinθ. B reflecting these points through the -ais, -ais and origin, we obtain the result for all non-quadrantal angles θ, and we leave it to the reader to verif these formulas hold for the quadrantal angles. r θ Q, P, r P, Q, = r cosθ, r sinθ θ O A, 0 B, 0 Not onl can we describe the coordinates of Q in terms of cosθ and sinθ but since the radius of the circle is r = +, we can also epress cosθ and sinθ in terms of the coordinates of Q. These results are summarized in the following theorem. Theorem 0.. If Q, is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle + = r then = r cosθ and = r sinθ. Moreover, cosθ = r = and sinθ = + r = + 0 Do ou remember wh?

0. The Unit Circle: Cosine and Sine 7 Note that in the case of the Unit Circle we have r = + =, so Theorem 0. reduces to our definitions of cosθ and sinθ. Eample 0..6.. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q,. Find sinθ and cosθ.. In Eample 0..5 in Section 0., we approimated the radius of the earth at.68 north latitude to be 960 miles. Justif this approimation if the radius of the Earth at the Equator is approimatel 960 miles. Solution.. Using Theorem 0. with = and =, we find r = + = 0 = 5 so that cosθ = r = = 5 5 5 and sinθ = r = = 5 5 5.. Assuming the Earth is a sphere, a cross-section through the poles produces a circle of radius 960 miles. Viewing the Equator as the -ais, the value we seek is the -coordinate of the point Q, indicated in the figure below. 960 Q,.68 Q, 960 The terminal side of θ contains Q, A point on the Earth at.68 N Using Theorem 0., we get = 960 cos.68. Using a calculator in degree mode, we find 960 cos.68 960. Hence, the radius of the Earth at North Latitude.68 is approimatel 960 miles.

7 Foundations of Trigonometr Theorem 0. gives us what we need to describe the position of an object traveling in a circular path of radius r with constant angular velocit ω. Suppose that at time t, the object has swept out an angle measuring θ radians. If we assume that the object is at the point r, 0 when t = 0, the angle θ is in standard position. B definition, ω = θ t which we rewrite as θ = ωt. According to Theorem 0., the location of the object Q, on the circle is found using the equations = r cosθ = r cosωt and = r sinθ = r sinωt. Hence, at time t, the object is at the point r cosωt, r sinωt. We have just argued the following. Equation 0.. Suppose an object is traveling in a circular path of radius r centered at the origin with constant angular velocit ω. If t = 0 corresponds to the point r, 0, then the and coordinates of the object are functions of t and are given b = r cosωt and = r sinωt. Here, ω > 0 indicates a counter-clockwise direction and ω < 0 indicates a clockwise direction. r Q, = r cosωt, r sinωt θ = ωt r Equations for Circular Motion Eample 0..7. Suppose we are in the situation of Eample 0..5. Find the equations of motion of Lakeland Communit College as the earth rotates. Solution. From Eample 0..5, we take r = 960 miles and and ω = hours. Hence, the equations of motion are = r cosωt = 960 cos t and = r sinωt = 960 sin t, where and are measured in miles and t is measured in hours. In addition to circular motion, Theorem 0. is also the ke to developing what is usuall called right triangle trigonometr. As we shall see in the sections to come, man applications in trigonometr involve finding the measures of the angles in, and lengths of the sides of, right triangles. Indeed, we made good use of some properties of right triangles to find the eact values of the cosine and sine of man of the angles in Eample 0.., so the following development shouldn t be that much of a surprise. Consider the generic right triangle below with corresponding acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the triangle opposite θ; and the remaining side of length c the side opposite the You ma have been eposed to this in High School.

0. The Unit Circle: Cosine and Sine 7 right angle is called the hpotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ ling along the positive -ais. c P a, b θ c b θ c a According to the Pthagorean Theorem, a + b = c, so that the point P a, b lies on a circle of radius c. Theorem 0. tells us that cosθ = a c and sinθ = b c, so we have determined the cosine and sine of θ in terms of the lengths of the sides of the right triangle. Thus we have the following theorem. Theorem 0.. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hpotenuse is c, then cosθ = a c and sinθ = b c. Eample 0..8. Find the measure of the missing angle and the lengths of the missing sides of: 0 Solution. The first and easiest task is to find the measure of the missing angle. Since the sum of angles of a triangle is 80, we know that the missing angle has measure 80 0 90 = 60. We now proceed to find the lengths of the remaining two sides of the triangle. Let c denote the length of the hpotenuse of the triangle. B Theorem 0., we have cos 0 = 7 c, or c = 7 7 cos0. Since cos 0 =, we have, after the usual fraction gmnastics, c =. At this point, we have two was to proceed to find the length of the side opposite the 0 angle, which we ll denote b. We know the length of the adjacent side is 7 and the length of the hpotenuse is, so we

7 Foundations of Trigonometr could use the Pthagorean Theorem to find the missing side and solve 7 + b = for b. Alternativel, we could use Theorem 0., namel that sin 0 = b c. Choosing the latter, we find b = c sin 0 = = 7. The triangle with all of its data is recorded below. c = 60 b = 7 0 7 We close this section b noting that we can easil etend the functions cosine and sine to real numbers b identifing a real number t with the angle θ = t radians. Using this identification, we define cost = cosθ and sint = sinθ. In practice this means epressions like cos and sin can be found b regarding the inputs as angles in radian measure or real numbers; the choice is the reader s. If we trace the identification of real numbers t with angles θ in radian measure to its roots on page 70, we can spell out this correspondence more precisel. For each real number t, we associate an oriented arc t units in length with initial point, 0 and endpoint P cost, sint. P cost, sint θ = t t θ = t In the same wa we studied polnomial, rational, eponential, and logarithmic functions, we will stud the trigonometric functions ft = cost and gt = sint. The first order of business is to find the domains and ranges of these functions. Whether we think of identifing the real number t with the angle θ = t radians, or think of wrapping an oriented arc around the Unit Circle to find coordinates on the Unit Circle, it should be clear that both the cosine and sine functions are defined for all real numbers t. In other words, the domain of ft = cost and of gt = sint is,. Since cost and sint represent - and -coordinates, respectivel, of points on the Unit Circle, the both take on all of the values between an, inclusive. In other words, the range of ft = cost and of gt = sint is the interval [, ]. To summarize:

0. The Unit Circle: Cosine and Sine 75 Theorem 0.5. Domain and Range of the Cosine and Sine Functions: The function ft = cost The function gt = sint has domain, has domain, has range [, ] has range [, ] Suppose, as in the Eercises, we are asked to solve an equation such as sint =. As we have alread mentioned, the distinction between t as a real number and as an angle θ = t radians is often blurred. Indeed, we solve sint = in the eact same manner as we did in Eample 0..5 number. Our solution is onl cosmeticall different in that the variable used is t rather than θ: t = 7 6 + k or t = 6 + k for integers, k. We will stud the cosine and sine functions in greater detail in Section 0.5. Until then, keep in mind that an properties of cosine and sine developed in the following sections which regard them as functions of angles in radian measure appl equall well if the inputs are regarded as real numbers. Well, to be pedantic, we would be technicall using reference numbers or reference arcs instead of reference angles but the idea is the same.

76 Foundations of Trigonometr 0.. Eercises In Eercises - 0, find the eact value of the cosine and sine of the given angle.. θ = 0. θ =. θ =. θ = 5. θ = 6. θ = 7. θ = 8. θ = 7 6 9. θ = 5 0. θ =. θ =. θ = 5. θ = 7. θ = 6 5. θ = 6. θ = 6 7. θ = 8. θ = 6 9. θ = 0 0. θ = 7 In Eercises - 0, use the results developed throughout the section to find the requested value.. If sinθ = 7 5 with θ in Quadrant IV, what is cosθ?. If cosθ = 9 with θ in Quadrant I, what is sinθ?. If sinθ = 5 with θ in Quadrant II, what is cosθ?. If cosθ = with θ in Quadrant III, what is sinθ? 5. If sinθ = with θ in Quadrant III, what is cosθ? 6. If cosθ = 8 5 with θ in Quadrant IV, what is sinθ? 7. If sinθ = 5 5 and < θ <, what is cosθ? 8. If cosθ = 0 0 and < θ < 5, what is sinθ? 9. If sinθ = 0. and < θ <, what is cosθ? 0. If cosθ = 0.98 and < θ <, what is sinθ?

0 0. The Unit Circle: Cosine and Sine 77 In Eercises - 9, find all of the angles which satisf the given equation.. sinθ =. cosθ =. sinθ = 0. cosθ = 5. sinθ = 6. cosθ = 7. sinθ = 8. cosθ = 9. cosθ =.00 In Eercises 0-8, solve the equation for t. See the comments following Theorem 0.5. 0. cost = 0. sint =. cost =. sint =. cost = 5. sint = 6. cost = 7. sint = 8. cost = In Eercises 9-5, use our calculator to approimate the given value to three decimal places. Make sure our calculator is in the proper angle measurement mode! 9. sin78.95 50. cos.0 5. sin9.99 5. cos07 5. sin 5. cose In Eercises 55-58, find the measurement of the missing angle and the lengths of the missing sides. See Eample 0..8 55. Find θ, b, and c. 56. Find θ, a, and c. c θ b a 5 c θ

78 Foundations of Trigonometr 57. Find α, a, and b. b α a 8 58. Find β, a, and c. a 8 c β 6 In Eercises 59-6, assume that θ is an acute angle in a right triangle and use Theorem 0. to find the requested side. 59. If θ = and the side adjacent to θ has length, how long is the hpotenuse? 60. If θ = 78. and the hpotenuse has length 580, how long is the side adjacent to θ? 6. If θ = 59 and the side opposite θ has length 7., how long is the hpotenuse? 6. If θ = 5 and the hpotenuse has length 0, how long is the side opposite θ? 6. If θ = 5 and the hpotenuse has length 0, how long is the side adjacent to θ? 6. If θ = 7.5 and the side opposite θ has length 06, how long is the side adjacent to θ? In Eercises 65-68, let θ be the angle in standard position whose terminal side contains the given point then compute cosθ and sinθ. 65. P 7, 66. Q, 67. R5, 9 68. T, In Eercises 69-7, find the equations of motion for the given scenario. Assume that the center of the motion is the origin, the motion is counter-clockwise and that t = 0 corresponds to a position along the positive -ais. See Equation 0. and Eample 0..5. 69. A point on the edge of the spinning o-o in Eercise 50 from Section 0.. Recall: The diameter of the o-o is.5 inches and it spins at 500 revolutions per minute. 70. The o-o in eercise 5 from Section 0.. Recall: The radius of the circle is 8 inches and it completes one revolution in seconds. 7. A point on the edge of the hard drive in Eercise 5 from Section 0.. Recall: The diameter of the hard disk is.5 inches and it spins at 700 revolutions per minute.

0. The Unit Circle: Cosine and Sine 79 7. A passenger on the Big Wheel in Eercise 55 from Section 0.. Recall: The diameter is 8 feet and completes revolutions in minutes, 7 seconds. 7. Consider the numbers: 0,,,,. Take the square root of each of these numbers, then divide each b. The resulting numbers should look hauntingl familiar. See the values in the table on 7. 7. Let α and β be the two acute angles of a right triangle. Thus α and β are complementar angles. Show that sinα = cosβ and sinβ = cosα. The fact that co-functions of complementar angles are equal in this case is not an accident and a more general result will be given in Section 0.. 75. In the scenario of Equation 0., we assumed that at t = 0, the object was at the point r, 0. If this is not the case, we can adjust the equations of motion b introducing a time dela. If t 0 > 0 is the first time the object passes through the point r, 0, show, with the help of our classmates, the equations of motion are = r cosωt t 0 and = r sinωt t 0.

70 Foundations of Trigonometr 0.. Answers. cos0 =, sin0 = 0. cos =, sin =. cos =, sin =. cos = 0, sin = 5. cos =, sin = 6. cos =, sin = 7 7 7. cos =, sin = 0 8. cos = 6, sin = 6 5 5 9. cos =, sin = 0. cos =, sin = 5. cos = 0, sin =. cos = 5, sin = 7 7. cos =, sin =. cos = 6, sin = 6 5. cos = 0, sin = 6. cos = 6, sin = 6 7. cos =, sin = 8. cos = 6, sin = 6 0 9. cos = 0, sin = 0. cos7 =, sin7 = 0. If sinθ = 7 5 with θ in Quadrant IV, then cosθ = 5.. If cosθ = 9 with θ in Quadrant I, then sinθ = 65 9.. If sinθ = 5 with θ in Quadrant II, then cosθ =.. If cosθ = with θ in Quadrant III, then sinθ = 7. 5. If sinθ = with θ in Quadrant III, then cosθ = 5. 6. If cosθ = 8 5 with θ in Quadrant IV, then sinθ = 5 5.

0. The Unit Circle: Cosine and Sine 7 7. If sinθ = 5 5 8. If cosθ = and < θ <, then cosθ = 5 5. 0 0 and < θ < 5 0, then sinθ = 0. 9. If sinθ = 0. and < θ <, then cosθ = 0.86 0.9075. 0. If cosθ = 0.98 and < θ <, then sinθ = 0.096 0.990.. sinθ = when θ = 6 + k or θ = 5 6 + k for an integer k.. cosθ = when θ = 5 6 + k or θ = 7 6. sinθ = 0 when θ = k for an integer k. + k for an integer k.. cosθ = when θ = + k or θ = 7 + k for an integer k. 5. sinθ = when θ = + k or θ = + k for an integer k. 6. cosθ = when θ = k + for an integer k. 7. sinθ = when θ = + k for an integer k. 8. cosθ = when θ = + k or θ = + k for an integer k. 6 6 9. cosθ =.00 never happens 0. cost = 0 when t = + k for an integer k.. sint = when t = 5 + k or t = 7. cost = never happens.. sint = when t = 7 6 + k or t = 6 + k for an integer k. + k for an integer k.. cost = when t = + k or t = 5 + k for an integer k. 5. sint = never happens 6. cost = when t = k for an integer k.

7 Foundations of Trigonometr 7. sint = when t = + k for an integer k. 8. cost = when t = + k or t = 5 + k for an integer k. 9. sin78.95 0.98 50. cos.0 0.5 5. sin9.99 0.9 5. cos07 0.89 5. sin 0.055 5. cose 0.9 55. θ = 60, b =, c = 56. θ = 5, a =, c = 57. α = 57, a = 8 cos 6.709, b = 8 sin.57 58. β =, c = 59. The hpotenuse has length 6 sin8 8.07, a = c 6 5.0 cos.089. 60. The side adjacent to θ has length 580 cos78. 086.68. 6. The hpotenuse has length 7. sin59 6.99. 6. The side opposite θ has length 0 sin5 0.87. 6. The side adjacent to θ has length 0 cos5 9.96. 6. The hpotenuse has length c = c 06 98.797. 65. cosθ = 7, sinθ = 5 5 06 sin7.5 50.660, so the side adjacent to θ has length 66. cosθ = 5, sinθ = 5 67. cosθ = 5 06 06, sinθ = 9 06 06 68. cosθ = 5 5, sinθ = 5 5 69. r =.5 inches, ω = 9000 radians minute, =.5 cos9000 t, =.5 sin9000 t. Here and are measured in inches and t is measured in minutes.

0. The Unit Circle: Cosine and Sine 7 radians 70. r = 8 inches, ω = second, = 8 cos t, = 8 sin t. Here and are measured in inches and t is measured in seconds. 7. r =.5 inches, ω = 00 radians minute, =.5 cos00 t, =.5 sin00 t. Here and are measured in inches and t is measured in minutes. 7. r = 6 feet, ω = radians 7 second, = 6 cos 7 t, = 6 sin 7 t. Here and are measured in feet and t is measured in seconds

7 Foundations of Trigonometr 0. The Si Circular Functions and Fundamental Identities In section 0., we defined cosθ and sinθ for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions. It turns out that cosine and sine are just two of the si commonl used circular functions which we define below. Definition 0.. The Circular Functions: Suppose θ is an angle plotted in standard position and P, is the point on the terminal side of θ which lies on the Unit Circle. ˆ The cosine of θ, denoted cosθ, is defined b cosθ =. ˆ The sine of θ, denoted sinθ, is defined b sinθ =. ˆ The secant of θ, denoted secθ, is defined b secθ =, provided 0. ˆ The cosecant of θ, denoted cscθ, is defined b cscθ =, provided 0. ˆ The tangent of θ, denoted tanθ, is defined b tanθ =, provided 0. ˆ The cotangent of θ, denoted cotθ, is defined b cotθ =, provided 0. While we left the histor of the name sine as an interesting research project in Section 0., the names tangent and secant can be eplained using the diagram below. Consider the acute angle θ below in standard position. Let P, denote, as usual, the point on the terminal side of θ which lies on the Unit Circle and let Q, denote the point on the terminal side of θ which lies on the vertical line =. Q, =, tanθ P, θ O A, 0 B, 0 In Theorem 0. we also showed cosine and sine to be functions of an angle residing in a right triangle so we could just as easil call them trigonometric functions. In later sections, ou will find that we do indeed use the phrase trigonometric function interchangeabl with the term circular function.

0. The Si Circular Functions and Fundamental Identities 75 The word tangent comes from the Latin meaning to touch, and for this reason, the line = is called a tangent line to the Unit Circle since it intersects, or touches, the circle at onl one point, namel, 0. Dropping perpendiculars from P and Q creates a pair of similar triangles OP A and OQB. Thus = which gives = = tanθ, where this last equalit comes from appling Definition 0.. We have just shown that for acute angles θ, tanθ is the -coordinate of the point on the terminal side of θ which lies on the line = which is tangent to the Unit Circle. Now the word secant means to cut, so a secant line is an line that cuts through a circle at two points. The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. With the point P ling on the Unit Circle, the length of the hpotenuse of OP A is. If we let h denote the length of the hpotenuse of OQB, we have from similar triangles that h =, or h = = secθ. Hence for an acute angle θ, secθ is the length of the line segment which lies on the secant line determined b the terminal side of θ and cuts off the tangent line =. Not onl do these observations help eplain the names of these functions, the serve as the basis for a fundamental inequalit needed for Calculus which we ll eplore in the Eercises. Of the si circular functions, onl cosine and sine are defined for all angles. Since cosθ = and sinθ = in Definition 0., it is customar to rephrase the remaining four circular functions in terms of cosine and sine. The following theorem is a result of simpl replacing with cosθ and with sinθ in Definition 0.. Theorem 0.6. Reciprocal and Quotient Identities: ˆ secθ =, provided cosθ 0; if cosθ = 0, secθ is undefined. cosθ ˆ cscθ =, provided sinθ 0; if sinθ = 0, cscθ is undefined. sinθ ˆ ˆ tanθ = sinθ, provided cosθ 0; if cosθ = 0, tanθ is undefined. cosθ cotθ = cosθ, provided sinθ 0; if sinθ = 0, cotθ is undefined. sinθ It is high time for an eample. Eample 0... Find the indicated value, if it eists.. sec 60. csc 7. cot. tan θ, where θ is an angle coterminal with. 5. cos θ, where cscθ = 5 and θ is a Quadrant IV angle. 6. sin θ, where tanθ = and < θ <. Compare this with the definition given in Section..

76 Foundations of Trigonometr Solution.. According to Theorem 0.6, sec 60 = cos60. Hence, sec 60 = / =.. Since sin 7 =, csc 7 = sin 7 = = / =.. Since θ = radians is not one of the common angles from Section 0., we resort to the calculator for a decimal approimation. Ensuring that the calculator is in radian mode, we find cot = cos sin 7.05.. If θ is coterminal with, then cosθ = cos = 0 and sinθ = sin =. Attempting to compute tanθ = sinθ cosθ results in 0, so tanθ is undefined. 5. We are given that cscθ = sinθ = 5 so sinθ = 5 = 5 5. As we saw in Section 0., we can use the Pthagorean Identit, cos θ + sin θ =, to find cosθ b knowing sinθ. Substituting, we get cos θ+ θ is a Quadrant IV angle, cosθ > 0, so cosθ = 5 5. 5 5 =, which gives cos θ = 5, or cosθ = ± 5 5. Since 6. If tanθ =, then sinθ cosθ =. Be careful - this does NOT mean we can take sinθ = and cosθ =. Instead, from sinθ cosθ = we get: sinθ = cosθ. To relate cosθ and sinθ, we once again emplo the Pthagorean Identit, cos θ + sin θ =. Solving sinθ = cosθ for cosθ, we find cosθ = sinθ. Substituting this into the Pthagorean Identit, we find sin θ + sinθ =. Solving, we get sin θ = 9 0 so sinθ = ± 0 0. Since < θ <, θ is a Quadrant III angle. This means sinθ < 0, so our final answer is sinθ = 0 0. While the Reciprocal and Quotient Identities presented in Theorem 0.6 allow us to alwas reduce problems involving secant, cosecant, tangent and cotangent to problems involving cosine and sine, it is not alwas convenient to do so. It is worth taking the time to memorize the tangent and cotangent values of the common angles summarized below. As we shall see shortl, when solving equations involving secant and cosecant, we usuall convert back to cosines and sines. However, when solving for tangent or cotangent, we usuall stick with what we re dealt.

0. The Si Circular Functions and Fundamental Identities 77 Tangent and Cotangent Values of Common Angles θdegrees θradians tanθ cotθ 0 0 0 undefined 0 6 5 60 90 undefined 0 Coupling Theorem 0.6 with the Reference Angle Theorem, Theorem 0., we get the following. Theorem 0.7. Generalized Reference Angle Theorem. The values of the circular functions of an angle, if the eist, are the same, up to a sign, of the corresponding circular functions of its reference angle. More specificall, if α is the reference angle for θ, then: cosθ = ± cosα, sinθ = ± sinα, secθ = ± secα, cscθ = ± cscα, tanθ = ± tanα and cotθ = ± cotα. The choice of the ± depends on the quadrant in which the terminal side of θ lies. We put Theorem 0.7 to good use in the following eample. Eample 0... Find all angles which satisf the given equation.. secθ =. tanθ =. cotθ =. Solution.. To solve secθ =, we convert to cosines and get cosθ = or cosθ =. This is the eact same equation we solved in Eample 0..5, number, so we know the answer is: θ = +k or θ = 5 + k for integers k.. From the table of common values, we see tan =. According to Theorem 0.7, we know the solutions to tanθ = must, therefore, have a reference angle of. Our net task is to determine in which quadrants the solutions to this equation lie. Since tangent is defined as the ratio of points, on the Unit Circle with 0, tangent is positive when and have the same sign i.e., when the are both positive or both negative. This happens in Quadrants I and III. In Quadrant I, we get the solutions: θ = + k for integers k, and for Quadrant III, we get θ = + k for integers k. While these descriptions of the solutions are correct, the can be combined into one list as θ = + k for integers k. The latter form of the solution is best understood looking at the geometr of the situation in the diagram below. See Eample 0..5 number in Section 0. for another eample of this kind of simplification of the solution.

78 Foundations of Trigonometr. From the table of common values, we see that has a cotangent of, which means the solutions to cotθ = have a reference angle of. To find the quadrants in which our solutions lie, we note that cotθ = for a point, on the Unit Circle where 0. If cotθ is negative, then and must have different signs i.e., one positive and one negative. Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = + k, and for Quadrant IV, we get θ = 7 +k for integers k. Can these lists be combined? Indeed the can - one such wa to capture all the solutions is: θ = + k for integers k. We have alread seen the importance of identities in trigonometr. Our net task is to use use the Reciprocal and Quotient Identities found in Theorem 0.6 coupled with the Pthagorean Identit found in Theorem 0. to derive new Pthagorean-like identities for the remaining four circular functions. Assuming cosθ 0, we ma start with cos θ + sin θ = and divide both sides b cos θ to obtain + sin θ cos θ =. Using properties of eponents along with the Reciprocal cos θ and Quotient Identities, this reduces to + tan θ = sec θ. If sinθ 0, we can divide both sides of the identit cos θ + sin θ = b sin θ, appl Theorem 0.6 once again, and obtain cot θ + = csc θ. These three Pthagorean Identities are worth memorizing and the, along with some of their other common forms, are summarized in the following theorem.

0. The Si Circular Functions and Fundamental Identities 79 Theorem 0.8. The Pthagorean Identities:. cos θ + sin θ =. Common Alternate Forms: ˆ ˆ sin θ = cos θ cos θ = sin θ. + tan θ = sec θ, provided cosθ 0. Common Alternate Forms: ˆ sec θ tan θ = ˆ sec θ = tan θ. + cot θ = csc θ, provided sinθ 0. Common Alternate Forms: ˆ csc θ cot θ = ˆ csc θ = cot θ Trigonometric identities pla an important role in not just Trigonometr, but in Calculus as well. We ll use them in this book to find the values of the circular functions of an angle and solve equations and inequalities. In Calculus, the are needed to simplif otherwise complicated epressions. In the net eample, we make good use of the Theorems 0.6 and 0.8. Eample 0... Verif the following identities. Assume that all quantities are defined.. cscθ = sinθ. secθ tanθsecθ + tanθ =. 5. 6 secθ tanθ = sinθ + sinθ. tanθ = sinθ secθ 6. secθ tanθ = cosθ sinθ sinθ cosθ = + cosθ sinθ Solution. In verifing identities, we tpicall start with the more complicated side of the equation and use known identities to transform it into the other side of the equation.. To verif cscθ = sinθ, we start with the left side. Using cscθ = sinθ, we get: which is what we were tring to prove. cscθ = sinθ = sinθ,

750 Foundations of Trigonometr. Starting with the right hand side of tanθ = sinθ secθ, we use secθ = cosθ sinθ secθ = sinθ cosθ = sinθ cosθ = tanθ, where the last equalit is courtes of Theorem 0.6. and find:. Epanding the left hand side of the equation gives: secθ tanθsecθ + tanθ = sec θ tan θ. According to Theorem 0.8, sec θ tan θ =. Putting it all together, secθ tanθsecθ + tanθ = sec θ tan θ =.. While both sides of our last identit contain fractions, the left side affords us more opportunities to use our identities. 5 Substituting secθ = sinθ cosθ and tanθ = cosθ, we get: secθ tanθ = = = cosθ sinθ cosθ cosθ = sinθ cosθ cosθ sinθ, cosθ sinθ cosθ cosθ cosθ which is eactl what we had set out to show. cosθ cosθ = cosθ sinθ cosθ cosθ 5. The right hand side of the equation seems to hold more promise. We get common denominators and add: sinθ + sinθ = + sinθ sinθ + sinθ sinθ + sinθ sinθ = + sinθ sin θ sinθ sin θ = = + sinθ sinθ sin θ 6 sinθ sin θ 5 Or, to put to another wa, earn more partial credit if this were an eam question!

0. The Si Circular Functions and Fundamental Identities 75 At this point, it is worth pausing to remind ourselves of our goal. We wish to transform this epression into 6 secθ tanθ. Using a reciprocal and quotient identit, we find 6 secθ tanθ = 6 sinθ cosθ cosθ. In other words, we need to get cosines in our denominator. Theorem 0.8 tells us sin θ = cos θ so we get: sinθ + sinθ 6 sinθ = sin θ = 6 sinθ cos θ sinθ = 6 = 6 secθ tanθ cosθ cosθ 6. It is debatable which side of the identit is more complicated. One thing which stands out is that the denominator on the left hand side is cosθ, while the numerator of the right hand side is + cosθ. This suggests the strateg of starting with the left hand side and multipling the numerator and denominator b the quantit + cosθ: sinθ cosθ = sinθ + cosθ cosθ + cosθ = sinθ + cosθ cosθ + cosθ = sinθ + cosθ cos θ = sinθ + cosθ sinθ sinθ = sinθ + cosθ sin θ = + cosθ sinθ In Eample 0.. number 6 above, we see that multipling cosθ b + cosθ produces a difference of squares that can be simplified to one term using Theorem 0.8. This is eactl the same kind of phenomenon that occurs when we multipl epressions such as b + or i b + i. Can ou recall instances from Algebra where we did such things? For this reason, the quantities cosθ and + cosθ are called Pthagorean Conjugates. Below is a list of other common Pthagorean Conjugates. ˆ ˆ ˆ Pthagorean Conjugates cosθ and + cosθ: cosθ + cosθ = cos θ = sin θ sinθ and + sinθ: sinθ + sinθ = sin θ = cos θ secθ and secθ + : secθ secθ + = sec θ = tan θ ˆ secθ tanθ and secθ+tanθ: secθ tanθsecθ+tanθ = sec θ tan θ = ˆ cscθ and cscθ + : cscθ cscθ + = csc θ = cot θ ˆ cscθ cotθ and cscθ+cotθ: cscθ cotθcscθ+cotθ = csc θ cot θ =

75 Foundations of Trigonometr Verifing trigonometric identities requires a health mi of tenacit and inspiration. You will need to spend man hours struggling with them just to become proficient in the basics. Like man things in life, there is no short-cut here there is no complete algorithm for verifing identities. Nevertheless, a summar of some strategies which ma be helpful depending on the situation is provided below and ample practice is provided for ou in the Eercises. ˆ Strategies for Verifing Identities Tr working on the more complicated side of the identit. ˆ Use the Reciprocal and Quotient Identities in Theorem 0.6 to write functions on one side of the identit in terms of the functions on the other side of the identit. Simplif the resulting comple fractions. ˆ Add rational epressions with unlike denominators b obtaining common denominators. ˆ ˆ ˆ Use the Pthagorean Identities in Theorem 0.8 to echange sines and cosines, secants and tangents, cosecants and cotangents, and simplif sums or differences of squares to one term. Multipl numerator and denominator b Pthagorean Conjugates in order to take advantage of the Pthagorean Identities in Theorem 0.8. If ou find ourself stuck working with one side of the identit, tr starting with the other side of the identit and see if ou can find a wa to bridge the two parts of our work. 0.. Beond the Unit Circle In Section 0., we generalized the cosine and sine functions from coordinates on the Unit Circle to coordinates on circles of radius r. Using Theorem 0. in conjunction with Theorem 0.8, we generalize the remaining circular functions in kind. Theorem 0.9. Suppose Q, is the point on the terminal side of an angle θ plotted in standard position which lies on the circle of radius r, + = r. Then: ˆ secθ = r = +, provided 0. ˆ cscθ = r = +, provided 0. ˆ tanθ =, provided 0. ˆ cotθ =, provided 0.

0. The Si Circular Functions and Fundamental Identities 75 Eample 0.... Suppose the terminal side of θ, when plotted in standard position, contains the point Q,. Find the values of the si circular functions of θ.. Suppose θ is a Quadrant IV angle with cotθ =. Find the values of the five remaining circular functions of θ. Solution.. Since = and =, r = + = + = 5 = 5. Theorem 0.9 tells us cosθ = 5, sinθ = 5, secθ = 5, cscθ = 5, tanθ = and cotθ =.. In order to use Theorem 0.9, we need to find a point Q, which lies on the terminal side of θ, when θ is plotted in standard position. We have that cotθ = =, and since θ is a Quadrant IV angle, we also know > 0 and < 0. Viewing =, we ma choose6 = and = so that r = + = + = 7. Appling Theorem 0.9 once more, we find cosθ = 7 = 7 7, sinθ = 7 = 7 7, secθ = 7, cscθ = 7 and tanθ =. We ma also specialize Theorem 0.9 to the case of acute angles θ which reside in a right triangle, as visualized below. c b θ a Theorem 0.0. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hpotenuse is c, then tanθ = b a secθ = c a cscθ = c b cotθ = a b The following eample uses Theorem 0.0 as well as the concept of an angle of inclination. The angle of inclination or angle of elevation of an object refers to the angle whose initial side is some kind of base-line sa, the ground, and whose terminal side is the line-of-sight to an object above the base-line. This is represented schematicall below. 6 We ma choose an values and so long as > 0, < 0 and =. For eample, we could choose = 8 and =. The fact that all such points lie on the terminal side of θ is a consequence of the fact that the terminal side of θ is the portion of the line with slope which etends from the origin into Quadrant IV.

75 Foundations of Trigonometr object θ Eample 0..5. base line The angle of inclination from the base line to the object is θ. The angle of inclination from a point on the ground 0 feet awa to the top of Lakeland s Armington Clocktower 7 is 60. Find the height of the Clocktower to the nearest foot.. In order to determine the height of a California Redwood tree, two sightings from the ground, one 00 feet directl behind the other, are made. If the angles of inclination were 5 and 0, respectivel, how tall is the tree to the nearest foot? Solution.. We can represent the problem situation using a right triangle as shown below. If we let h denote the height of the tower, then Theorem 0.0 gives tan 60 = h 0. From this we get h = 0 tan 60 = 0 5.96. Hence, the Clocktower is approimatel 5 feet tall. h ft. 60 0 ft. Finding the height of the Clocktower. Sketching the problem situation below, we find ourselves with two unknowns: the height h of the tree and the distance from the base of the tree to the first observation point. 7 Named in honor of Ramond Q. Armington, Lakeland s Clocktower has been a part of campus since 97.

0. The Si Circular Functions and Fundamental Identities 755 h ft. 0 5 00 ft. ft. Finding the height of a California Redwood Using Theorem 0.0, we get a pair of equations: tan 5 = h and tan 0 = +00. Since tan 5 =, the first equation gives h =, or = h. Substituting this into the second h equation gives h+00 = tan 0 =. Clearing fractions, we get h = h + 00. The result is a linear equation for h, so we proceed to epand the right hand side and gather all the terms involving h to one side. h = h + 00 h = h + 00 h h = 00 h = 00 Hence, the tree is approimatel 7 feet tall. h = 00 7.0 As we did in Section 0.., we ma consider all si circular functions as functions of real numbers. At this stage, there are three equivalent was to define the functions sect, csct, tant and cott for real numbers t. First, we could go through the formalit of the wrapping function on page 70 and define these functions as the appropriate ratios of and coordinates of points on the Unit Circle; second, we could define them b associating the real number t with the angle θ = t radians so that the value of the trigonometric function of t coincides with that of θ; lastl, we could simpl define them using the Reciprocal and Quotient Identities as combinations of the functions ft = cost and gt = sint. Presentl, we adopt the last approach. We now set about determining the domains and ranges of the remaining four circular functions. Consider the function F t = sect defined as F t = sect = cost. We know F is undefined whenever cost = 0. From Eample 0..5 number, we know cost = 0 whenever t = + k for integers k. Hence, our domain for F t = sect, in set builder notation is {t : t + k, for integers k}. To get a better understanding what set of real numbers we re dealing with, it pas to write out and graph this set. Running through a few values of k, we find the domain to be {t : t ±, ±, ± 5,...}. Graphing this set on the number line we get h

756 Foundations of Trigonometr 5 0 5 Using interval notation to describe this set, we get... 5,,,,, 5... This is cumbersome, to sa the least! In order to write this in a more compact wa, we note that from the set-builder description of the domain, the kth point ecluded from the domain, which we ll call k, can be found b the formula k = +k. We are using sequence notation from Chapter 9. Getting a common denominator and factoring out the in the numerator, we get k = k+. The domain consists of the intervals determined b successive points k : k, k + = k+, k+. In order to capture all of the intervals in the domain, k must run through all of the integers, that is, k = 0, ±, ±,.... The wa we denote taking the union of infinitel man intervals like this is to use what we call in this tet etended interval notation. The domain of F t = sect can now be written as k= k +, k + The reader should compare this notation with summation notation introduced in Section 9., in particular the notation used to describe geometric series in Theorem 9.. In the same wa the inde k in the series k= ar k can never equal the upper limit, but rather, ranges through all of the natural numbers, the inde k in the union k + k +, k= can never actuall be or, but rather, this conves the idea that k ranges through all of the integers. Now that we have painstakingl determined the domain of F t = sect, it is time to discuss the range. Once again, we appeal to the definition F t = sect = cost. The range of ft = cost is [, ], and since F t = sect is undefined when cost = 0, we split our discussion into two cases: when 0 < cost and when cost < 0. If 0 < cost, then we can divide the inequalit cost b cost to obtain sect = cost. Moreover, using the notation introduced in Section., we have that as cost 0 +, sect = cost ver big +. ver small + In other words, as cost 0 +, sect. If, on the other hand, if cost < 0, then dividing b cost causes a reversal of the inequalit so that sect = sect. In this case, as cost 0, sect = cost ver big, so that as cost ver small 0, we get sect. Since

0. The Si Circular Functions and Fundamental Identities 757 ft = cost admits all of the values in [, ], the function F t = sect admits all of the values in, ] [,. Using set-builder notation, the range of F t = sect can be written as {u : u or u }, or, more succinctl, 8 as {u : u }. 9 Similar arguments can be used to determine the domains and ranges of the remaining three circular functions: csct, tant and cott. The reader is encouraged to do so. See the Eercises. For now, we gather these facts into the theorem below. Theorem 0.. Domains and Ranges of the Circular Functions The function ft = cost The function gt = sint has domain, has domain, has range [, ] has range [, ] The function F t = sect = cost has domain {t : t + k, for integers k} = has range {u : u } =, ] [, The function Gt = csct = sint has domain {t : t k, for integers k} = has range {u : u } =, ] [, k= k= The function Jt = tant = sint cost has domain {t : t + k, for integers k} = has range, The function Kt = cott = cost sint has domain {t : t k, for integers k} = has range, k= k +, k, k + k= k +, k, k + k + k + 8 Using Theorem. from Section.. 9 Notice we have used the variable u as the dumm variable to describe the range elements. While there is no mathematical reason to do this we are describing a set of real numbers, and, as such, could use t again we choose u to help solidif the idea that these real numbers are the outputs from the inputs, which we have been calling t.

758 Foundations of Trigonometr We close this section with a few notes about solving equations which involve the circular functions. First, the discussion on page 75 in Section 0.. concerning solving equations applies to all si circular functions, not just ft = cost and gt = sint. In particular, to solve the equation cott = for real numbers t, we can use the same thought process we used in Eample 0.., number to solve cotθ = for angles θ in radian measure we just need to remember to write our answers using the variable t as opposed to θ. Net, it is critical that ou know the domains and ranges of the si circular functions so that ou know which equations have no solutions. For eample, sect = has no solution because is not in the range of secant. Finall, ou will need to review the notions of reference angles and coterminal angles so that ou can see wh csct = has an infinite set of solutions in Quadrant III and another infinite set of solutions in Quadrant IV.

0. The Si Circular Functions and Fundamental Identities 759 0.. Eercises In Eercises - 0, find the eact value or state that it is undefined.. tan 5. tan 6. sec 6 9. tan 7 0. sec. tan 7. tan 6. sec 5. sec. csc 5 6 7. csc. cot 8. cot. csc. cot 5 5. csc 7 8. sec 7 9. csc 7 6. cot 6 0. cot In Eercises -, use the given the information to find the eact values of the remaining circular functions of θ.. sinθ = 5. cscθ = 5 5. cscθ = 0 9 9 with θ in Quadrant II. tanθ = 5 with θ in Quadrant III with θ in Quadrant I. secθ = 7 with θ in Quadrant IV with θ in Quadrant III 6. cotθ = with θ in Quadrant II 7. tanθ = with θ in Quadrant IV. 8. secθ = with θ in Quadrant II. 9. cotθ = 5 with θ in Quadrant III. 0. cosθ = with θ in Quadrant I.. cotθ = with 0 < θ <.. cscθ = 5 with < θ <.. tanθ = 0 with < θ <.. secθ = 5 with < θ <. In Eercises 5 -, use our calculator to approimate the given value to three decimal places. Make sure our calculator is in the proper angle measurement mode! 5. csc78.95 6. tan.0 7. cot9.99 8. sec07 9. csc5.90 0. tan9.67. cot. sec0.5

760 Foundations of Trigonometr In Eercises - 57, find all of the angles which satisf the equation.. tanθ =. secθ = 5. cscθ = 6. cotθ = 7. tanθ = 0 8. secθ = 9. cscθ = 50. cotθ = 0 5. tanθ = 5. secθ = 0 5. cscθ = 5. secθ = 55. tanθ = 56. cscθ = 57. cotθ = In Eercises 58-65, solve the equation for t. Give eact values. 58. cott = 59. tant = 6. cott = 6. tant = 60. sect = 6. sect = In Eercises 66-69, use Theorem 0.0 to find the requested quantities. 6. csct = 0 65. csct = 66. Find θ, a, and c. 67. Find α, b, and c. b θ c 9 60 a α c 68. Find θ, a, and c. 69. Find β, b, and c. b β a 7 c θ c 50.5 6

0. The Si Circular Functions and Fundamental Identities 76 In Eercises 70-75, use Theorem 0.0 to answer the question. Assume that θ is an angle in a right triangle. 70. If θ = 0 and the side opposite θ has length, how long is the side adjacent to θ? 7. If θ = 5 and the hpotenuse has length 0, how long is the side opposite θ? 7. If θ = 87 and the side adjacent to θ has length, how long is the side opposite θ? 7. If θ = 8. and the side opposite θ has lengh, how long is the hpoteneuse? 7. If θ =.05 and the hpotenuse has length.98, how long is the side adjacent to θ? 75. If θ = and the side adjacent to θ has length, how long is the side opposite θ? 76. A tree standing verticall on level ground casts a 0 foot long shadow. The angle of elevation from the end of the shadow to the top of the tree is.. Find the height of the tree to the nearest foot. With the help of our classmates, research the term umbra versa and see what it has to do with the shadow in this problem. 77. The broadcast tower for radio station WSAZ Home of Algebra in the Morning with Carl and Jeff has two enormous flashing red lights on it: one at the ver top and one a few feet below the top. From a point 5000 feet awa from the base of the tower on level ground the angle of elevation to the top light is 7.970 and to the second light is 7.5. Find the distance between the lights to the nearest foot. 78. On page 75 we defined the angle of inclination also known as the angle of elevation and in this eercise we introduce a related angle - the angle of depression also known as the angle of declination. The angle of depression of an object refers to the angle whose initial side is a horizontal line above the object and whose terminal side is the line-of-sight to the object below the horizontal. This is represented schematicall below. horizontal θ observer object The angle of depression from the horizontal to the object is θ a Show that if the horizontal is above and parallel to level ground then the angle of depression from observer to object and the angle of inclination from object to observer will be congruent because the are alternate interior angles.

76 Foundations of Trigonometr b From a firetower 00 feet above level ground in the Sasquatch National Forest, a ranger spots a fire off in the distance. The angle of depression to the fire is.5. How far awa from the base of the tower is the fire? c The ranger in part 78b sees a Sasquatch running directl from the fire towards the firetower. The ranger takes two sightings. At the first sighting, the angle of depression from the tower to the Sasquatch is 6. The second sighting, taken just 0 seconds later, gives the the angle of depression as 6.5. How far did the Saquatch travel in those 0 seconds? Round our answer to the nearest foot. How fast is it running in miles per hour? Round our answer to the nearest mile per hour. If the Sasquatch keeps up this pace, how long will it take for the Sasquatch to reach the firetower from his location at the second sighting? Round our answer to the nearest minute. 79. When I stand 0 feet awa from a tree at home, the angle of elevation to the top of the tree is 50 and the angle of depression to the base of the tree is 0. What is the height of the tree? Round our answer to the nearest foot. 80. From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of Lake Ippizuti, a lifeguard spots a boat out on the lake sailing directl toward the lighthouse. The first sighting had an angle of depression of 8. and the second sighting had an angle of depression of 5.9. How far had the boat traveled between the sightings? 8. A gu wire 000 feet long is attached to the top of a tower. When pulled taut it makes a angle with the ground. How tall is the tower? How far awa from the base of the tower does the wire hit the ground? In Eercises 8-8, verif the identit. Assume that all quantities are defined. 8. cosθ secθ = 8. tanθ cosθ = sinθ 8. sinθ cscθ = 85. tanθ cotθ = 86. cscθ cosθ = cotθ 87. 88. 90. 9. cosθ sin = cscθ cotθ θ 89. cosθ sinθ = cscθ cotθ 9. sinθ cos = secθ tanθ θ + sinθ cosθ = secθ + tanθ cosθ sin θ = secθ sinθ cos θ = cscθ 9. secθ + tan θ = cosθ

0. The Si Circular Functions and Fundamental Identities 76 9. 96. cscθ + cot θ = sinθ 95. tanθ sec θ = cotθ cotθ csc θ = tanθ 97. cos θ + sin θ = 98. 9 cos θ sin θ = 8 99. tan θ = tanθ sec θ tanθ 00. sin 5 θ = cos θ sinθ 0. sec 0 θ = + tan θ sec θ 0. cos θ tan θ = tanθ sinθ cosθ 0. sec θ sec θ = tan θ + tan θ 0. cosθ + cosθ = + secθ secθ 05. sinθ + sinθ = + cscθ cscθ 06. cotθ + cotθ = tanθ tanθ + 07. tanθ cosθ sinθ = + tanθ cosθ + sinθ 08. tanθ + cotθ = secθ cscθ 09. cscθ sinθ = cotθ cosθ 0. cosθ secθ = tanθ sinθ. cosθtanθ + cotθ = cscθ. sinθtanθ + cotθ = secθ.. 6. 8. 0... 6. 8. cosθ + + cosθ = csc θ secθ + + secθ = cscθ cotθ 5. cscθ + + = secθ tanθ cscθ cscθ cotθ cscθ + cotθ = cotθ 7. cosθ tanθ + sinθ = sinθ + cosθ cotθ secθ + tanθ = secθ tanθ 9. = secθ + tanθ secθ tanθ cscθ cotθ = cscθ + cotθ. = cscθ cotθ cscθ + cotθ sinθ = sec θ + secθ tanθ. cosθ = csc θ + cscθ cotθ 5. cosθ + sinθ = sinθ cosθ sinθ = secθ tanθ + sinθ + sinθ = sec θ secθ tanθ + cosθ = csc θ cscθ cotθ 7. cscθ cotθ = sinθ + cosθ

76 Foundations of Trigonometr In Eercises 9 -, verif the identit. You ma need to consult Sections. and 6. for a review of the properties of absolute value and logarithms before proceeding. 9. ln secθ = ln cosθ 0. ln cscθ = ln sinθ. ln secθ tanθ = ln secθ + tanθ. ln cscθ + cotθ = ln cscθ cotθ. Verif the domains and ranges of the tangent, cosecant and cotangent functions as presented in Theorem 0... As we did in Eercise 7 in Section 0., let α and β be the two acute angles of a right triangle. Thus α and β are complementar angles. Show that secα = cscβ and tanα = cotβ. The fact that co-functions of complementar angles are equal in this case is not an accident and a more general result will be given in Section 0.. 5. We wish to establish the inequalit cosθ < sinθ < for 0 < θ <. Use the diagram from θ the beginning of the section, partiall reproduced below, to answer the following. Q P θ O B, 0 a Show that triangle OP B has area sinθ. b Show that the circular sector OP B with central angle θ has area θ. c Show that triangle OQB has area tanθ. d Comparing areas, show that sinθ < θ < tanθ for 0 < θ <. e Use the inequalit sinθ < θ to show that sinθ θ f Use the inequalit θ < tanθ to show that cosθ < sinθ θ with the previous part to complete the proof. < for 0 < θ <. for 0 < θ <. Combine this

0. The Si Circular Functions and Fundamental Identities 765 6. Show that cosθ < sinθ θ < also holds for < θ < 0. 7. Eplain wh the fact that tanθ = = does not mean sinθ = and cosθ =? See the solution to number 6 in Eample 0...

766 Foundations of Trigonometr 0.. Answers. tan =. sec = 6. cot = 5. tan = 6 7. csc = 8. cot 0. sec 5 5. csc = 6 6. sec is undefined = 0 9. tan 7 = 0 =. csc is undefined. cot 5 is undefined. tan is undefined. sec = 5. csc 7 = 7 6. cot = 7. tan = 8. sec 7 = 6 9. csc = 0. cot =. sinθ = 5, cosθ = 5, tanθ =, cscθ = 5, secθ = 5, cotθ =. sinθ =, cosθ = 5, tanθ = 5, cscθ =, secθ = 5, cotθ = 5. sinθ = 5, cosθ = 7 5, tanθ = 5 7, cscθ =, secθ = 5 7, cotθ = 7. sinθ = 7, cosθ = 7, tanθ =, cscθ = 7, secθ = 7, cotθ = 5. sinθ = 9 6. sinθ = 0, cosθ = 0, tanθ = 9, cscθ = 0 9 9, secθ = 0, cotθ = 9 9 50 50, cosθ = 50 50, tanθ =, cscθ = 50, secθ = 7. sinθ = 5 5, cosθ = 5 5, tanθ =, cscθ = 5, secθ = 5, cotθ = 5 50, cotθ = 8. sinθ =, cosθ =, tanθ = 5, cscθ = 5 5, secθ =, cotθ = 5 5 9. sinθ = 6 6, cosθ = 0 6, tanθ = 5 5, cscθ = 6, secθ = 0 5, cotθ = 5 0. sinθ =, cosθ =, tanθ =, cscθ =, secθ =, cotθ =. sinθ = 5 5, cosθ = 5 5, tanθ =, cscθ = 5, secθ = 5, cotθ =. sinθ = 5, cosθ = 6 5, tanθ = 6, cscθ = 5, secθ = 5 6, cotθ = 6. sinθ = 0 0 0, secθ =, cotθ =, cosθ =, tanθ = 0, cscθ =. sinθ = 95 0, cosθ = 5 0, tanθ = 9, cscθ = 95 9, secθ = 5, cotθ = 9 9 0 0

0. The Si Circular Functions and Fundamental Identities 767 5. csc78.95.09 6. tan.0.9 7. cot9.99.9 8. sec07. 9. csc5.90.688 0. tan9.67 0.89. cot 9.08. sec0.5.. tanθ = when θ = + k for an integer k. secθ = when θ = + k or θ = 5 + k for an integer k 5. cscθ = when θ = + k for an integer k. 6. cotθ = when θ = + k for an integer k 7. tanθ = 0 when θ = k for an integer k 8. secθ = when θ = k for an integer k 9. cscθ = when θ = 6 + k or θ = 5 6 + k for an integer k. 50. cotθ = 0 when θ = + k for an integer k 5. tanθ = when θ = + k for an integer k 5. secθ = 0 never happens 5. cscθ = never happens 5. secθ = when θ = + k = k + for an integer k 55. tanθ = when θ = + k for an integer k 56. cscθ = when θ = 7 6 + k or θ = 6 + k for an integer k 57. cotθ = when θ = + k for an integer k 58. cott = when t = + k for an integer k 59. tant = when t = + k for an integer k 6

768 Foundations of Trigonometr 60. sect = 6. csct = 0 never happens when t = 5 6 + k or t = 7 6 + k for an integer k 6. cott = when t = 5 6 + k for an integer k 6. tant = 6. sect = 65. csct = when t = 5 6 when t = 6 + k for an integer k + k or t = 6 when t = + k or t = 66. θ = 0, a =, c = 08 = 6 67. α = 56, b = tan = 8.09, c = sec = 68. θ =, a = 6 cot7 = + k for an integer k + k for an integer k 6 tan7 5.595, c = 6 csc7 = cos.75 6 sin7 8.0 69. β = 0, b =.5 tan50.979, c =.5 sec50 =.5 cos50.889 70. The side adjacent to θ has length 6.98 7. The side opposite θ has length 0 sin5.588 7. The side opposite θ is tan87 8.6 7. The hpoteneuse has length csc8. = sin8..69 7. The side adjacent to θ has length.98 cos.05.977 75. The side opposite θ has length tan 7.9 76. The tree is about 7 feet tall. 77. The lights are about 75 feet apart. 78. b The fire is about 58 feet from the base of the tower. c The Sasquatch ran 00 cot6 00 cot6.5 7 feet in those 0 seconds. This translates to 0 miles per hour. At the scene of the second sighting, the Sasquatch was 755 feet from the tower, which means, if it keeps up this pace, it will reach the tower in about minutes.

0. The Si Circular Functions and Fundamental Identities 769 79. The tree is about feet tall. 80. The boat has traveled about feet. 8. The tower is about 68 feet tall. The gu wire hits the ground about 7 feet awa from the base of the tower.

770 Foundations of Trigonometr 0. Trigonometric Identities In Section 0., we saw the utilit of the Pthagorean Identities in Theorem 0.8 along with the Quotient and Reciprocal Identities in Theorem 0.6. Not onl did these identities help us compute the values of the circular functions for angles, the were also useful in simplifing epressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beond. Our first set of identities is the Even / Odd identities. Theorem 0.. Even / Odd Identities: For all applicable angles θ, ˆ cos θ = cosθ ˆ sin θ = sinθ ˆ tan θ = tanθ ˆ sec θ = secθ ˆ csc θ = cscθ ˆ cot θ = cotθ In light of the Quotient and Reciprocal Identities, Theorem 0.6, it suffices to show cos θ = cosθ and sin θ = sinθ. The remaining four circular functions can be epressed in terms of cosθ and sinθ so the proofs of their Even / Odd Identities are left as eercises. Consider an angle θ plotted in standard position. Let θ 0 be the angle coterminal with θ with 0 θ 0 <. We can construct the angle θ 0 b rotating counter-clockwise from the positive -ais to the terminal side of θ as pictured below. Since θ and θ 0 are coterminal, cosθ = cosθ 0 and sinθ = sinθ 0. θ 0 θ 0 P cosθ 0, sinθ 0 θ Qcos θ 0, sin θ 0 θ 0 We now consider the angles θ and θ 0. Since θ is coterminal with θ 0, there is some integer k so that θ = θ 0 + k. Therefore, θ = θ 0 k = θ 0 + k. Since k is an integer, so is k, which means θ is coterminal with θ 0. Hence, cos θ = cos θ 0 and sin θ = sin θ 0. Let P and Q denote the points on the terminal sides of θ 0 and θ 0, respectivel, which lie on the Unit Circle. B definition, the coordinates of P are cosθ 0, sinθ 0 and the coordinates of Q are cos θ 0, sin θ 0. Since θ 0 and θ 0 sweep out congruent central sectors of the Unit Circle, it As mentioned at the end of Section 0., properties of the circular functions when thought of as functions of angles in radian measure hold equall well if we view these functions as functions of real numbers. Not surprisingl, the Even / Odd properties of the circular functions are so named because the identif cosine and secant as even functions, while the remaining four circular functions are odd. See Section.6.

0. Trigonometric Identities 77 follows that the points P and Q are smmetric about the -ais. Thus, cos θ 0 = cosθ 0 and sin θ 0 = sinθ 0. Since the cosines and sines of θ 0 and θ 0 are the same as those for θ and θ, respectivel, we get cos θ = cosθ and sin θ = sinθ, as required. The Even / Odd Identities are readil demonstrated using an of the common angles noted in Section 0.. Their true utilit, however, lies not in computation, but in simplifing epressions involving the circular functions. In fact, our net batch of identities makes heav use of the Even / Odd Identities. Theorem 0.. Sum and Difference Identities for Cosine: For all angles α and β, ˆ ˆ cosα + β = cosα cosβ sinα sinβ cosα β = cosα cosβ + sinα sinβ We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles α and β to angles α 0 and β 0, coterminal with α and β, respectivel, each of which measure between 0 and radians. Since α and α 0 are coterminal, as are β and β 0, it follows that α β is coterminal with α 0 β 0. Consider the case below where α 0 β 0. P cosα 0, sinα 0 α 0 β 0 Qcosβ 0, sinβ 0 Acosα 0 β 0, sinα 0 β 0 α 0 β 0 α 0 β 0 O O B, 0 Since the angles P OQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B. The distance formula, Equation., ields cosα0 cosβ 0 + sinα 0 sinβ 0 = cosα 0 β 0 + sinα 0 β 0 0 Squaring both sides, we epand the left hand side of this equation as cosα 0 cosβ 0 + sinα 0 sinβ 0 = cos α 0 cosα 0 cosβ 0 + cos β 0 + sin α 0 sinα 0 sinβ 0 + sin β 0 = cos α 0 + sin α 0 + cos β 0 + sin β 0 cosα 0 cosβ 0 sinα 0 sinβ 0 In the picture we ve drawn, the triangles P OQ and AOB are congruent, which is even better. However, α 0 β 0 could be 0 or it could be, neither of which makes a triangle. It could also be larger than, which makes a triangle, just not the one we ve drawn. You should think about those three cases.

77 Foundations of Trigonometr From the Pthagorean Identities, cos α 0 + sin α 0 = and cos β 0 + sin β 0 =, so cosα 0 cosβ 0 + sinα 0 sinβ 0 = cosα 0 cosβ 0 sinα 0 sinβ 0 Turning our attention to the right hand side of our equation, we find cosα 0 β 0 + sinα 0 β 0 0 = cos α 0 β 0 cosα 0 β 0 + + sin α 0 β 0 = + cos α 0 β 0 + sin α 0 β 0 cosα 0 β 0 Once again, we simplif cos α 0 β 0 + sin α 0 β 0 =, so that cosα 0 β 0 + sinα 0 β 0 0 = cosα 0 β 0 Putting it all together, we get cosα 0 cosβ 0 sinα 0 sinβ 0 = cosα 0 β 0, which simplifies to: cosα 0 β 0 = cosα 0 cosβ 0 + sinα 0 sinβ 0. Since α and α 0, β and β 0 and α β and α 0 β 0 are all coterminal pairs of angles, we have cosα β = cosα cosβ + sinα sinβ. For the case where α 0 β 0, we can appl the above argument to the angle β 0 α 0 to obtain the identit cosβ 0 α 0 = cosβ 0 cosα 0 + sinβ 0 sinα 0. Appling the Even Identit of cosine, we get cosβ 0 α 0 = cos α 0 β 0 = cosα 0 β 0, and we get the identit in this case, too. To get the sum identit for cosine, we use the difference formula along with the Even/Odd Identities cosα + β = cosα β = cosα cos β + sinα sin β = cosα cosβ sinα sinβ We put these newfound identities to good use in the following eample. Eample 0.... Find the eact value of cos 5.. Verif the identit: cos θ = sinθ. Solution.. In order to use Theorem 0. to find cos 5, we need to write 5 as a sum or difference of angles whose cosines and sines we know. One wa to do so is to write 5 = 5 0. cos 5 = cos 5 0 = cos 5 cos 0 + sin 5 sin 0 = + = 6 +

0. Trigonometric Identities 77. In a straightforward application of Theorem 0., we find cos θ = cos cos θ + sin sin θ = 0 cosθ + sinθ = sinθ The identit verified in Eample 0.., namel, cos θ = sinθ, is the first of the celebrated cofunction identities. These identities were first hinted at in Eercise 7 in Section 0.. From sinθ = cos θ, we get: [ ] sin θ = cos θ = cosθ, which sas, in words, that the co sine of an angle is the sine of its co mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as eercises. Theorem 0.. Cofunction Identities: For all applicable angles θ, ˆ cos θ = sinθ ˆ sec θ = cscθ ˆ tan θ = cotθ ˆ sin θ = cosθ ˆ csc θ = secθ ˆ cot θ = tanθ With the Cofunction Identities in place, we are now in the position to derive the sum and difference formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identit, then epand using the difference formula for cosine sinα + β = cos α + β [ ] = cos α β = cos α cosβ + sin α sinβ = sinα cosβ + cosα sinβ We can derive the difference formula for sine b rewriting sinα β as sinα + β and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader. Theorem 0.5. Sum and Difference Identities for Sine: For all angles α and β, ˆ sinα + β = sinα cosβ + cosα sinβ ˆ sinα β = sinα cosβ cosα sinβ

77 Foundations of Trigonometr Eample 0.... Find the eact value of sin 9. If α is a Quadrant II angle with sinα = 5, and β is a Quadrant III angle with tanβ =, find sinα β.. Derive a formula for tanα + β in terms of tanα and tanβ. Solution.. As in Eample 0.., we need to write the angle 9 as a sum or difference of common angles. The denominator of suggests a combination of angles with denominators and. One such combination is 9 = +. Appling Theorem 0.5, we get 9 sin = sin + = sin cos + cos sin = + = 6. In order to find sinα β using Theorem 0.5, we need to find cosα and both cosβ and sinβ. To find cosα, we use the Pthagorean Identit cos α + sin α =. Since sinα = 5, we have cos α + 5 =, or cosα = ±. Since α is a Quadrant II angle, cosα =. We now set about finding cosβ and sinβ. We have several was to proceed, but the Pthagorean Identit + tan β = sec β is a quick wa to get secβ, and hence, cosβ. With tanβ =, we get + = sec β so that secβ = ± 5. Since β is a Quadrant III angle, we choose secβ = 5 so cosβ = secβ = = 5 5 5. We now need to determine sinβ. We could use The Pthagorean Identit cos β + sin β =, but we opt instead to use a quotient identit. From tanβ = sinβ cosβ, we have sinβ = tanβ cosβ so we get sinβ = 5 5 = 5 5. We now have all the pieces needed to find sinα β: sinα β = sinα cosβ cosα sinβ 5 5 = 5 5 5 = 9 5 65

0. Trigonometric Identities 775. We can start epanding tanα + β using a quotient identit and our sum formulas tanα + β = = sinα + β cosα + β sinα cosβ + cosα sinβ cosα cosβ sinα sinβ Since tanα = sinα sinβ cosα and tanβ = cosβ, it looks as though if we divide both numerator and denominator b cosα cosβ we will have what we want tanα + β = sinα cosβ + cosα sinβ cosα cosβ sinα sinβ cosα cosβ cosα cosβ = = = sinα cosβ cosα sinβ + cosα cosβ cosα cosβ cosα cosβ sinα sinβ cosα cosβ cosα cosβ sinα cosβ cosα cosβ + cosα sinβ cosα cosβ cosα cosβ sinα sinβ cosα cosβ cosα cosβ tanα + tanβ tanα tanβ Naturall, this formula is limited to those cases where all of the tangents are defined. The formula developed in Eercise 0.. for tanα+β can be used to find a formula for tanα β b rewriting the difference as a sum, tanα+ β, and the reader is encouraged to fill in the details. Below we summarize all of the sum and difference formulas for cosine, sine and tangent. Theorem 0.6. Sum and Difference Identities: For all applicable angles α and β, ˆ ˆ ˆ cosα ± β = cosα cosβ sinα sinβ sinα ± β = sinα cosβ ± cosα sinβ tanα ± β = tanα ± tanβ tanα tanβ In the statement of Theorem 0.6, we have combined the cases for the sum + and difference of angles into one formula. The convention here is that if ou want the formula for the sum + of

776 Foundations of Trigonometr two angles, ou use the top sign in the formula; for the difference,, use the bottom sign. For eample, tanα tanβ tanα β = + tanα tanβ If we specialize the sum formulas in Theorem 0.6 to the case when α = β, we obtain the following Double Angle Identities. Theorem 0.7. Double Angle Identities: For all applicable angles θ, cos θ sin θ ˆ cosθ = cos θ sin θ ˆ sinθ = sinθ cosθ ˆ tanθ = tanθ tan θ The three different forms for cosθ can be eplained b our abilit to echange squares of cosine and sine via the Pthagorean Identit cos θ + sin θ = and we leave the details to the reader. It is interesting to note that to determine the value of cosθ, onl one piece of information is required: either cosθ or sinθ. To determine sinθ, however, it appears that we must know both sinθ and cosθ. In the net eample, we show how we can find sinθ knowing just one piece of information, namel tanθ. Eample 0.... Suppose P, lies on the terminal side of θ when θ is plotted in standard position. Find cosθ and sinθ and determine the quadrant in which the terminal side of the angle θ lies when it is plotted in standard position.. If sinθ = for θ, find an epression for sinθ in terms of.. Verif the identit: sinθ = tanθ + tan θ.. Epress cosθ as a polnomial in terms of cosθ. Solution.. Using Theorem 0. from Section 0. with = and =, we find r = + = 5. Hence, cosθ = 5 and sinθ = 5. Appling Theorem 0.7, we get cosθ = cos θ sin θ = 5 5 = 7 5, and sinθ = sinθ cosθ = 5 5 = 5. Since both cosine and sine of θ are negative, the terminal side of θ, when plotted in standard position, lies in Quadrant III.

0. Trigonometric Identities 777. If our first reaction to sinθ = is No it s not, cosθ =! then ou have indeed learned something, and we take comfort in that. However, contet is everthing. Here, is just a variable - it does not necessaril represent the -coordinate of the point on The Unit Circle which lies on the terminal side of θ, assuming θ is drawn in standard position. Here, represents the quantit sinθ, and what we wish to know is how to epress sinθ in terms of. We will see more of this kind of thing in Section 0.6, and, as usual, this is something we need for Calculus. Since sinθ = sinθ cosθ, we need to write cosθ in terms of to finish the problem. We substitute = sinθ into the Pthagorean Identit, cos θ + sin θ =, to get cos θ + =, or cosθ = ±. Since θ, cosθ 0, and thus cosθ =. Our final answer is sinθ = sinθ cosθ =.. We start with the right hand side of the identit and note that + tan θ = sec θ. From this point, we use the Reciprocal and Quotient Identities to rewrite tanθ and secθ in terms of cosθ and sinθ: sinθ tanθ + tan = tanθ sinθ θ sec θ = = cos θ cosθ = cosθ cos θ sinθ cosθ cosθ cosθ = sinθ cosθ = sinθ. In Theorem 0.7, one of the formulas for cosθ, namel cosθ = cos θ, epresses cosθ as a polnomial in terms of cosθ. We are now asked to find such an identit for cosθ. Using the sum formula for cosine, we begin with cosθ = cosθ + θ = cosθ cosθ sinθ sinθ Our ultimate goal is to epress the right hand side in terms of cosθ onl. We substitute cosθ = cos θ and sinθ = sinθ cosθ which ields cosθ = cosθ cosθ sinθ sinθ = cos θ cosθ sinθ cosθ sinθ = cos θ cosθ sin θ cosθ Finall, we echange sin θ for cos θ courtes of the Pthagorean Identit, and get and we are done. cosθ = cos θ cosθ sin θ cosθ = cos θ cosθ cos θ cosθ = cos θ cosθ cosθ + cos θ = cos θ cosθ

778 Foundations of Trigonometr In the last problem in Eample 0.., we saw how we could rewrite cosθ as sums of powers of cosθ. In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identit cosθ = cos θ for cos θ and the identit cosθ = sin θ for sin θ results in the aptl-named Power Reduction formulas below. Theorem 0.8. Power Reduction Formulas: For all angles θ, ˆ ˆ cos θ = + cosθ sin θ = cosθ Eample 0... Rewrite sin θ cos θ as a sum and difference of cosines to the first power. Solution. We begin with a straightforward application of Theorem 0.8 sin θ cos θ = cosθ + cosθ = cos θ = cos θ Net, we appl the power reduction formula to cos θ to finish the reduction sin θ cos θ = cos θ + cosθ = = 8 8 cosθ = 8 8 cosθ Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we appl the Power Reduction Formula to cos θ cos θ = + cos θ = + cosθ. We can obtain a formula for cos θ b etracting square roots. In a similar fashion, we ma obtain a half angle formula for sine, and b using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below.

0. Trigonometric Identities 779 Theorem 0.9. Half Angle Formulas: For all applicable angles θ, ˆ ˆ ˆ θ + cosθ cos = ± θ cosθ sin = ± θ cosθ tan = ± + cosθ where the choice of ± depends on the quadrant in which the terminal side of θ lies. Eample 0..5.. Use a half angle formula to find the eact value of cos 5.. Suppose θ 0 with cosθ = 5. Find sin θ.. Use the identit given in number of Eample 0.. to derive the identit θ tan = sinθ + cosθ Solution.. To use the half angle formula, we note that 5 = 0 and since 5 is a Quadrant I angle, its cosine is positive. Thus we have + cos 0 cos 5 + = + = = + = + + = Back in Eample 0.., we found cos 5 b using the difference formula for cosine. In that case, we determined cos 5 = 6+. The reader is encouraged to prove that these two epressions are equal.. If θ 0, then θ 0, which means sin θ < 0. Theorem 0.9 gives θ cos θ 5 sin = = + 5 = 5 8 5 5 = 0 = 5

780 Foundations of Trigonometr. Instead of our usual approach to verifing identities, namel starting with one side of the equation and tring to transform it into the other, we will start with the identit we proved in number of Eample 0.. and manipulate it into the identit we are asked to prove. The identit we are asked to start with is sinθ = tanθ. If we are to use this to derive an identit for tan θ +tan θ, it seems reasonable to proceed b replacing each occurrence of θ with θ sin θ tan θ = + tan θ tan θ sinθ = + tan θ We now have the sinθ we need, but we somehow need to get a factor of + cosθ involved. To get cosines involved, recall that + tan θ = sec θ. We continue to manipulate our given identit b converting secants to cosines and using a power reduction formula tan θ sinθ = + tan θ sinθ = tan θ sec θ sinθ = tan θ cos θ sinθ = tan θ + cos θ sinθ = tan θ + cosθ θ sinθ tan = + cosθ Our net batch of identities, the Product to Sum Formulas, are easil verified b epanding each of the right hand sides in accordance with Theorem 0.6 and as ou should epect b now we leave the details as eercises. The are of particular use in Calculus, and we list them here for reference. Theorem 0.0. Product to Sum Formulas: For all angles α and β, ˆ ˆ ˆ cosα cosβ = [cosα β + cosα + β] sinα sinβ = [cosα β cosα + β] sinα cosβ = [sinα β + sinα + β] These are also known as the Prosthaphaeresis Formulas and have a rich histor. The authors recommend that ou conduct some research on them as our schedule allows.

0. Trigonometric Identities 78 Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section 0.7. These are easil verified using the Product to Sum Formulas, and as such, their proofs are left as eercises. Theorem 0.. Sum to Product Formulas: For all angles α and β, α + β α β ˆ cosα + cosβ = cos cos α + β α β ˆ cosα cosβ = sin sin α ± β α β ˆ sinα ± sinβ = sin cos Eample 0..6.. Write cosθ cos6θ as a sum.. Write sinθ sinθ as a product. Solution.. Identifing α = θ and β = 6θ, we find cosθ cos6θ = [cosθ 6θ + cosθ + 6θ] = cos θ + cos8θ = cosθ + cos8θ, where the last equalit is courtes of the even identit for cosine, cos θ = cosθ.. Identifing α = θ and β = θ ields θ θ sinθ sinθ = sin cos = sin θ cos θ = sin θ cos θ, θ + θ where the last equalit is courtes of the odd identit for sine, sin θ = sinθ. The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles in radian measure appl equall well to the circular trigonometric functions regarded as functions of real numbers. In Eercises 8 - in Section 0.5, we see how some of these identities manifest themselves geometricall as we stud the graphs of the these functions. In the upcoming Eercises, however, ou need to do all of our work analticall without graphs.

78 Foundations of Trigonometr 0.. Eercises In Eercises - 6, use the Even / Odd Identities to verif the identit. Assume all quantities are defined.. sin θ = sinθ. cos 5t = cos 5t +. tan t + = tant. csc θ 5 = cscθ + 5 5. sec 6t = sec6t 6. cot9 7θ = cot7θ 9 In Eercises 7 -, use the Sum and Difference Identities to find the eact value. You ma have need of the Quotient, Reciprocal or Even / Odd Identities as well. 7. cos75 8. sec65 9. sin05 0. csc95. cot55. tan75. cos. sin 7 7 6. cos 7. tan 5 9. cot 0. csc 5. If α is a Quadrant IV angle with cosα =, and sinβ = 5 5. tan 8. sin. sec 0 0, where a cosα + β b sinα + β c tanα + β d cosα β e sinα β f tanα β < β <, find. If cscα =, where 0 < α <, and β is a Quadrant II angle with tanβ = 7, find a cosα + β b sinα + β c tanα + β d cosα β e sinα β f tanα β. If sinα = 5, where 0 < α <, and cosβ = where < β <, find a sinα + β b cosα β c tanα β

0. Trigonometric Identities 78 5. If secα = 5, where < α <, and tanβ = 7, where < β <, find a cscα β b secα + β c cotα + β In Eercises 6-8, verif the identit. 6. cosθ = cosθ 7. sin θ = sinθ 8. tan θ + = cotθ 9. sinα + β + sinα β = sinα cosβ 0. sinα + β sinα β = cosα sinβ. cosα + β + cosα β = cosα cosβ. cosα + β cosα β = sinα sinβ. sinα + β + cotα tanβ = sinα β cotα tanβ. 6. 7. 8. cosα + β tanα tanβ = cosα β + tanα tanβ sint + h sint h cost + h cost h tant + h tant h tanα + β 5. tanα β sinh cosh = cost + sint h h cosh = cost h sint tanh sec t = h tant tanh sinh h = sinα cosα + sinβ cosβ sinα cosα sinβ cosβ In Eercises 9-8, use the Half Angle Formulas to find the eact value. You ma have need of the Quotient, Reciprocal or Even / Odd Identities as well. 9. cos75 compare with Eercise 7 0. sin05 compare with Eercise 9. cos67.5. sin57.5 7. tan.5. cos compare with Eercise 6 5. sin compare with Eercise 8 6. cos 8 7. sin 5 8 8. tan 7 8

78 Foundations of Trigonometr In Eercises 9-58, use the given information about θ to find the eact values of ˆ ˆ sinθ θ sin 9. sinθ = 7 5 where ˆ ˆ cosθ θ cos ˆ ˆ tanθ θ tan < θ < 50. cosθ = 8 5 where 0 < θ < 5. tanθ = 5 where < θ < 5. cscθ = where < θ < 5. cosθ = 5 where 0 < θ < 5. sinθ = 5 where < θ < 55. cosθ = where < θ < 56. sinθ = 5 where < θ < 57. secθ = 5 where < θ < 58. tanθ = where < θ < In Eercises 59-7, verif the identit. Assume all quantities are defined. 59. cosθ + sinθ = + sinθ 60. cosθ sinθ = sinθ 6. tanθ = tanθ + tanθ 6. cscθ = cotθ + tanθ 6. 8 sin θ = cosθ cosθ + 6. 8 cos θ = cosθ + cosθ + 65. sinθ = sinθ sin θ 66. sinθ = sinθ cos θ sin θ cosθ 67. sin θ cos θ = + cosθ cosθ cos6θ 68. sin θ cos θ = cosθ cosθ + cos6θ 69. cosθ = 8 cos θ 8 cos θ + 70. cos8θ = 8 cos 8 θ 56 cos 6 θ + 60 cos θ cos θ + HINT: Use the result to 69. 7. secθ = 7. 7. cosθ cosθ + sinθ + sinθ cosθ sinθ cosθ sinθ + cosθ + sinθ = cosθ cosθ cosθ sinθ cosθ + sinθ = sinθ cosθ

0. Trigonometric Identities 785 In Eercises 7-79, write the given product as a sum. You ma need to use an Even/Odd Identit. 7. cosθ cos5θ 75. sinθ sin7θ 76. sin9θ cosθ 77. cosθ cos6θ 78. sinθ sinθ 79. cosθ sinθ In Eercises 80-85, write the given sum as a product. You ma need to use an Even/Odd or Cofunction Identit. 80. cosθ + cos5θ 8. sinθ sin7θ 8. cos5θ cos6θ 8. sin9θ sin θ 8. sinθ + cosθ 85. cosθ sinθ 86. Suppose θ is a Quadrant I angle with sinθ =. Verif the following formulas a cosθ = b sinθ = c cosθ = 87. Discuss with our classmates how each of the formulas, if an, in Eercise 86 change if we change assume θ is a Quadrant II, III, or IV angle. 88. Suppose θ is a Quadrant I angle with tanθ =. Verif the following formulas a cosθ = + b sinθ = + c sinθ = + d cosθ = + 89. Discuss with our classmates how each of the formulas, if an, in Eercise 88 change if we change assume θ is a Quadrant II, III, or IV angle. 90. If sinθ = for < θ <, find an epression for cosθ in terms of. 9. If tanθ = 7 for < θ <, find an epression for sinθ in terms of. 9. If secθ = for 0 < θ <, find an epression for ln secθ + tanθ in terms of. 9. Show that cos θ sin θ = cos θ = sin θ for all θ. 9. Let θ be a Quadrant III angle with cosθ =. Show that this is not enough information to 5 θ determine the sign of sin b first assuming < θ < 7 and then assuming < θ < θ and computing sin in both cases.

786 Foundations of Trigonometr 95. Without using our calculator, show that + 6 + = 96. In part of Eample 0.., we wrote cosθ as a polnomial in terms of cosθ. In Eercise 69, we had ou verif an identit which epresses cosθ as a polnomial in terms of cosθ. Can ou find a polnomial in terms of cosθ for cos5θ? cos6θ? Can ou find a pattern so that cosnθ could be written as a polnomial in cosine for an natural number n? 97. In Eercise 65, we has ou verif an identit which epresses sinθ as a polnomial in terms of sinθ. Can ou do the same for sin5θ? What about for sinθ? If not, what goes wrong? 98. Verif the Even / Odd Identities for tangent, secant, cosecant and cotangent. 99. Verif the Cofunction Identities for tangent, secant, cosecant and cotangent. 00. Verif the Difference Identities for sine and tangent. 0. Verif the Product to Sum Identities. 0. Verif the Sum to Product Identities.

0. Trigonometric Identities 787 0.. Answers 6 7. cos75 = 6 + 9. sin05 = 8. sec65 = = 6 + 6 0. csc95 = 6 = + 6. cot55 =. cos + =. tan75 = 6 + =. sin = + = 6 5. tan = + = 7 6 6. cos = 7 7. tan = + 6 8. sin = 9. cot = + 5 0. csc = 6. sec = 6. a cosα + β = c tanα + β = 7 0 b sinα + β = 7 0 d cosα β = e sinα β = f tanα β =. a cosα + β = + 7 0 c tanα + β = 8 + + 7 6 00 = b sinα + β = 8 0 d cosα β = + 7 0 e sinα β = 8 + 0. a sinα + β = 6 65 b cosα β = 65 f tanα β = 8 + 7 = 6 + 00 c tanα β = 56

788 Foundations of Trigonometr 5. a cscα β = 5 b secα + β = 5 7 c cotα + β = 7 9. cos75 = 0. sin05 = +. cos67.5 =. tan.5 = + =. cos. sin57.5 = 7 = 5. sin = 5 + 7. sin = 8 + 6. cos = 8 8. tan 7 = 8 + = 9. ˆ sinθ = 6 65 ˆ sin θ = 0 50. ˆ sinθ = 50 809 ˆ sin θ 5 06 = 06 5. ˆ sinθ = 0 69 ˆ sin θ = 5 5. ˆ sinθ = 8 ˆ sin θ 8 + 5 = 5. ˆ sinθ = 5 ˆ sin θ 5 = 5 ˆ cosθ = 57 65 ˆ cos θ 7 = 0 ˆ cosθ = 809 ˆ cos θ 9 06 = 06 ˆ cosθ = 9 69 ˆ cos θ = ˆ cosθ = 7 8 ˆ cos θ 8 5 = ˆ cosθ = 7 5 ˆ cos θ 5 = 5 ˆ tanθ = 6 57 ˆ tan θ = 7 ˆ tanθ = 50 ˆ tan θ 5 = 9 ˆ tanθ = 0 9 ˆ tan θ = 5 ˆ tanθ = 7 ˆ tan θ 8 + 5 = 8 5 tan θ = + 5 ˆ tanθ = 7 ˆ tan θ =

0. Trigonometric Identities 789 5. ˆ sinθ = 5 ˆ sin θ 5 = 5 55. ˆ sinθ = 0 69 ˆ sin θ 6 = 6 56. ˆ sinθ = 0 69 ˆ sin θ 5 6 = 6 ˆ cosθ = 7 5 ˆ cos θ 5 = 5 ˆ cosθ = 9 69 ˆ cos θ = 5 6 6 ˆ cosθ = 9 69 ˆ cos θ 6 = 6 ˆ tanθ = 7 ˆ tan θ = ˆ tanθ = 0 9 ˆ tan θ = 5 ˆ tanθ = 0 9 ˆ tan θ = 5 57. ˆ sinθ = 5 ˆ cosθ = 5 ˆ tanθ = ˆ sin θ 50 0 5 = 0 ˆ cos θ 50 + 0 5 = 0 ˆ tan θ 5 5 = 5 + 5 tan θ = 5 5 5 0 58. ˆ sinθ = 5 ˆ cosθ = 5 ˆ tanθ = ˆ sin θ 50 + 0 5 = 0 ˆ cos θ 50 0 5 = 0 ˆ tan θ 5 + 5 = 5 5 tan θ = 5 + 5 5 0 7. 77. cosθ + cos8θ cosθ + cos8θ 80. cosθ cosθ 75. 78. cos5θ cos9θ cosθ cos5θ 9 8. cos θ 8. cosθ sin5θ 8. cos θ 5 sin θ 76. sin8θ + sin0θ sinθ + sinθ 79. 8. sin θ sin θ 85. sin θ 90. 9. + 9 9. ln + + 6 ln

790 Foundations of Trigonometr 0.5 Graphs of the Trigonometric Functions In this section, we return to our discussion of the circular trigonometric functions as functions of real numbers and pick up where we left off in Sections 0.. and 0... As usual, we begin our stud with the functions ft = cost and gt = sint. 0.5. Graphs of the Cosine and Sine Functions From Theorem 0.5 in Section 0.., we know that the domain of ft = cost and of gt = sint is all real numbers,,, and the range of both functions is [, ]. The Even / Odd Identities in Theorem 0. tell us cos t = cost for all real numbers t and sin t = sint for all real numbers t. This means ft = cost is an even function, while gt = sint is an odd function. Another important propert of these functions is that for coterminal angles α and β, cosα = cosβ and sinα = sinβ. Said differentl, cost+k = cost and sint+k = sint for all real numbers t and an integer k. This last propert is given a special name. Definition 0.. Periodic Functions: A function f is said to be periodic if there is a real number c so that ft + c = ft for all real numbers t in the domain of f. The smallest positive number p for which ft + p = ft for all real numbers t in the domain of f, if it eists, is called the period of f. We have alread seen a famil of periodic functions in Section.: the constant functions. However, despite being periodic a constant function has no period. We ll leave that odd gem as an eercise for ou. Returning to the circular functions, we see that b Definition 0., ft = cost is periodic, since cost + k = cost for an integer k. To determine the period of f, we need to find the smallest real number p so that ft + p = ft for all real numbers t or, said differentl, the smallest positive real number p such that cost + p = cost for all real numbers t. We know that cost + = cost for all real numbers t but the question remains if an smaller real number will do the trick. Suppose p > 0 and cost + p = cost for all real numbers t. Then, in particular, cos0 + p = cos0 so that cosp =. From this we know p is a multiple of and, since the smallest positive multiple of is itself, we have the result. Similarl, we can show gt = sint is also periodic with as its period. Having period essentiall means that we can completel understand everthing about the functions ft = cost and gt = sint b studing one interval of length, sa [0, ]. One last propert of the functions ft = cost and gt = sint is worth pointing out: both of these functions are continuous and smooth. Recall from Section. that geometricall this means the graphs of the cosine and sine functions have no jumps, gaps, holes in the graph, asmptotes, See section.6 for a review of these concepts. Alternativel, we can use the Cofunction Identities in Theorem 0. to show that gt = sint is periodic with period since gt = sint = cos t = f t. Technicall, we should stud the interval [0,, since whatever happens at t = is the same as what happens at t = 0. As we will see shortl, t = gives us an etra check when we go to graph these functions. In some advanced tets, the interval of choice is [,.

0.5 Graphs of the Trigonometric Functions 79 corners or cusps. As we shall see, the graphs of both ft = cost and gt = sint meander nicel and don t cause an trouble. We summarize these facts in the following theorem. Theorem 0.. Properties of the Cosine and Sine Functions The function f = cos The function g = sin has domain, has domain, has range [, ] has range [, ] is continuous and smooth is continuous and smooth is even is odd has period has period In the chart above, we followed the convention established in Section.6 and used as the independent variable and as the dependent variable. 5 This allows us to turn our attention to graphing the cosine and sine functions in the Cartesian Plane. To graph = cos, we make a table as we did in Section.6 using some of the common values of in the interval [0, ]. This generates a portion of the cosine graph, which we call the fundamental ccle of = cos. cos, cos 0 0,, 0, 0,, 5, 5 0 7, 0 7,, The fundamental ccle of = cos. 5 7 A few things about the graph above are worth mentioning. First, this graph represents onl part of the graph of = cos. To get the entire graph, we imagine coping and pasting this graph end to end infinitel in both directions left and right on the -ais. Secondl, the vertical scale here has been greatl eaggerated for clarit and aesthetics. Below is an accurate-to-scale graph of = cos showing several ccles with the fundamental ccle plotted thicker than the others. The 5 The use of and in this contet is not to be confused with the - and -coordinates of points on the Unit Circle which define cosine and sine. Using the term trigonometric function as opposed to circular function can help with that, but one could then ask, He, where s the triangle?

79 Foundations of Trigonometr graph of = cos is usuall described as wavelike indeed, man of the applications involving the cosine and sine functions feature modeling wavelike phenomena. An accuratel scaled graph of = cos. We can plot the fundamental ccle of the graph of = sin similarl, with similar results. sin, sin 0 0 0, 0,,, 0, 0 5, 5 7, 7, 0, 0 The fundamental ccle of = sin. 5 7 As with the graph of = cos, we provide an accuratel scaled graph of = sin below with the fundamental ccle highlighted. An accuratel scaled graph of = sin. It is no accident that the graphs of = cos and = sin are so similar. Using a cofunction identit along with the even propert of cosine, we have sin = cos = cos = cos Recalling Section.7, we see from this formula that the graph of = sin is the result of shifting the graph of = cos to the right units. A visual inspection confirms this. Now that we know the basic shapes of the graphs of = cos and = sin, we can use Theorem.7 in Section.7 to graph more complicated curves. To do so, we need to keep track of

0.5 Graphs of the Trigonometric Functions 79 the movement of some ke points on the original graphs. We choose to track the values = 0,,, and. These quarter marks correspond to quadrantal angles, and as such, mark the location of the zeros and the local etrema of these functions over eactl one period. Before we begin our net eample, we need to review the concept of the argument of a function as first introduced in Section.. For the function f = 5 cos, the argument of f is. We shall have occasion, however, to refer to the argument of the cosine, which in this case is. Loosel stated, the argument of a trigonometric function is the epression inside the function. Eample 0.5.. Graph one ccle of the following functions. State the period of each.. f = cos +. g = sin + Solution.. We set the argument of the cosine,, equal to each of the values: 0,,,, and solve for. We summarize the results below. a 0 = a = 0 = = = = 5 Net, we substitute each of these values into f = cos + to determine the corresponding -values and connect the dots in a pleasing wavelike fashion. f, f,,,, 5 5, 5 One ccle of = f. One ccle is graphed on [, 5] so the period is the length of that interval which is.. Proceeding as above, we set the argument of the sine,, equal to each of our quarter marks and solve for.

79 Foundations of Trigonometr a = a 0 = 0 = = 0 = = We now find the corresponding -values on the graph b substituting each of these -values into g = sin +. Once again, we connect the dots in a wavelike fashion. g, g, 0, 0,,, One ccle of = g. One ccle was graphed on the interval [, ] so the period is =. The functions in Eample 0.5. are eamples of sinusoids. Roughl speaking, a sinusoid is the result of taking the basic graph of f = cos or g = sin and performing an of the transformations 6 mentioned in Section.7. Sinusoids can be characterized b four properties: period, amplitude, phase shift and vertical shift. We have alread discussed period, that is, how long it takes for the sinusoid to complete one ccle. The standard period of both f = cos and g = sin is, but horizontal scalings will change the period of the resulting sinusoid. The amplitude of the sinusoid is a measure of how tall the wave is, as indicated in the figure below. The amplitude of the standard cosine and sine functions is, but vertical scalings can alter this. 6 We have alread seen how the Even/Odd and Cofunction Identities can be used to rewrite g = sin as a transformed version of f = cos, so of course, the reverse is true: f = cos can be written as a transformed version of g = sin. The authors have seen some instances where sinusoids are alwas converted to cosine functions while in other disciplines, the sinusoids are alwas written in terms of sine functions. We will discuss the applications of sinusoids in greater detail in Chapter. Until then, we will keep our options open.

0.5 Graphs of the Trigonometric Functions 795 amplitude baseline period The phase shift of the sinusoid is the horizontal shift eperienced b the fundamental ccle. We have seen that a phase horizontal shift of to the right takes f = cos to g = sin since cos = sin. As the reader can verif, a phase shift of to the left takes g = sin to f = cos. The vertical shift of a sinusoid is eactl the same as the vertical shifts in Section.7. In most contets, the vertical shift of a sinusoid is assumed to be 0, but we state the more general case below. The following theorem, which is reminiscent of Theorem.7 in Section.7, shows how to find these four fundamental quantities from the formula of the given sinusoid. Theorem 0.. For ω > 0, the functions C = A cosω + φ + B and S = A sinω + φ + B ˆ have period ω ˆ have phase shift φ ω ˆ have amplitude A ˆ have vertical shift B We note that in some scientific and engineering circles, the quantit φ mentioned in Theorem 0. is called the phase of the sinusoid. Since our interest in this book is primaril with graphing sinusoids, we focus our attention on the horizontal shift φ ω induced b φ. The proof of Theorem 0. is a direct application of Theorem.7 in Section.7 and is left to the reader. The parameter ω, which is stipulated to be positive, is called the angular frequenc of the sinusoid and is the number of ccles the sinusoid completes over a interval. We can alwas ensure ω > 0 using the Even/Odd Identities. 7 We now test out Theorem 0. using the functions f and g featured in Eample 0.5.. First, we write f in the form prescribed in Theorem 0., f = cos + = cos + +, 7 Tr using the formulas in Theorem 0. applied to C = cos + to see wh we need ω > 0.

796 Foundations of Trigonometr so that A =, ω =, φ = and B =. According to Theorem 0., the period of f is ω = / =, the amplitude is A = =, the phase shift is φ ω = / / = indicating a shift to the right unit and the vertical shift is B = indicating a shift up unit. All of these match with our graph of = f. Moreover, if we start with the basic shape of the cosine graph, shift it unit to the right, unit up, stretch the amplitude to and shrink the period to, we will have reconstructed one period of the graph of = f. In other words, instead of tracking the five quarter marks through the transformations to plot = f, we can use five other pieces of information: the phase shift, vertical shift, amplitude, period and basic shape of the cosine curve. Turning our attention now to the function g in Eample 0.5., we first need to use the odd propert of the sine function to write it in the form required b Theorem 0. g = sin + = sin + = sin + = sin + + We find A =, ω =, φ = and B =. The period is then =, the amplitude is =, the phase shift is = indicating a shift right units and the vertical shift is up. Note that, in this case, all of the data match our graph of = g with the eception of the phase shift. Instead of the graph starting at =, it ends there. Remember, however, that the graph presented in Eample 0.5. is onl one portion of the graph of = g. Indeed, another complete ccle begins at =, and this is the ccle Theorem 0. is detecting. The reason for the discrepanc is that, in order to appl Theorem 0., we had to rewrite the formula for g using the odd propert of the sine function. Note that whether we graph = g using the quarter marks approach or using the Theorem 0., we get one complete ccle of the graph, which means we have completel determined the sinusoid. Eample 0.5.. Below is the graph of one complete ccle of a sinusoid = f., 5 5, 5, 7, 5, One ccle of = f.. Find a cosine function whose graph matches the graph of = f.

0.5 Graphs of the Trigonometric Functions 797. Find a sine function whose graph matches the graph of = f. Solution.. We fit the data to a function of the form C = A cosω + φ + B. Since one ccle is graphed over the interval [, 5], its period is 5 = 6. According to Theorem 0., 6 = ω, so that ω =. Net, we see that the phase shift is, so we have φ ω =, or φ = ω =. To find the amplitude, note that the range of the sinusoid is [, 5 ]. As a result, the amplitude A = [ 5 ] = =. Finall, to determine the vertical shift, we average the endpoints of the range to find B = [ 5 + ] = =. Our final answer is C = cos + +.. Most of the work to fit the data to a function of the form S = A sinω + φ + B is done. The period, amplitude and vertical shift are the same as before with ω =, A = and B =. The trickier part is finding the phase shift. To that end, we imagine etending the graph of the given sinusoid as in the figure below so that we can identif a ccle beginning at 7,. Taking the phase shift to be 7, we get φ ω = 7, or φ = 7 ω = 7 = 7 6. Hence, our answer is S = sin 7 6 +. 5, 5 7,, 5 6 7 8 9 0 9, 5 8, Etending the graph of = f. Note that each of the answers given in Eample 0.5. is one choice out of man possible answers. For eample, when fitting a sine function to the data, we could have chosen to start at, taking A =. In this case, the phase shift is so φ = 6 for an answer of S = sin 6 +. Alternativel, we could have etended the graph of = f to the left and considered a sine function starting at 5,, and so on. Each of these formulas determine the same sinusoid curve and their formulas are all equivalent using identities. Speaking of identities, if we use the sum identit for cosine, we can epand the formula to ield C = A cosω + φ + B = A cosω cosφ A sinω sinφ + B.

798 Foundations of Trigonometr Similarl, using the sum identit for sine, we get S = A sinω + φ + B = A sinω cosφ + A cosω sinφ + B. Making these observations allows us to recognize and graph functions as sinusoids which, at first glance, don t appear to fit the forms of either C or S. Eample 0.5.. Consider the function f = cos sin. Find a formula for f:. in the form C = A cosω + φ + B for ω > 0. in the form S = A sinω + φ + B for ω > 0 Check our answers analticall using identities and graphicall using a calculator. Solution.. The ke to this problem is to use the epanded forms of the sinusoid formulas and match up corresponding coefficients. Equating f = cos sin with the epanded form of C = A cosω + φ + B, we get cos sin = A cosω cosφ A sinω sinφ + B It should be clear that we can take ω = and B = 0 to get cos sin = A cos cosφ A sin sinφ To determine A and φ, a bit more work is involved. We get started b equating the coefficients of the trigonometric functions on either side of the equation. On the left hand side, the coefficient of cos is, while on the right hand side, it is A cosφ. Since this equation is to hold for all real numbers, we must have 8 that A cosφ =. Similarl, we find b equating the coefficients of sin that A sinφ =. What we have here is a sstem of nonlinear equations! We can temporaril eliminate the dependence on φ b using the Pthagorean Identit. We know cos φ + sin φ =, so multipling this b A gives A cos φ+a sin φ = A. Since A cosφ = and A sinφ =, we get A = + = or A = ±. Choosing A =, we have cosφ = and sinφ = or, after some rearrangement, cosφ = and sinφ =. One such angle φ which satisfies this criteria is φ =. Hence, one wa to write f as a sinusoid is f = cos +. We can easil check our answer using the sum formula for cosine f = cos + = [ cos cos sin sin ] [ = cos ] sin = cos sin 8 This should remind ou of equation coefficients of like powers of in Section 8.6.

0.5 Graphs of the Trigonometric Functions 799. Proceeding as before, we equate f = cos sin with the epanded form of S = A sinω + φ + B to get cos sin = A sinω cosφ + A cosω sinφ + B Once again, we ma take ω = and B = 0 so that cos sin = A sin cosφ + A cos sinφ We equate 9 the coefficients of cos on either side and get A sinφ = and A cosφ =. Using A cos φ + A sin φ = A as before, we get A = ±, and again we choose A =. This means sinφ =, or sinφ =, and cosφ =, which means cosφ =. One such angle which meets these criteria is φ = 5 6. Hence, we have f = sin + 5 6. Checking our work analticall, we have f = sin + 5 6 = [ sin cos 5 6 + cos sin 5 ] [ 6 = sin + cos ] = cos sin Graphing the three formulas for f result in the identical curve, verifing our analtic work. It is important to note that in order for the technique presented in Eample 0.5. to fit a function into one of the forms in Theorem 0., the arguments of the cosine and sine function much match. That is, while f = cos sin is a sinusoid, g = cos sin is not. 0 It is also worth mentioning that, had we chosen A = instead of A = as we worked through Eample 0.5., our final answers would have looked different. The reader is encouraged to rework Eample 0.5. using A = to see what these differences are, and then for a challenging eercise, use identities to show that the formulas are all equivalent. The general equations to fit a function of the form f = a cosω + b sinω + B into one of the forms in Theorem 0. are eplored in Eercise 5. 9 Be careful here! 0 This graph does, however, ehibit sinusoid-like characteristics! Check it out!

800 Foundations of Trigonometr 0.5. Graphs of the Secant and Cosecant Functions We now turn our attention to graphing = sec. Since sec = cos, we can use our table of values for the graph of = cos and take reciprocals. We know from Section 0.. that the domain of F = sec ecludes all odd multiples of, and sure enough, we run into trouble at = and = since cos = 0 at these values. Using the notation introduced in Section., we have that as, cos 0 +, so sec. See Section 0.. for a more detailed analsis. Similarl, we find that as +, sec ; as, sec ; and as +, sec. This means we have a pair of vertical asmptotes to the graph of = sec, = and =. Since cos is periodic with period, it follows that sec is also. Below we graph a fundamental ccle of = sec along with a more complete graph obtained b the usual coping and pasting. cos sec, sec 0 0,, 0 undefined,, 5 5, 0 undefined 7, 7, The fundamental ccle of = sec. 5 7 The graph of = sec. Provided secα and secβ are defined, secα = secβ if and onl if cosα = cosβ. Hence, sec inherits its period from cos. In Section 0.., we argued the range of F = sec is, ] [,. We can now see this graphicall.

0.5 Graphs of the Trigonometric Functions 80 As one would epect, to graph = csc we begin with = sin and take reciprocals of the corresponding -values. Here, we encounter issues at = 0, = and =. Proceeding with the usual analsis, we graph the fundamental ccle of = csc below along with the dotted graph of = sin for reference. Since = sin and = cos are merel phase shifts of each other, so too are = csc and = sec. sin csc, csc 0 0 undefined,,, 0 undefined 5,, 7, 5 7 0 undefined 5 7 The fundamental ccle of = csc. Once again, our domain and range work in Section 0.. is verified geometricall in the graph of = G = csc. The graph of = csc. Note that, on the intervals between the vertical asmptotes, both F = sec and G = csc are continuous and smooth. In other words, the are continuous and smooth on their domains. The following theorem summarizes the properties of the secant and cosecant functions. Note that Just like the rational functions in Chapter are continuous and smooth on their domains because polnomials are continuous and smooth everwhere, the secant and cosecant functions are continuous and smooth on their domains since the cosine and sine functions are continuous and smooth everwhere.

80 Foundations of Trigonometr all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectivel. Theorem 0.. Properties of the Secant and Cosecant Functions ˆ The function F = sec has domain { : + k, k is an integer} = has range { : } =, ] [, is continuous and smooth on its domain is even has period k= k +, k + ˆ The function G = csc has domain { : k, k is an integer} = has range { : } =, ] [, is continuous and smooth on its domain is odd has period k= k, k + In the net eample, we discuss graphing more general secant and cosecant curves. Eample 0.5.. Graph one ccle of the following functions. State the period of each.. f = sec Solution.. g = csc 5. To graph = sec, we follow the same procedure as in Eample 0.5.. First, we set the argument of secant,, equal to the quarter marks 0,,, and and solve for. a = a 0 = 0 0 = = = =

0.5 Graphs of the Trigonometric Functions 80 Net, we substitute these values into f. If f eists, we have a point on the graph; otherwise, we have found a vertical asmptote. In addition to these points and asmptotes, we have graphed the associated cosine curve in this case = cos dotted in the picture below. Since one ccle is graphed over the interval [0, ], the period is 0 =. f, f 0 0, undefined, undefined, One ccle of = sec.. Proceeding as before, we set the argument of cosecant in g = csc 5 equal to the quarter marks and solve for. a = a 0 = 0 = = 0 = = Substituting these -values into g, we generate the graph below and find the period to be =. The associated sine curve, = sin 5, is dotted in as a reference. g, g undefined 0 undefined,, undefined One ccle of = csc 5.

80 Foundations of Trigonometr Before moving on, we note that it is possible to speak of the period, phase shift and vertical shift of secant and cosecant graphs and use even/odd identities to put them in a form similar to the sinusoid forms mentioned in Theorem 0.. Since these quantities match those of the corresponding cosine and sine curves, we do not spell this out eplicitl. Finall, since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves. 0.5. Graphs of the Tangent and Cotangent Functions Finall, we turn our attention to the graphs of the tangent and cotangent functions. When constructing a table of values for the tangent function, we see that J = tan is undefined at = and =, in accordance with our findings in Section 0... As, sin and cos 0 +, so that tan = sin cos producing a vertical asmptote at =. Using a similar analsis, we get that as +, tan ; as, tan ; and as +, tan. Plotting this information and performing the usual cop and paste produces: tan, tan 0 0 0, 0, undefined, 0, 0 5 5, undefined 7 7, 0, 0 5 7 The graph of = tan over [0, ]. The graph of = tan.

0.5 Graphs of the Trigonometric Functions 805 From the graph, it appears as if the tangent function is periodic with period. To prove that this is the case, we appeal to the sum formula for tangents. We have: tan + = tan + tan tan tan = tan + 0 tan0 = tan, which tells us the period of tan is at most. To show that it is eactl, suppose p is a positive real number so that tan + p = tan for all real numbers. For = 0, we have tanp = tan0 + p = tan0 = 0, which means p is a multiple of. The smallest positive multiple of is itself, so we have established the result. We take as our fundamental ccle for = tan the interval,, and use as our quarter marks =,, 0, and. From the graph, we see confirmation of our domain and range work in Section 0... It should be no surprise that K = cot behaves similarl to J = tan. Plotting cot over the interval [0, ] results in the graph below. cot, cot 0 undefined 0 undefined 5 0 7 undefined,, 0, 5,, 0 7, 5 7 The graph of = cot over [0, ]. From these data, it clearl appears as if the period of cot is, and we leave it to the reader to prove this. We take as one fundamental ccle the interval 0, with quarter marks: = 0,,, and. A more complete graph of = cot is below, along with the fundamental ccle highlighted as usual. Once again, we see the domain and range of K = cot as read from the graph matches with what we found analticall in Section 0... Certainl, mimicking the proof that the period of tan is an option; for another approach, consider transforming tan to cot using identities.

806 Foundations of Trigonometr The graph of = cot. The properties of the tangent and cotangent functions are summarized below. As with Theorem 0., each of the results below can be traced back to properties of the cosine and sine functions and the definition of the tangent and cotangent functions as quotients thereof. Theorem 0.5. Properties of the Tangent and Cotangent Functions ˆ The function J = tan has domain { : + k, k is an integer} = has range, is continuous and smooth on its domain is odd has period k= k +, k + ˆ The function K = cot has domain { : k, k is an integer} = k, k + has range, is continuous and smooth on its domain is odd has period k=

0.5 Graphs of the Trigonometric Functions 807 Eample 0.5.5. Graph one ccle of the following functions. Find the period.. f = tan.. g = cot + +. Solution.. We proceed as we have in all of the previous graphing eamples b setting the argument of tangent in f = tan, namel, equal to each of the quarter marks,, 0, and, and solving for. a = a = = 0 = 0 0 = = Substituting these -values into f, we find points on the graph and the vertical asmptotes. f, f undefined, 0 0,, 0 0 undefined We see that the period is =. One ccle of = tan.. The quarter marks for the fundamental ccle of the cotangent curve are 0,,, and. To graph g = cot + +, we begin b setting + equal to each quarter mark and solving for.

808 Foundations of Trigonometr a 0 + = a + = 0 + = + = + = + = 0 We now use these -values to generate our graph. g, g undefined,,, 0 undefined We find the period to be 0 =. One ccle of = cot + +. As with the secant and cosecant functions, it is possible to etend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 0.. Since the number of classical applications involving sinusoids far outnumber those involving tangent and cotangent functions, we omit this. The ambitious reader is invited to formulate such a theorem, however.

0.5 Graphs of the Trigonometric Functions 809 0.5. Eercises In Eercises -, graph one ccle of the given function. State the period, amplitude, phase shift and vertical shift of the function.. = sin. = sin. = cos. = cos 5. = sin + 6. = sin 7. = cos + 8. = cos + 9. = sin 0. = cos +. = cos +. = sin + In Eercises -, graph one ccle of the given function. State the period of the function.. = tan. = tan 5. = tan + 6. = sec 7. = csc + 8. = sec + 9. = csc 0. = sec +. = csc. = cot +. = cot 6 5. = cot + + In Eercises 5 -, use Eample 0.5. as a guide to show that the function is a sinusoid b rewriting it in the forms C = A cosω + φ + B and S = A sinω + φ + B for ω > 0 and 0 φ <. 5. f = sin + cos + 6. f = sin cos 7. f = sin + cos 8. f = sin cos 9. f = cos sin 0. f = cos sin + 6. f = cos5 sin5. f = 6 cos 6 sin

80 Foundations of Trigonometr. f = 5 sin 5 cos. f = sin cos 6 6 5. In Eercises 5 -, ou should have noticed a relationship between the phases φ for the S and C. Show that if f = A sinω + α + B, then f = A cosω + β + B where β = α. 6. Let φ be an angle measured in radians and let P a, b be a point on the terminal side of φ when it is drawn in standard position. Use Theorem 0. and the sum identit for sine in Theorem 0.5 to show that f = a sinω + b cosω + B with ω > 0 can be rewritten as f = a + b sinω + φ + B. 7. With the help of our classmates, epress the domains of the functions in Eamples 0.5. and 0.5.5 using etended interval notation. We will revisit this in Section 0.7. In Eercises 8 -, verif the identit b graphing the right and left hand sides on a calculator. 8. sin + cos = 9. sec tan = 0. cos = sin. tan + = tan. sin = sin cos. tan = sin + cos In Eercises - 50, graph the function with the help of our calculator and discuss the given questions with our classmates.. f = cos + sin. Is this function periodic? If so, what is the period? 5. f = sin. What appears to be the horizontal asmptote of the graph? 6. f = sin. Graph = ± on the same set of aes and describe the behavior of f. 7. f = sin. What s happening as 0? 8. f = tan. Graph = on the same set of aes and describe the behavior of f. 9. f = e 0. cos + sin. Graph = ±e 0. on the same set of aes and describe the behavior of f. 50. f = e 0. cos + sin. Graph = ±e 0. on the same set of aes and describe the behavior of f. 5. Show that a constant function f is periodic b showing that f + 7 = f for all real numbers. Then show that f has no period b showing that ou cannot find a smallest number p such that f + p = f for all real numbers. Said another wa, show that f + p = f for all real numbers for ALL values of p > 0, so no smallest value eists to satisf the definition of period.

0.5 Graphs of the Trigonometric Functions 8 0.5.5 Answers. = sin Period: Amplitude: Phase Shift: 0 Vertical Shift: 0. = sin Period: Amplitude: Phase Shift: 0 Vertical Shift: 0 6. = cos Period: Amplitude: Phase Shift: 0 Vertical Shift: 0. = cos Period: Amplitude: Phase Shift: Vertical Shift: 0 5

8 Foundations of Trigonometr 5. = sin + Period: Amplitude: Phase Shift: Vertical Shift: 0 6 7 6 5 6. = sin Period: Amplitude: Phase Shift: Vertical Shift: 0 5 7. = cos + Period: Amplitude: Phase Shift: Vertical Shift: 0 7 0 8. = cos + Period: Amplitude: Phase Shift: Vertical Shift: 5 5 6 7 6

0.5 Graphs of the Trigonometric Functions 8 9. = sin Period: Amplitude: Phase Shift: You need to use = sin + to find this. 5 9 7 5 5 7 Vertical Shift: 0. = cos + Period: 5 Amplitude: Phase Shift: You need to use 8 = cos + to find this. 6 Vertical Shift: 8 8 8 8 5 8. = cos + Period: Amplitude: Phase Shift: 6 6 7 5 6 Vertical Shift:. = sin + Period: Amplitude: Phase Shift: You need to use = sin to find this. 7 Vertical Shift: 0 5 5 Two ccles of the graph are shown to illustrate the discrepanc discussed on page 796. 6 Again, we graph two ccles to illustrate the discrepanc discussed on page 796. 7 This will be the last time we graph two ccles to illustrate the discrepanc discussed on page 796.

8 Foundations of Trigonometr. = tan Period: 6 7 5 6. = tan Period: 5 5. = tan + is equivalent to = tan + + via the Even / Odd identit for tangent. Period: 5 8 8

0.5 Graphs of the Trigonometric Functions 85 6. = sec Start with = cos Period: 5 7. = csc + Start with = sin + Period: 6 7 6 5 8. = sec + Start with = cos + Period: 7 0

86 Foundations of Trigonometr 9. = csc Start with = sin Period: 5 0. = sec + Start with = cos + Period: 5 5 6 7 6. = csc Start with = sin Period: 5 7

0.5 Graphs of the Trigonometric Functions 87. = cot + 6 Period: 6 7 5 6. = cot 5 Period: 5 5 5 5 5. = cot + + Period: 5 8 8

88 Foundations of Trigonometr 5. f = sin + cos + = sin + + = cos + 7 6. f = sin cos = 6 sin + 6 7. f = sin + cos = sin + 8. f = sin cos = sin + 9. f = cos sin = sin 0. f = cos. f = cos5 + = 6 cos + + = cos + sin + 6 = sin + 5 6 sin5 = sin 5 + 7 6. f = 6 cos 6 sin = sin. f = 5 sin 5 cos = 5 sin + 7. f = sin cos = 6 sin 6 6 6 + 5 = cos + 5 6 = cos + 6 + + 6 = cos + + 6 = cos 5 + = cos + 5 6 = 5 cos + 5 = 6 cos 6 + 7 6

0.6 The Inverse Trigonometric Functions 89 0.6 The Inverse Trigonometric Functions As the title indicates, in this section we concern ourselves with finding inverses of the circular trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the si circular functions is one-to-one. To remed this, we restrict the domains of the circular functions in the same wa we restricted the domain of the quadratic function in Eample 5.. in Section 5. to obtain a one-to-one function. We first consider f = cos. Choosing the interval [0, ] allows us to keep the range as [, ] as well as the properties of being smooth and continuous. Restricting the domain of f = cos to [0, ]. Recall from Section 5. that the inverse of a function f is tpicall denoted f. For this reason, some tetbooks use the notation f = cos for the inverse of f = cos. The obvious pitfall here is our convention of writing cos as cos, cos as cos and so on. It is far too eas to confuse cos with cos = sec so we will not use this notation in our tet. Instead, we use the notation f = arccos, read arc-cosine of. To understand the arc in arccosine, recall that an inverse function, b definition, reverses the process of the original function. The function ft = cost takes a real number input t, associates it with the angle θ = t radians, and returns the value cosθ. Digging deeper, we have that cosθ = cost is the -coordinate of the terminal point on the Unit Circle of an oriented arc of length t whose initial point is, 0. Hence, we ma view the inputs to ft = cost as oriented arcs and the outputs as -coordinates on the Unit Circle. The function f, then, would take -coordinates on the Unit Circle and return oriented arcs, hence the arc in arccosine. Below are the graphs of f = cos and f = arccos, where we obtain the latter from the former b reflecting it across the line =, in accordance with Theorem 5.. f = cos, 0 reflect across = switch and coordinates f = arccos. But be aware that man books do! As alwas, be sure to check the contet! See page 70 if ou need a review of how we associate real numbers with angles in radian measure.

80 Foundations of Trigonometr We restrict g = sin in a similar manner, although the interval of choice is [, ]. Restricting the domain of f = sin to [, ]. It should be no surprise that we call g = arcsin, which is read arc-sine of. g = sin,. reflect across = switch and coordinates g = arcsin. We list some important facts about the arccosine and arcsine functions in the following theorem. Theorem 0.6. Properties of the Arccosine and Arcsine Functions ˆ Properties of F = arccos Domain: [, ] Range: [0, ] arccos = t if and onl if 0 t and cost = cosarccos = provided arccoscos = provided 0 ˆ Properties of G = arcsin Domain: [, ] Range: [, ] arcsin = t if and onl if t and sint = sinarcsin = provided arcsinsin = provided additionall, arcsine is odd

0.6 The Inverse Trigonometric Functions 8 Everthing in Theorem 0.6 is a direct consequence of the facts that f = cos for 0 and F = arccos are inverses of each other as are g = sin for and G = arcsin. It s about time for an eample. Eample 0.6... Find the eact values of the following. a arccos c arccos e arccos cos 6 g cos arccos 5 b arcsin d arcsin f arccos cos 6 h sin arccos 5. Rewrite the following as algebraic epressions of and state the domain on which the equivalence is valid. a tan arccos b cos arcsin Solution.. a To find arccos, we need to find the real number t or, equivalentl, an angle measuring t radians which lies between 0 and with cost =. We know t = meets these criteria, so arccos =. b The value of arcsin is a real number t between and with sint =. The number we seek is t =. Hence, arcsin =. lies in the interval [0, ] with cost =. Our answer c The number t = arccos is arccos =. d To find arcsin [, we seek the number t in the interval, ] with sint =. The answer is t = 6 so that arcsin = 6. e Since 0 6, we could simpl invoke Theorem 0.6 to get arccos cos 6 = 6. However, in order to make sure we understand wh this is the case, we choose to work the eample through using the definition of arccosine. Working from the inside out, arccos cos = arccos. Now, arccos is the real number t with 0 t and cost = 6. We find t = 6, so that arccos cos 6 = 6.

8 Foundations of Trigonometr f Since 6 does not fall between 0 and, Theorem 0.6 does not appl. We are forced to work through from the inside out starting with arccos cos 6 = arccos. From the previous problem, we know arccos = 6. Hence, arccos cos 6 = 6. g One wa to simplif cos arccos 5 is to use Theorem 0.6 directl. Since 5 is between and, we have that cos arccos 5 = 5 and we are done. However, as before, to reall understand wh this cancellation occurs, we let t = arccos 5. Then, b definition, cost = 5. Hence, cos arccos 5 = cost = 5, and we are finished in nearl the same amount of time. h As in the previous eample, we let t = arccos 5 so that cost = 5 for some t where 0 t. Since cost < 0, we can narrow this down a bit and conclude that < t <, so that t corresponds to an angle in Quadrant II. In terms of t, then, we need to find sin arccos 5 = sint. Using the Pthagorean Identit cos t + sin t =, we get 5 + sin t = or sint = ± 5. Since t corresponds to a Quadrants II angle, we choose sint = 5. Hence, sin arccos 5 =. a We begin this problem in the same manner we began the previous two problems. To help us see the forest for the trees, we let t = arccos, so our goal is to find a wa to epress tan arccos = tant in terms of. Since t = arccos, we know cost = where 0 t, but since we are after an epression for tant, we know we need to throw out t = from consideration. Hence, either 0 t < or < t so that, geometricall, t corresponds to an angle in Quadrant I or Quadrant II. One approach to finding tant is to use the quotient identit tant = sint cost. Substituting cost = into the Pthagorean Identit cos t + sin t = gives + sin t =, from which we get sint = ±. Since t corresponds to angles in Quadrants I and II, sint 0, so we choose sint =. Thus, tant = sint cost = To determine the values of for which this equivalence is valid, we consider our substitution t = arccos. Since the domain of arccos is [, ], we know we must restrict. Additionall, since we had to discard t =, we need to discard = cos = 0. Hence, tan arccos = is valid for in [, 0 0, ]. b We proceed as in the previous problem b writing t = arcsin so that t lies in the interval [, ] with sint =. We aim to epress cos arcsin = cost in terms of. Since cost is defined everwhere, we get no additional restrictions on t as we did in the previous problem. We have three choices for rewriting cost: cos t sin t, cos t and sin t. Since we know = sint, it is easiest to use the last form: cos arcsin = cost = sin t = Alternativel, we could use the identit: + tan t = sec t. Since = cost, sect = cost =. The reader is invited to work through this approach to see what, if an, difficulties arise. 5.

0.6 The Inverse Trigonometric Functions 8 To find the restrictions on, we once again appeal to our substitution t = arcsin. Since arcsin is defined onl for, the equivalence cos arcsin = is valid onl on [, ]. A few remarks about Eample 0.6. are in order. Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that the are one-to-one. One instance of this phenomenon is the fact that arccos cos 6 = 6 as opposed to 6. This is the eact same phenomenon discussed in Section 5. when we saw = as opposed to. Additionall, even though the epression we arrived at in part b above, namel, is defined for all real numbers, the equivalence cos arcsin = is valid for onl. This is akin to the fact that while the epression is defined for all real numbers, the equivalence = is valid onl for 0. For this reason, it pas to be careful when we determine the intervals where such equivalences are valid. The net pair of functions we wish to discuss are the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectivel. First, we restrict f = tan to its fundamental ccle on, to obtain f = arctan. Among other things, note that the vertical asmptotes = and = of the graph of f = tan become the horizontal asmptotes = and = of the graph of f = arctan. f = tan, < <. reflect across = switch and coordinates f = arctan. Net, we restrict g = cot to its fundamental ccle on 0, to obtain g = arccot. Once again, the vertical asmptotes = 0 and = of the graph of g = cot become the horizontal asmptotes = 0 and = of the graph of g = arccot. We show these graphs on the net page and list some of the basic properties of the arctangent and arccotangent functions.

8 Foundations of Trigonometr g = cot, 0 < <. reflect across = switch and coordinates g = arccot. Theorem 0.7. Properties of the Arctangent and Arccotangent Functions ˆ Properties of F = arctan Domain:, Range:, as, arctan + ; as, arctan arctan = t if and onl if < t < and tant = arctan = arccot for > 0 tan arctan = for all real numbers arctantan = provided < < additionall, arctangent is odd ˆ Properties of G = arccot Domain:, Range: 0, as, arccot ; as, arccot 0 + arccot = t if and onl if 0 < t < and cott = arccot = arctan for > 0 cot arccot = for all real numbers arccotcot = provided 0 < <

0.6 The Inverse Trigonometric Functions 85 Eample 0.6... Find the eact values of the following. a arctan b arccot c cotarccot 5 d sin arctan. Rewrite the following as algebraic epressions of and state the domain on which the equivalence is valid. a tan arctan b cosarccot Solution.. a We know arctan is the real number t between and with tant =. We find t =, so arctan =. b The real number t = arccot lies in the interval 0, with cott =. We get arccot = 5 6. c We can appl Theorem 0.7 directl and obtain cotarccot 5 = 5. However, working it through provides us with et another opportunit to understand wh this is the case. Letting t = arccot 5, we have that t belongs to the interval 0, and cott = 5. Hence, cotarccot 5 = cott = 5. d We start simplifing sin arctan b letting t = arctan. Then tant = for some < t <. Since tant < 0, we know, in fact, < t < 0. One wa to proceed is to use The Pthagorean Identit, +cot t = csc t, since this relates the reciprocals of tant and sint and is valid for all t under consideration. From tant =, we get cott =. Substituting, we get + = csc t so that csct = ± 5. Since < t < 0, we choose csct = 5, so sint = 5. Hence, sin arctan = 5.. a If we let t = arctan, then < t < and tant =. We look for a wa to epress tan arctan = tant in terms of. Before we get started using identities, we note that tant is undefined when t = + k for integers k. Dividing both sides of this equation b tells us we need to eclude values of t where t = + k, where k is an integer. The onl members of this famil which lie in, are t = ±, which means the values of t under consideration are,,,. Returning to arctant, we note the double angle identit tant = tant, is valid for all the tan t values of t under consideration, hence we get tan arctan = tant = tant tan t = It s alwas a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Eample 0.6.. Were the identities we used there valid for all t under consideration? A pedantic point, to be sure, but what else do ou epect from this book?

86 Foundations of Trigonometr To find where this equivalence is valid we check back with our substitution t = arctan. Since the domain of arctan is all real numbers, the onl eclusions come from the values of t we discarded earlier, t = ±. Since = tant, this means we eclude = tan ± = ±. Hence, the equivalence tan arctan = holds for all in,,,. b To get started, we let t = arccot so that cott = where 0 < t <. In terms of t, cosarccot = cost, and our goal is to epress the latter in terms of. Since cost is alwas defined, there are no additional restrictions on t, so we can begin using identities to relate cott to cost. The identit cott = cost sint is valid for t in 0,, so our strateg is to obtain sint in terms of, then write cost = cott sint. The identit + cot t = csc t holds for all t in 0, and relates cott and csct = sint. Substituting cott =, we get + = csc t, or csct = ± +. Since t is between 0 and, csct > 0, so csct = + which gives sint =. Hence, + cosarccot = cost = cott sint = + Since arccot is defined for all real numbers and we encountered no additional restrictions on t, we have cos arccot = for all real numbers. + The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were first discussed in Subsection 0.5., are given below with the fundamental ccles highlighted. The graph of = sec. The graph of = csc. It is clear from the graph of secant that we cannot find one single continuous piece of its graph which covers its entire range of, ] [, and restricts the domain of the function so that it is one-to-one. The same is true for cosecant. Thus in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namel [,, and another piece to cover the bottom, namel, ]. There are two generall accepted was make these choices which restrict the domains of these functions so that the are one-to-one. One approach simplifies the Trigonometr associated with the inverse functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometr less so. We present both points of view.

0.6 The Inverse Trigonometric Functions 87 0.6. Inverses of Secant and Cosecant: Trigonometr Friendl Approach In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectivel. For f = sec, we restrict the domain to [ 0,, ] f = sec on [ 0, reflect across =, ] switch and coordinates f = arcsec and we restrict g = csc to [, 0 0, ]. g = csc on [, 0 0, ] reflect across = switch and coordinates g = arccsc Note that for both arcsecant and arccosecant, the domain is, ] [,. Taking a page from Section., we can rewrite this as { : }. This is often done in Calculus tetbooks, so we include it here for completeness. Using these definitions, we get the following properties of the arcsecant and arccosecant functions.

88 Foundations of Trigonometr Theorem 0.8. Properties of the Arcsecant and Arccosecant Functions a ˆ Properties of F = arcsec Domain: { : } =, ] [, Range: [ 0,, ] as, arcsec + ; as, arcsec arcsec = t if and onl if 0 t < or < t and sect = arcsec = arccos provided sec arcsec = provided arcsecsec = provided 0 < or < ˆ Properties of G = arccsc Domain: { : } =, ] [, Range: [, 0 0, ] as, arccsc 0 ; as, arccsc 0 + arccsc = t if and onl if t < 0 or 0 < t and csct = arccsc = arcsin provided csc arccsc = provided arccsccsc = provided < 0 or 0 < additionall, arccosecant is odd a... assuming the Trigonometr Friendl ranges are used. Eample 0.6... Find the eact values of the following. a arcsec b arccsc c arcsec sec 5 d cot arccsc. Rewrite the following as algebraic epressions of and state the domain on which the equivalence is valid. a tanarcsec b cosarccsc

0.6 The Inverse Trigonometric Functions 89 Solution.. a Using Theorem 0.8, we have arcsec = arccos =. b Once again, Theorem 0.8 comes to our aid giving arccsc = arcsin = c Since 5 doesn t fall between 0 and or and, we cannot use the inverse propert stated in Theorem 0.8. We can, nevertheless, begin b working inside out which ields arcsec sec 5 = arcsec = arccos =. d One wa to begin to simplif cot arccsc is to let t = arccsc. Then, csct = and, since this is negative, we have that t lies in the interval [, 0. We are after cot arccsc = cott, so we use the Pthagorean Identit + cot t = csc t. Substituting, we have + cot t =, or cott = ± 8 = ±. Since t < 0, cott < 0, so we get cot arccsc =.. a We begin simplifing tanarcsec b letting t = arcsec. Then, sect = for t in [ 0,, ], and we seek a formula for tant. Since tant is defined for all t values under consideration, we have no additional restrictions on t. To relate sect to tant, we use the identit +tan t = sec t. This is valid for all values of t under consideration, and when we substitute sect =, we get + tan t =. Hence, tant = ±. If t belongs to [ 0, then tant 0; if, on the the other hand, t belongs to, ] then tant 0. As a result, we get a piecewise defined function for tant 6. tant = {, if 0 t <, if < t Now we need to determine what these conditions on t mean for. Since = sect, when 0 t <,, and when < t,. Since we encountered no further restrictions on t, the equivalence below holds for all in, ] [,. tanarcsec = {, if, if b [ To simplif cosarccsc, we start b letting t = arccsc. Then csct = for t in, 0 0, ], and we now set about finding an epression for cosarccsc = cost. Since cost is defined for all t, we do not encounter an additional restrictions on t. From csct =, we get sint =, so to find cost, we can make use if the identit cos t + sin t =. Substituting sint = gives cos t + 6 cost = ± 6 6 = ± =. Solving, we get Since t belongs to [, 0 0, ], we know cost 0, so we choose cost = 6. The absolute values here are necessar, since could be negative. To find the values for

80 Foundations of Trigonometr which this equivalence is valid, we look back at our original substution, t = arccsc. Since the domain of arccsc requires its argument to satisf, the domain of arccsc requires. Using Theorem., we rewrite this inequalit and solve to get or. Since we had no additional restrictions on t, the equivalence cosarccsc = 6 holds for all in, [ ],. 0.6. Inverses of Secant and Cosecant: Calculus Friendl Approach In this subsection, we restrict f = sec to [ 0, [, f = sec on [ 0, [, and we restrict g = csc to 0, ] ],. reflect across = switch and coordinates f = arcsec g = csc on 0, ] ], reflect across = switch and coordinates g = arccsc Using these definitions, we get the following result.

0.6 The Inverse Trigonometric Functions 8 Theorem 0.9. Properties of the Arcsecant and Arccosecant Functions a ˆ Properties of F = arcsec Domain: { : } =, ] [, Range: [ 0, [, as, arcsec ; as, arcsec arcsec = t if and onl if 0 t < or t < and sect = arcsec = arccos for onl b sec arcsec = provided arcsecsec = provided 0 < or < ˆ Properties of G = arccsc Domain: { : } =, ] [, Range: 0, ] ], as, arccsc + ; as, arccsc 0 + arccsc = t if and onl if 0 < t or < t arccsc = arcsin for onl c csc arccsc = provided arccsccsc = provided 0 < or < a... assuming the Calculus Friendl ranges are used. b Compare this with the similar result in Theorem 0.8. c Compare this with the similar result in Theorem 0.8. and csct = Our net eample is a duplicate of Eample 0.6.. The interested reader is invited to compare and contrast the solution to each. Eample 0.6... Find the eact values of the following. a arcsec b arccsc c arcsec sec 5 d cot arccsc. Rewrite the following as algebraic epressions of and state the domain on which the equivalence is valid. a tanarcsec b cosarccsc

8 Foundations of Trigonometr Solution.. a Since, we ma invoke Theorem 0.9 to get arcsec = arccos =. b Unfortunatel, is not greater to or equal to, so we cannot appl Theorem 0.9 to arccsc and convert this into an arcsine problem. Instead, we appeal to the definition. The real number t = arccsc lies in 0, ] ], and satisfies csct =. The t we re after is t = 7 6, so arccsc = 7 6. c Since 5 lies between and, we ma appl Theorem 0.9 directl to simplif arcsec sec 5 = 5. We encourage the reader to work this through using the definition as we have done in the previous eamples to see how it goes. d To simplif cot arccsc we let t = arccsc so that cot arccsc = cott. We know csct =, and since this is negative, t lies in, ]. Using the identit + cot t = csc t, we find + cot t = so that cott = ± 8 = ±. Since t is in the interval, ], we know cott > 0. Our answer is cot arccsc =.. a [ We begin [ simplifing tanarcsec b letting t = arcsec. Then, sect = for t in 0,,, and we seek a formula for tant. Since tant is defined for all t values under consideration, we have no additional restrictions on t. To relate sect to tant, we use the identit +tan t = sec t. This is valid for all values of t under consideration, and when we substitute sect =, we get + tan t =. Hence, tant = ±. Since t lies in [ 0, [,, tant 0, so we choose tant =. Since we found no additional restrictions on t, the equivalence tanarcsec = holds for all in the domain of t = arcsec, namel, ] [,. b To simplif ] cosarccsc, ] we start b letting t = arccsc. Then csct = for t in 0,,, and we now set about finding an epression for cosarccsc = cost. Since cost is defined for all t, we do not encounter an additional restrictions on t. From csct =, we get sint =, so to find cost, we can make use if the identit cos t + sin t =. Substituting sint = gives cos t + =. Solving, we get 6 cost = ± 6 6 = ± If t lies in 0, ], then cost 0, and we choose cost = 6. Otherwise, t belongs to, ] in which case cost 0, so, we choose cost = 6 This leads us to a momentaril piecewise defined function for cost 6, if 0 t cost = 6, if < t

0.6 The Inverse Trigonometric Functions 8 We now see what these restrictions mean in terms of. Since = csct, we get that for 0 t,, or. In this case, we can simplif = so cost = 6 = 6 Similarl, for < t, we get, or. In this case, =, so we also get 6 cost = 6 = 6 = 6 Hence, in all cases, cosarccsc =, and this equivalence is valid for all in the domain of t = arccsc, namel, ] [, 0.6. Calculators and the Inverse Circular Functions. In the sections to come, we will have need to approimate the values of the inverse circular functions. On most calculators, onl the arcsine, arccosine and arctangent functions are available and the are usuall labeled as sin, cos and tan, respectivel. If we are asked to approimate these values, it is a simple matter to punch up the appropriate decimal on the calculator. If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to emplo some ingenuit, as our net eample illustrates. Eample 0.6.5.. Use a calculator to approimate the following values to four decimal places. a arccot b arcsec5 c arccot d arccsc. Find the domain and range of the following functions. Check our answers using a calculator. a f = arccos b f = arctan. c f = arccot + 5 Solution.. a Since > 0, we can use the propert listed in Theorem 0.7 to rewrite arccot as arccot = arctan. In radian mode, we find arccot = arctan 0.66. b Since 5, we can use the propert from either Theorem 0.8 or Theorem 0.9 to write arcsec5 = arccos 5.69.

8 Foundations of Trigonometr c Since the argument is negative, we cannot directl appl Theorem 0.7 to help us find arccot. Let t = arccot. Then t is a real number such that 0 < t < and cott =. Moreover, since cott < 0, we know < t <. Geometricall, this means t corresponds to a Quadrant II angle θ = t radians. This allows us to proceed using a reference angle approach. Consider α, the reference angle for θ, as pictured below. B definition, α is an acute angle so 0 < α <, and the Reference Angle Theorem, Theorem 0., tells us that cotα =. This means α = arccot radians. Since the argument of arccotangent is now a positive, we can use Theorem 0.7 to get α = arccot = arctan radians. Since θ = α = arctan.6779 radians, we get arccot.6779. θ = arccot radians α Another wa to attack the problem is to use arctan. B definition, the real number t = arctan satisfies tant = with < t <. Since tant < 0, we know more specificall that < t < 0, so t corresponds to an angle β in Quadrant IV. To find the value of arccot, we once again visualize the angle θ = arccot radians and note that it is a Quadrant II angle with tanθ =. This means it is eactl units awa from β, and we get θ = + β = + arctan.6779 radians. Hence, as before, arccot.6779.

0.6 The Inverse Trigonometric Functions 85 θ = arccot radians β d If the range of arccosecant is taken to be [, 0 0, ], we can use Theorem 0.8 to get arccsc = arcsin 0.797. If, on the other hand, the range of arccosecant is taken to be 0, ] ],, then we proceed as in the previous problem b letting t = arccsc. Then t is a real number with csct =. Since csct < 0, we have that < θ, so t corresponds to a Quadrant III angle, θ. As above, we let α be the reference angle for θ. Then 0 < α < and cscα =, which means α = arccsc radians. Since the argument of arccosecant is now positive, we ma use Theorem 0.9 to get α = arccsc = arcsin radians. Since θ = + α = + arcsin.87 radians, arccsc.87. θ = arccsc radians α

86 Foundations of Trigonometr. a Since the domain of F = arccos is, we can find the domain of f = arccos 5 b setting the argument of the arccosine, in this case 5, between and. Solving 5 gives 5 5, so the domain is [ 5, 5]. To determine the range of f, we take a cue from Section.7. Three ke points on the graph of F = arccos are,, 0, and, 0. Following the procedure outlined in Theorem.7, we track these points to 5,, 0, 0 and 5,. Plotting these values tells us that the range 5 of f is [, ]. Our graph confirms our results. b To find the domain and range of f = arctan, we note that since the domain of F = arctan is all real numbers, the onl restrictions, if an, on the domain of f = arctan come from the argument of the arctangent, in this case,. Since is defined for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f, we can, once again, appeal to Theorem.7. Choosing our ke point to be 0, 0 and tracking the horizontal asmptotes = and =, we find that the graph of = f = arctan differs from the graph of = F = arctan b a horizontal compression b a factor of and a vertical stretch b a factor of. It is the latter which affects the range, producing a range of,. We confirm our findings on the calculator below. = f = arccos 5 = f = arctan c To find the domain of g = arccot +, we proceed as above. Since the domain of G = arccot is,, and is defined for all, we get that the domain of g is, as well. As for the range, we note that the range of G = arccot, like that of F = arctan, is limited b a pair of horizontal asmptotes, in this case = 0 and =. Following Theorem.7, we graph = g = arccot + starting with = G = arccot and first performing a horizontal epansion b a factor of and following that with a vertical shift upwards b. This latter transformation is the one which affects the range, making it now,. To check this graphicall, we encounter a bit of a problem, since on man calculators, there is no shortcut button corresponding to the arccotangent function. Taking a cue from number c, we attempt to rewrite g = arccot + in terms of the arctangent function. Using Theorem 0.7, we have that arccot = arctan when > 0, or, in this case, when > 0. Hence, for > 0, we have g = arctan +. When < 0, we can use the same argument in number c that gave us arccot = + arctan to give us arccot = + arctan. 5 It also confirms our domain!

0.6 The Inverse Trigonometric Functions 87 Hence, for < 0, g = + arctan + = arctan +. What about = 0? We know g0 = arccot0 + =, and neither of the formulas for g involving arctangent will produce this result. 6 Hence, in order to graph = g on our calculators, we need to write it as a piecewise defined function: g = arccot + = We show the input and the result below. arctan +, if < 0, if = 0 arctan +, if > 0 = g in terms of arctangent = g = arccot + The inverse trigonometric functions are tpicall found in applications whenever the measure of an angle is required. One such scenario is presented in the following eample. Eample 0.6.6. 7 The roof on the house below has a 6/ pitch. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Epress our answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we find that the roof line forms the hpotenuse of a right triangle with legs of length 6 feet and feet. Using Theorem 0.0, we 6 Without Calculus, of course... 7 The authors would like to thank Dan Stitz for this problem and associated graphics.

88 Foundations of Trigonometr find the angle of inclination, labeled θ below, satisfies tanθ = 6 =. Since θ is an acute angle, we can use the arctangent function and we find θ = arctan radians 6.56. 6 feet θ feet 0.6. Solving Equations Using the Inverse Trigonometric Functions. In Sections 0. and 0., we learned how to solve equations like sinθ = for angles θ and tant = for real numbers t. In each case, we ultimatel appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of common angles listed on page 7. If, on the other hand, we had been asked to find all angles with sinθ = or solve tant = for real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation = is a lot like sinθ = in that it has friendl, common value answers = ±. The equation = 7, on the other hand, is a lot like sinθ =. We know8 there are answers, but we can t epress them using friendl numbers. 9 To solve = 7, we make use of the square root function and write = ± 7. We can certainl approimate these answers using a calculator, but as far as eact answers go, we leave them as = ± 7. In the same wa, we will use the arcsine function to solve sinθ =, as seen in the following eample. Eample 0.6.7. Solve the following equations.. Find all angles θ for which sinθ =.. Find all real numbers t for which tant =. Solve sec = 5 for. Solution.. If sinθ =, then the terminal side of θ, when plotted in standard position, intersects the Unit Circle at =. Geometricall, we see that this happens at two places: in Quadrant I and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions 8 How do we know this again? 9 This is all, of course, a matter of opinion. For the record, the authors find ± 7 just as nice as ±.

0.6 The Inverse Trigonometric Functions 89 to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all of the solutions to this equation in Quadrant II. α = arcsin radians α Since isn t the sine of an of the common angles discussed earlier, we use the arcsine functions to epress our answers. The real number t = arcsin is defined so it satisfies 0 < t < with sint =. Hence, α = arcsin radians. Since the solutions in Quadrant I are all coterminal with α, we get part of our solution to be θ = α + k = arcsin + k for integers k. Turning our attention to Quadrant II, we get one solution to be α. Hence, the Quadrant II solutions are θ = α + k = arcsin + k, for integers k.. We ma visualize the solutions to tant = as angles θ with tanθ =. Since tangent is negative onl in Quadrants II and IV, we focus our efforts there. β = arctan radians β Since isn t the tangent of an of the common angles, we need to use the arctangent function to epress our answers. The real number t = arctan satisfies tant = and < t < 0. If we let β = arctan radians, we see that all of the Quadrant IV solutions

80 Foundations of Trigonometr to tanθ = are coterminal with β. Moreover, the solutions from Quadrant II differ b eactl units from the solutions in Quadrant IV, so all the solutions to tanθ = are of the form θ = β + k = arctan + k for some integer k. Switching back to the variable t, we record our final answer to tant = as t = arctan + k for integers k.. The last equation we are asked to solve, sec = 5, poses two immediate problems. First, we are not told whether or not represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Second, as we have mentioned, there is no universall accepted range of the arcsecant function. For that reason, we adopt the advice given in Section 0. and convert this to the cosine problem cos = 5. Adopting an angle approach, we consider the equation cosθ = 5 and note that our solutions lie in Quadrants II and III. Since 5 isn t the cosine of an of the common angles, we ll need to epress our solutions in terms of the arccosine function. The real number t = arccos 5 is defined so that < t < with cost = 5. If we let β = arccos 5 radians, we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, we ma simpl use β = arccos 5. Since all angle solutions are coterminal with β or β, we get our solutions to cosθ = 5 to be θ = β + k = arccos 5 + k or θ = β + k = arccos 5 + k for integers k. Switching back to the variable, we record our final answer to sec = 5 as = arccos 5 + k or = arccos 5 + k for integers k. β = arccos 5 radians β = arccos 5 radians β = arccos 5 radians The reader is encouraged to check the answers found in Eample 0.6.7 - both analticall and with the calculator see Section 0.6.. With practice, the inverse trigonometric functions will become as familiar to ou as the square root function. Speaking of practice...

0.6 The Inverse Trigonometric Functions 8 0.6.5 Eercises In Eercises - 0, find the eact value.. arcsin. arcsin 5. arcsin 0 6. arcsin. arcsin 7. arcsin 9. arcsin 0. arccos. arccos. arccos 7. arccos. arctan. arccos 0 5. arccos. arcsin 8. arcsin. arccos 6. arccos 8. arccos 9. arctan 0. arctan. arctan 0. arctan. arctan 5. arctan 6. arccot 7. arccot 8. arccot 9. arccot 0 0. arccot. arccot. arccot. arcsec. arccsc 5. arcsec 6. arccsc 7. arcsec 8. arccsc 9. arcsec 0. arccsc In Eercises - 8, assume that the range of arcsecant is [ 0, [, and that the range of arccosecant is 0, ] ], when finding the eact value.. arcsec. arcsec. arcsec. arcsec 5. arccsc 6. arccsc 7. arccsc 8. arccsc

8 Foundations of Trigonometr In Eercises 9-56, assume that the range of arcsecant is [ 0,, ] and that the range of arccosecant is [, 0 0, ] when finding the eact value. 9. arcsec 50. arcsec 5. arcsec 5. arcsec 5. arccsc 5. arccsc 55. arccsc 56. arccsc In Eercises 57-86, find the eact value or state that it is undefined. 57. sin arcsin 58. sin arcsin 5 60. sin arcsin 0. 6. sin arcsin 6. cos arccos 5 6. cos arccos 59. sin arcsin 6. cos arccos 5 65. cos arccos 0.998 66. cos arccos 67. tan arctan 68. tan arctan 5 69. tan arctan 70. tan arctan 0.965 7. tan arctan 7. cot arccot 7. cot arccot 7. cot arccot 7 7 75. cot arccot 0.00 76. cot arccot 77. sec arcsec 78. sec arcsec 79. sec arcsec 8. sec arcsec 7 8. csc arccsc 8. csc 8. csc arccsc 80. sec arcsec 0.75 arccsc 85. csc arccsc.000 86. csc arccsc In Eercises 87-06, find the eact value or state that it is undefined. 87. arcsin sin 6 88. arcsin sin 89. arcsin sin

0.6 The Inverse Trigonometric Functions 8 90. arcsin sin 6 9. arccos cos 5 96. arccos cos 9. arcsin sin 9. arccos cos 97. arctan tan 99. arctan tan 00. arctan tan 0. arccot cot 05. arccot cot 0. arccot cot 06. arccot cot 9. arccos cos 95. arccos cos 6 98. arctan tan 0. arctan tan 0. arccot cot In Eercises 07-8, assume that the range of arcsecant is [ 0, [, and that the range of arccosecant is 0, ] ], when finding the eact value. 07. arcsec sec 0. arcsec sec 5. arccsc csc 6. arccsc csc 6 08. arcsec sec 5. arcsec sec. arccsc csc 7. arcsec sec 09. arcsec sec 5 6. arccsc csc 6 5. arccsc csc 8. arccsc csc 9 8 In Eercises 9-0, assume that the range of arcsecant is [ 0,, ] and that the range of arccosecant is [, 0 0, ] when finding the eact value. 9. arcsec sec. arcsec sec 5 5. arccsc csc 8. arccsc csc 6 0. arcsec sec 5. arcsec sec 6. arccsc csc 9. arcsec sec 5. arcsec sec 6. arccsc csc 6 7. arccsc csc 0. arccsc csc 9 8

8 Foundations of Trigonometr In Eercises - 5, find the eact value or state that it is undefined.. sin arccos. sin arccos 5. sin arctan. sin arccot 5 5. sin arccsc 6. cos arcsin 5 7. cos arctan 7 8. cos arccot 9. cos arcsec 5 0. tan arcsin 5 5. tan arccos. tan arccot. cot arcsin 5. tan arcsec 5. cot 6. cot arccsc 5 7. cot arctan 0.5 8. sec 9. sec arcsin 5. csc arccot 9 5. csc arcsin 5 arccos arccos 0 50. sec arctan 0 5. sec arccot 0 In Eercises 55-6, find the eact value or state that it is undefined. 5 55. sin arcsin + 57. tan arctan + arccos 5 59. sin arccsc 5 6. cos arcsin 5 5. csc arctan 56. cos arcsec + arctan 58. sin arcsin 5 60. sin arctan 5 6. cos arcsec 7 6. cos arccot 5 arctan 6. sin

0.6 The Inverse Trigonometric Functions 85 In Eercises 65-8, rewrite the quantit as algebraic epressions of and state the domain on which the equivalence is valid. 65. sin arccos 66. cos arctan 67. tan arcsin 68. sec arctan 69. csc arccos 70. sin arctan 7. sin arccos 7. cos arctan 7. sinarccos 7. sin arccos 75. cos arcsin 76. cos arctan 5 77. sin arcsin7 78. sin arcsin 79. cos arcsin 80. secarctan tanarctan 8. sin arcsin + arccos 8. cos arcsin + arctan 8. tan arcsin 8. sin arctan 85. If sinθ = for < θ <, find an epression for θ + sinθ in terms of. 86. If tanθ = 7 for < θ <, find an epression for θ sinθ in terms of. 87. If secθ = for 0 < θ <, find an epression for tanθ θ in terms of. In Eercises 88-07, solve the equation using the techniques discussed in Eample 0.6.7 then approimate the solutions which lie in the interval [0, to four decimal places. 88. sin = 7 89. cos = 9 90. sin = 0.569 9. cos = 0.7 9. sin = 0.008 9. cos = 59 60 9. tan = 7 95. cot = 96. sec = 97. csc = 90 7 00. cos = 7 6 98. tan = 0 99. sin = 8 0. tan = 0.0 0. sin = 0.50

86 Foundations of Trigonometr 0. sin = 0.7 0. cos = 0.98 05. cos = 0.567 06. cot = 7 07. tan = 0.609 In Eercises 08-0, find the two acute angles in the right triangle whose sides have the given lengths. Epress our answers using degree measure rounded to two decimal places. 08., and 5 09. 5, and 0. 6, 57 and 65. A gu wire 000 feet long is attached to the top of a tower. When pulled taut it touches level ground 60 feet from the base of the tower. What angle does the wire make with the ground? Epress our answer using degree measure rounded to one decimal place.. At Cliffs of Insanit Point, The Great Sasquatch Canon is 77 feet deep. From that point, a fire is seen at a location known to be 0 miles awa from the base of the sheer canon wall. What angle of depression is made b the line of sight from the canon edge to the fire? Epress our answer using degree measure rounded to one decimal place.. Shelving is being built at the Utilit Muffin Research Librar which is to be inches deep. An 8-inch rod will be attached to the wall and the underside of the shelf at its edge awa from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall?. A parasailor is being pulled b a boat on Lake Ippizuti. The cable is 00 feet long and the parasailor is 00 feet above the surface of the water. What is the angle of elevation from the boat to the parasailor? Epress our answer using degree measure rounded to one decimal place. 5. A tag-and-release program to stud the Sasquatch population of the eponmous Sasquatch National Park is begun. From a 00 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directl towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the ground. If the range of the rifle with a tranquilizer dart is 00 feet, find the smallest value of θ for which the corresponding point on the ground is in range of the rifle. Round our answer to the nearest hundreth of a degree. In Eercises 6 -, rewrite the given function as a sinusoid of the form S = A sinω + φ using Eercises 5 and 6 in Section 0.5 for reference. Approimate the value of φ which is in radians, of course to four decimal places. 6. f = 5 sin + cos 7. f = cos + sin 8. f = cos sin 9. f = 7 sin0 cos0

0.6 The Inverse Trigonometric Functions 87 0. f = cos sin. f = sin cos In Eercises -, find the domain of the given function. Write our answers in interval notation.. f = arcsin5. f = arccos. f = arcsin 5. f = arccos 6. f = arctan 7. f = arccot 9 8. f = arctanln 9. f = arccot 0. f = arcsec. f = arccsc + 5. f = arcsec. f = arccsc e 8 [. Show that arcsec = arccos for as long as we use 0, ], as the range of f = arcsec. 5. Show that arccsc = arcsin of f = arccsc. 6. Show that arcsin + arccos = 7. Discuss with our classmates wh arcsin for as long as we use [, 0 0, ] as the range for. 0. 8. Use the following picture and the series of eercises on the net page to show that arctan + arctan + arctan = D, A0, α β γ O0, 0 B, 0 C, 0

88 Foundations of Trigonometr a Clearl AOB and BCD are right triangles because the line through O and A and the line through C and D are perpendicular to the -ais. Use the distance formula to show that BAD is also a right triangle with BAD being the right angle b showing that the sides of the triangle satisf the Pthagorean Theorem. b Use AOB to show that α = arctan c Use BAD to show that β = arctan d Use BCD to show that γ = arctan e Use the fact that O, B and C all lie on the -ais to conclude that α + β + γ =. Thus arctan + arctan + arctan =.

0.6 The Inverse Trigonometric Functions 89 0.6.6 Answers. arcsin =. arcsin =. arcsin =. arcsin = 5. arcsin 0 = 0 6. arcsin = 6 6 7. arcsin = 8. arcsin = 9. arcsin = 0. arccos =. arccos = 5. arccos = 6. arccos = 6. arccos =. arccos 0 = 7. arccos = 6 9. arctan = 0. arctan =. arctan 0 = 0. arctan = 6 5. arccos = 8. arccos = 0. arctan = 6. arctan = 5. arctan = 8. arccot = 6. arccot = 5 6 9. arccot 0 = 7. arccot = 0. arccot =. arccot =. arccot = 6. arcsec =. arccsc = 6 7. arcsec = 6 5. arcsec = 8. arccsc = 6. arccsc = 9. arcsec = 0 0. arccsc =. arcsec =. arcsec = 5

850 Foundations of Trigonometr. arcsec 6. arccsc = 5 9. arcsec = = 7 6. arcsec = 5. arccsc = 7 6 7. arccsc 50. arcsec = = 8. arccsc = 5. arcsec = 5 6 5. arcsec = 5. arccsc = 5. arccsc = 6 55. arccsc = 56. arccsc = 57. sin arcsin = 58. sin arcsin = 59. sin arcsin = 60. sin arcsin 0. = 0. 5 5 6. sin arcsin 5 is undefined. 6. cos arccos = 6. cos arccos = 5 6. cos arccos = 5 65. cos arccos 0.998 = 0.998 66. cos arccos is undefined. 67. tan arctan = 68. tan arctan = 5 69. tan arctan = 5 70. tan arctan 0.965 = 0.965 7. tan arctan = 7. cot arccot = 7. cot arccot = 7. cot arccot 7 = 7 75. cot arccot 0.00 = 0.00 7 76. cot arccot = 7 77. sec arcsec = 78. sec arcsec =

0.6 The Inverse Trigonometric Functions 85 79. sec arcsec is undefined. 80. sec arcsec 0.75 is undefined. 8. sec arcsec 7 = 7 8. csc arccsc = 8. csc arccsc = 8. csc arccsc is undefined. 85. csc arccsc.000 =.000 86. csc arccsc is undefined. 87. arcsin sin = 6 6 89. arcsin sin = 9. arcsin sin = 9. arccos cos = 95. arccos cos = 6 6 88. arcsin sin = 90. arcsin sin = 6 6 9. arccos cos = 9. arccos cos = 5 96. arccos cos = 97. arctan tan = 98. arctan tan = 99. arctan tan = 0 00. arctan tan is undefined 0. arctan tan = 0. arccot cot = 05. arccot cot = 07. arcsec sec = 5 09. arcsec sec = 7 6 6. arcsec sec 5 = 0. arccot cot = 0. arccot cot is undefined 06. arccot cot 08. arcsec sec = = 0. arcsec sec is undefined.. arccsc csc = 6 6

85 Foundations of Trigonometr 5. arccsc csc = 5 5. arccsc csc = 7. arcsec sec = 9. arcsec sec = 5. arcsec sec = 5 6 6 5. arcsec sec = 5. arccsc csc 5 = 7. arccsc csc = 9. arcsec sec =. sin arccos =. sin arctan = 5 5 5. sin arccsc =. arccsc csc 6. arccsc csc 8. arccsc csc 0. arcsec sec = = 7 6 6 9 = 9 8 8 =. arcsec sec is undefined.. arccsc csc = 6 6 6. arccsc csc = 8. arccsc csc = 6 6 9 0. arccsc csc = 8 8. sin arccos 5 = 5. sin arccot 5 6 = 6 6. cos arcsin 5 = 7. cos arctan 7 = 8. cos arccot = 0 0 9. cos arcsec 5 = 0. tan arcsin 5 = 5 5. tan arccos = 5. tan arcsec =. tan arccot =. cot arcsin = 5

0.6 The Inverse Trigonometric Functions 85 5. cot arccos = 6. cot arccsc 5 = 7. cot arctan 0.5 = 8. sec arccos 9. sec arcsin = 5 50. sec arctan 0 = 0 0 5. sec arccot = 0 5. csc arccot 9 = 8 5. csc arcsin = 5 5 5 55. sin arcsin 57. tan arctan + arccos 59. sin arccsc 6. cos arcsin + = 7 6 5 5 5 = 0 69 = 7 5 6. cos arccot 5 = = 65. sin arccos = for 66. cos arctan = 67. tan arcsin = + for all for < < 68. sec arctan = + for all 69. csc arccos = 70. sin arctan = + for < < for all 7. sin arccos = for = 5. csc arctan = 5 0 56. cos arcsec + arctan = 58. sin arcsin = 5 5 60. sin arctan = 5 5 6. cos arcsec = 57 7 65 arctan 5 5 6. sin = 0 5

85 Foundations of Trigonometr 7. cos arctan = for all + 7. sinarccos = for 5 7. sin arccos = for 5 5 5 5 75. cos arcsin = for 76. cos arctan = + 9 for all 77. sin arcsin7 = 9 for 7 7 78. sin arcsin = for 79. cos arcsin = for 80. secarctan tanarctan = + for all 8. sin arcsin + arccos = for 8. cos arcsin + arctan = for + 8. 0 tan arcsin = for in,, 8. sin arctan = + + + + for 0 for < 0 85. If sinθ = for < θ <, then θ + sinθ = arcsin 86. If tanθ = 7 for < θ <, then θ sinθ = arctan 7 +, 7 + 9 0 The equivalence for = ± can be verified independentl of the derivation of the formula, but Calculus is required to full understand what is happening at those values. You ll see what we mean when ou work through the details of the identit for tant. For now, we eclude = ± from our answer.

0.6 The Inverse Trigonometric Functions 855 87. If secθ = for 0 < θ <, then tanθ θ = 6 arcsec 7 88. = arcsin 89. = arccos 9 7 + k or = arcsin + k or = arccos 9 + k, in [0,, 0.6898,.58 + k, in [0,,.799,.88 90. = + arcsin0.569 + k or = arcsin0.569 + k, in [0,,.769, 5.6779 9. = arccos0.7 + k or = arccos0.7 + k, in [0,,.55,.897 9. = arcsin0.008 + k or = arcsin0.008 + k, in [0,, 0.0080,.6 59 59 9. = arccos + k or = arccos + k, in [0,, 0.076, 6.086 60 60 9. = arctan7 + k, in [0,,.565,.708 95. = arctan + k, in [0,,.0585, 6.000 96. = arccos 97. = + arcsin + k or = arccos 7 + k or = arcsin 90 98. = arctan 0 + k, in [0,,.877, 5.087 99. = arcsin + k or = arcsin 8 8 00. = arccos 7 + k or = arccos 7 6 6 0. = arctan0.0 + k, in [0,, 0.000,.76 + k, in [0,, 0.8, 5. 7 + k, in [0,,.6, 6.09 90 + k, in [0,, 0.8,.757 + k, in [0,,.06,.596 0. = arcsin0.50 + k or = arcsin0.50 + k, in [0,, 0.578,.78 0. = + arcsin0.7 + k or = arcsin0.7 + k, in [0,,.968, 5.780 0. = arccos0.98 + k or = arccos0.98 + k, in [0,, 0.879, 6.095 05. = arccos 0.567 + k or = arccos 0.567 + k, in [0,,.697,.5 06. = arctan7 + k, in [0,,.56,.708 07. = arctan 0.609 + k, in [0,,.59, 5.78

856 Foundations of Trigonometr 08. 6.87 and 5. 09..6 and 67.8 0..5 and 57.8. 68.9. 7.7. 5. 9.5 5..8 6. f = 5 sin + cos = sin + arcsin sin +.760 7. f = cos + sin = 5 sin + arcsin 5 sin + 0.65 5 8. f = cos sin = 0 sin + arccos 0 0 sin +.898 0 9. f = 7 sin0 cos0 = 5 sin 0 + arcsin 5 sin0.870 5 0. f = cos sin = sin + + arcsin sin +.8. f = sin cos = 5 5 sin + arcsin 5 sin 0.66 5.. [ 5, ] 5 [, ]. [, ] 5., 5] [, ] [ 5, 6., 7.,,, [ 8., 9., 0., ] [,., 6] [,., ] [,. [0,

0.7 Trigonometric Equations and Inequalities 857 0.7 Trigonometric Equations and Inequalities In Sections 0., 0. and most recentl 0.6, we solved some basic equations involving the trigonometric functions. Below we summarize the techniques we ve emploed thus far. Note that we use the neutral letter u as the argument of each circular function for generalit. ˆ ˆ Strategies for Solving Basic Equations Involving Trigonometric Functions To solve cosu = c or sinu = c for c, first solve for u in the interval [0, and add integer multiples of the period. If c < or of c >, there are no real solutions. To solve secu = c or cscu = c for c or c, convert to cosine or sine, respectivel, and solve as above. If < c <, there are no real solutions. ˆ To solve tanu = c for an real number c, first solve for u in the interval, and add integer multiples of the period. ˆ To solve cotu = c for c 0, convert to tangent and solve as above. If c = 0, the solution to cotu = 0 is u = + k for integers k. Using the above guidelines, we can comfortabl solve sin = and find the solution = 6 + k or = 5 6 + k for integers k. How do we solve something like sin =? Since this equation has the form sinu =, we know the solutions take the form u = 6 + k or u = 5 6 + k for integers k. Since the argument of sine here is, we have = 6 + k or = 5 6 + k for integers k. To solve for, we divide both sides of these equations b, and obtain = 8 + k or = 5 8 + k for integers k. This is the technique emploed in the eample below. Eample 0.7.. Solve the following equations and check our answers analticall. solutions which lie in the interval [0, and verif them using a graphing utilit.. cos =. csc =. cot = 0. sec = 5. tan = 6. sin = 0.87 Solution. List the. The solutions to cosu = are u = 5 6 + k or u = 7 6 + k for integers k. Since the argument of cosine here is, this means = 5 6 + k or = 7 6 + k for integers k. Solving for gives = 5 7 + k or = + k for integers k. To check these answers analticall, we substitute them into the original equation. For an integer k we have cos [ 5 + k] = cos 5 6 + k = cos 5 = See the comments at the beginning of Section 0.5 for a review of this concept. Don t forget to divide the k b as well! 6 the period of cosine is

858 Foundations of Trigonometr Similarl, we find cos [ 7 + k] = cos 7 6 + k = cos 7 6 =. To determine which of our solutions lie in [0,, we substitute integer values for k. The solutions we keep come from the values of k = 0 and k = and are = 5, 7, 7 9 and. Using a over [0, and eamine where these two calculator, we graph = cos and = graphs intersect. We see that the -coordinates of the intersection points correspond to the decimal representations of our eact answers.. Since this equation has the form cscu =, we rewrite this as sinu = and find u = + k or u = + k for integers k. Since the argument of cosecant here is, = + k or = + k To solve = + k, we first add to both sides = + k + A common error is to treat the k and terms as like terms and tr to combine them when the are not. We can, however, combine the and terms to get = 5 + k We now finish b multipling both sides b to get 5 = + k = 5 + 6k Solving the other equation, = + k produces = + 6k for integers k. To check the first famil of answers, we substitute, combine line terms, and simplif. csc [ 5 + 6k] = csc 5 + k = csc + k = csc = the period of cosecant is The famil = + 6k checks similarl. Despite having infinitel man solutions, we find that none of them lie in [0,. To verif this graphicall, we use a reciprocal identit to rewrite the cosecant as a sine and we find that = sin and = do not intersect at all over the interval [0,. Do ou see wh?

0.7 Trigonometric Equations and Inequalities 859 = cos and = = sin and =. Since cot = 0 has the form cotu = 0, we know u = + k, so, in this case, = + k for integers k. Solving for ields = 6 + k. Checking our answers, we get cot [ 6 + k] = cot + k = cot = 0 the period of cotangent is As k runs through the integers, we obtain si answers, corresponding to k = 0 through k = 5, which lie in [0, : = 6,, 5 6, 7 6, and 6. To confirm these graphicall, we must be careful. On man calculators, there is no function button for cotangent. We choose to use the quotient identit cot = cos cos sin. Graphing = sin and = 0 the -ais, we see that the -coordinates of the intersection points approimatel match our solutions.. The complication in solving an equation like sec = comes not from the argument of secant, which is just, but rather, the fact the secant is being squared. To get this equation to look like one of the forms listed on page 857, we etract square roots to get sec = ±. Converting to cosines, we have cos = ±. For cos =, we get = + k or = 5 + k for integers k. For cos =, we get = + k or = + k for integers k. If we take a step back and think of these families of solutions geometricall, we see we are finding the measures of all angles with a reference angle of. As a result, these solutions can be combined and we ma write our solutions as = + k and = + k for integers k. To check the first famil of solutions, we note that, depending on the integer k, sec + k doesn t alwas equal sec. However, it is true that for all integers k, sec + k = ± sec = ±. Can ou show this? As a result, sec + k = ± sec = ± = The same holds for the famil = + k. The solutions which lie in [0, come from the values k = 0 and k =, namel =,, and 5. To confirm graphicall, we use The reader is encouraged to see what happens if we had chosen the reciprocal identit cot = tan instead. The graph on the calculator appears identical, but what happens when ou tr to find the intersection points?

860 Foundations of Trigonometr a reciprocal identit to rewrite the secant as cosine. The -coordinates of the intersection points of = and = verif our answers. cos = cos sin and = 0 = cos and = 5. The equation tan = has the form tanu =, whose solution is u = arctan +k. Hence, = arctan + k, so = arctan + k for integers k. To check, we note tan arctan +k = tan arctan + k = tan arctan the period of tangent is = See Theorem 0.7 To determine which of our answers lie in the interval [0,, we first need to get an idea of the value of arctan. While we could easil find an approimation using a calculator, 5 we proceed analticall. Since < 0, it follows that < arctan < 0. Multipling through b gives < arctan < 0. We are now in a position to argue which of the solutions = arctan + k lie in [0,. For k = 0, we get = arctan < 0, so we discard this answer and all answers = arctan + k where k < 0. Net, we turn our attention to k = and get = arctan +. Starting with the inequalit < arctan < 0, we add and get < arctan + <. This means = arctan + lies in [0,. Advancing k to produces = arctan +. Once again, we get from < arctan < 0 that < arctan + <. Since this is outside the interval [0,, we discard = arctan + and all solutions of the form = arctan + k for k >. Graphicall, we see = tan and = intersect onl once on [0, at = arctan +.785. 6. To solve sin = 0.87, we first note that it has the form sinu = 0.87, which has the famil of solutions u = arcsin0.87 + k or u = arcsin0.87 + k for integers k. Since the argument of sine here is, we get = arcsin0.87 + k or = arcsin0.87 + k which gives = arcsin0.87 + k or = arcsin0.87 + k for integers k. To check, 5 Your instructor will let ou know if ou should abandon the analtic route at this point and use our calculator. But seriousl, what fun would that be?

0.7 Trigonometric Equations and Inequalities 86 sin [ arcsin0.87 + k] = sin arcsin0.87 + k = sin arcsin0.87 the period of sine is For the famil = arcsin0.87 + k, we get = 0.87 See Theorem 0.6 sin [ arcsin0.87 + k] = sin arcsin0.87 + k = sin arcsin0.87 the period of sine is = sin arcsin0.87 sin t = sint = 0.87 See Theorem 0.6 To determine which of these solutions lie in [0,, we first need to get an idea of the value of = arcsin0.87. Once again, we could use the calculator, but we adopt an analtic route here. B definition, 0 < arcsin0.87 < so that multipling through b gives us 0 < arcsin0.87 <. Starting with the famil of solutions = arcsin0.87 + k, we use the same kind of arguments as in our solution to number 5 above and find onl the solutions corresponding to k = 0 and k = lie in [0, : = arcsin0.87 and = arcsin0.87 +. Net, we move to the famil = arcsin0.87 + k for integers k. Here, we need to get a better estimate of arcsin0.87. From the inequalit 0 < arcsin0.87 <, we first multipl through b and then add to get > arcsin0.87 >, or < arcsin0.87 <. Proceeding with the usual arguments, we find the onl solutions which lie in [0, correspond to k = 0 and k =, namel = arcsin0.87 and = arcsin0.87. All told, we have found four solutions to sin = 0.87 in [0, : = arcsin0.87, = arcsin0.87 +, = arcsin0.87 and = arcsin0.87. B graphing = sin and = 0.87, we confirm our results. = tan and = = sin and = 0.87

86 Foundations of Trigonometr Each of the problems in Eample 0.7. featured one trigonometric function. If an equation involves two different trigonometric functions or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and Algebra to reduce the equation to the same form as those given on page 857. Eample 0.7.. Solve the following equations and list the solutions which lie in the interval [0,. Verif our solutions on [0, graphicall.. sin = sin. sec = tan +. cos = cos. cos = cos 5. cos = cos5 6. sin = cos 7. sin cos + cos sin = 8. cos sin = Solution.. We resist the temptation to divide both sides of sin = sin b sin What goes wrong if ou do? and instead gather all of the terms to one side of the equation and factor. sin = sin sin sin = 0 sin sin = 0 Factor out sin from both terms. We get sin = 0 or sin = 0. Solving for sin, we find sin = 0 or sin =. The solution to the first equation is = k, with = 0 and = being the two solutions which lie in [0,. To solve sin =, we use the arcsine function to get = arcsin +k or = arcsin + k for integers k. We find the two solutions here which lie in [0, to be = arcsin and = arcsin. To check graphicall, we plot = sin and = sin and find the -coordinates of the intersection points of these two curves. Some etra zooming is required near = 0 and = to verif that these two curves do in fact intersect four times. 6. Analsis of sec = tan + reveals two different trigonometric functions, so an identit is in order. Since sec = + tan, we get sec = tan + + tan = tan + Since sec = + tan. tan tan = 0 u u = 0 Let u = tan. u + u = 0 6 Note that we are not counting the point, 0 in our solution set since = is not in the interval [0,. In the forthcoming solutions, remember that while = ma be a solution to the equation, it isn t counted among the solutions in [0,.

0.7 Trigonometric Equations and Inequalities 86 This gives u = or u =. Since u = tan, we have tan = or tan =. From tan =, we get = + k for integers k. To solve tan =, we emplo the arctangent function and get = arctan + k for integers k. From the first set of solutions, we get = and = 7 as our answers which lie in [0,. Using the same sort of argument we saw in Eample 0.7., we get = arctan and = + arctan as answers from our second set of solutions which lie in [0,. Using a reciprocal identit, we rewrite the secant as a cosine and graph = and = tan + to find the -values of the points where cos the intersect. = sin and = sin = cos and = tan +. In the equation cos = cos, we have the same circular function, namel cosine, on both sides but the arguments differ. Using the identit cos = cos, we obtain a quadratic in disguise and proceed as we have done in the past. cos = cos cos = cos Since cos = cos. cos cos + = 0 u u + = 0 Let u = cos. u u = 0 This gives u = or u =. Since u = cos, we get cos = or cos =. Solving cos =, we get = + k or = 5 + k for integers k. From cos =, we get = k for integers k. The answers which lie in [0, are = 0,, and 5. Graphing = cos and = cos, we find, after a little etra effort, that the curves intersect in three places on [0,, and the -coordinates of these points confirm our results.. To solve cos = cos, we use the same technique as in the previous problem. From Eample 0.., number, we know that cos = cos cos. This transforms the equation into a polnomial in terms of cos. cos = cos cos cos = cos cos cos = 0 u u = 0 Let u = cos.

86 Foundations of Trigonometr To solve u u = 0, we need the techniques in Chapter to factor u u into u u + u +. We get either u = 0 or u +u+ = 0, and since the discriminant of the latter is negative, the onl real solution to u u = 0 is u =. Since u = cos, we get cos =, so = k for integers k. The onl solution which lies in [0, is = 0. Graphing = cos and = cos on the same set of aes over [0, shows that the graphs intersect at what appears to be 0,, as required. = cos and = cos = cos and = cos 5. While we could approach cos = cos5 in the same manner as we did the previous two problems, we choose instead to showcase the utilit of the Sum to Product Identities. From cos = cos5, we get cos5 cos = 0, and it is the presence of 0 on the right hand side that indicates a switch to a product would be a good move. 7 Using Theorem 0., we have that cos5 cos = sin 5+ sin 5 = sin sin. Hence, the equation cos5 = cos is equivalent to sin sin = 0. From this, we get sin = 0 or sin = 0. Solving sin = 0 gives = k for integers k, and the solution to sin = 0 is = k for integers k. The second set of solutions is contained in the first set of solutions, 8 so our final solution to cos5 = cos is = k for integers k. There are eight of these answers which lie in [0, : = 0,,,,, 5, and 7. Our plot of the graphs of = cos and = cos5 below after some careful zooming bears this out. 6. In eamining the equation sin = cos, not onl do we have different circular functions involved, namel sine and cosine, we also have different arguments to contend with, namel and. Using the identit sin = sin cos makes all of the arguments the same and we proceed as we would solving an nonlinear equation gather all of the nonzero terms on one side of the equation and factor. sin = cos sin cos = cos Since sin = sin cos. sin cos cos = 0 cos sin = 0 from which we get cos = 0 or sin =. From cos = 0, we obtain = + k for integers k. From sin =, we get = +k or = +k for integers k. The answers 7 As alwas, eperience is the greatest teacher here! 8 As alwas, when in doubt, write it out!

0.7 Trigonometric Equations and Inequalities 865 which lie in [0, are =,, and. We graph = sin and = cos and, after some careful zooming, verif our answers. = cos and = cos5 = sin and = cos 7. Unlike the previous problem, there seems to be no quick wa to get the circular functions or their arguments to match in the equation sin cos + cos sin =. If we stare at it long enough, however, we realize that the left hand side is the epanded form of the sum formula for sin +. Hence, our original equation is equivalent to sin =. Solving, we find = + k for integers k. Two of these solutions lie in [0, : = and = 5. Graphing = sin cos + cos sin and = validates our solutions. 8. With the absence of double angles or squares, there doesn t seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid. 9 To fit f = cos sin to the form A sinωt + φ + B, we use what we learned in Eample 0.5. and find A =, B = 0, ω = and φ = 5 6. Hence, we can rewrite the equation cos sin = as sin + 5 6 =, or sin + 5 6 =. Solving the latter, we get = + k for integers k. Onl one of these solutions, = 5, which corresponds to k =, lies in [0,. Geometricall, we see that = cos sin and = intersect just once, supporting our answer. = sin cos + cos sin and = = cos sin and = We repeat here the advice given when solving sstems of nonlinear equations in section 8.7 when it comes to solving equations involving the trigonometric functions, it helps to just tr something. 9 We are essentiall undoing the sum / difference formula for cosine or sine, depending on which form we use, so this problem is actuall closel related to the previous one!

866 Foundations of Trigonometr Net, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we ma use the sign diagram technique we ve used in the past to solve the inequalities. 0 Eample 0.7.. Solve the following inequalities on [0,. Epress our answers using interval notation and verif our answers graphicall.. sin. sin > cos. tan Solution.. We begin solving sin b collecting all of the terms on one side of the equation and zero on the other to get sin 0. Net, we let f = sin and note that our original inequalit is equivalent to solving f 0. We now look to see where, if ever, f is undefined and where f = 0. Since the domain of f is all real numbers, we can immediatel set about finding the zeros of f. Solving f = 0, we have sin = 0 or sin =. The solutions here are = 6 + k and = 5 6 + k for integers k. Since we are restricting our attention to [0,, onl = 6 and = 5 6 are of concern to us. Net, we choose test values in [0, other than the zeros and determine if f is positive or negative there. For = 0 we have f0 =, for = we get f = and for = we get f =. Since our original inequalit is equivalent to f 0, we are looking for where the function is negative or 0, and we get the intervals [ 0, ] [ 6 5 6,. We can confirm our answer graphicall b seeing where the graph of = sin crosses or is below the graph of =. 0 + 0 0 6 5 6 = sin and =. We first rewrite sin > cos as sin cos > 0 and let f = sin cos. Our original inequalit is thus equivalent to f > 0. The domain of f is all real numbers, so we can advance to finding the zeros of f. Setting f = 0 ields sin cos = 0, which, b wa of the double angle identit for sine, becomes sin cos cos = 0 or cos sin = 0. From cos = 0, we get = +k for integers k of which onl = and = lie in [0,. For sin = 0, we get sin = which gives = 6 + k or = 5 6 + k for integers k. Of those, onl = 6 and = 5 6 lie in [0,. Net, we choose 0 See page, Eample..5, page, page 99, Eample 6.. and Eample 6.. for discussion of this technique.

0.7 Trigonometric Equations and Inequalities 867 our test values. For = 0 we find f0 = ; when = we get f = = ; for = we get f = + = ; when = we have f =, and lastl, for = 7 we get f 7 = =. We see f > 0 on 6, 5 6,, so this is our answer. We can use the calculator to check that the graph of = sin is indeed above the graph of = cos on those intervals. 0 + 0 0 + 0 0 6 5 6 = sin and = cos. Proceeding as in the last two problems, we rewrite tan as tan 0 and let f = tan. We note that on [0,, f is undefined at = and, so those values will need the usual disclaimer on the sign diagram. Moving along to zeros, solving f = tan = 0 requires the arctangent function. We find = arctan + k for integers k and of these, onl = arctan and = arctan + lie in [0,. Since > 0, we know 0 < arctan < which allows us to position these zeros correctl on the sign diagram. To choose test values, we begin with = 0 and find f0 =. Finding a convenient test value in the interval arctan, is a bit more challenging. Keep in mind that the arctangent function is increasing and is bounded above b. This means that the number = arctan7 is guaranteed to lie between arctan and. We see that farctan7 = tanarctan7 =. For our net test value, we take = and find f =. To find our net test value, we note that since arctan < arctan7 <, it follows that arctan + < arctan7 + <. Evaluating f at = arctan7 + ields farctan7 + = tanarctan7 + = tanarctan7 =. We choose our last test value to be = 7 and find f 7 =. Since we want f 0, we see that our answer is [ arctan, [ arctan +,. Using the graphs of = tan and =, we see when the graph of the former is above or meets the graph of the latter. See page for a discussion of the non-standard character known as the interrobang. We could have chosen an value arctant where t >.... b adding through the inequalit...

868 Foundations of Trigonometr 0 + 0 + 0 arctan arctan + = tan and = Our net eample puts solving equations and inequalities to good use finding domains of functions. Eample 0.7.. Epress the domain of the following functions using etended interval notation.. f = csc + Solution.. f = sin cos. f = cot. To find the domain of f = csc +, we rewrite f in terms of sine as f = sin+. Since the sine function is defined everwhere, our onl concern comes from zeros in the denominator. Solving sin + = 0, we get = 6 + k for integers k. In set-builder notation, our domain is { : 6 + k for integers k}. To help visualize the domain, we follow the old mantra When in doubt, write it out! We get { : 6, 6, 6, 5 6, 7 6, 8 6,...}, where we have kept the denominators 6 throughout to help see the pattern. Graphing the situation on a numberline, we have 7 6 6 6 6 5 6 8 6 Proceeding as we did on page 756 in Section 0.., we let k denote the kth number ecluded from the domain and we have k = 6 + k = k 6 for integers k. The intervals which comprise the domain are of the form k, k + = k 6, k+ 6 as k runs through the integers. Using etended interval notation, we have that the domain is k= k, 6 k + 6 We can check our answer b substituting in values of k to see that it matches our diagram. See page 756 for details about this notation.

0.7 Trigonometric Equations and Inequalities 869. Since the domains of sin and cos are all real numbers, the onl concern when finding the domain of f = is division b zero so we set the denominator equal to zero and sin cos solve. From cos = 0 we get cos = so that = +k or = 5 +k for integers k. Using set-builder notation, the domain is { : + k and 5 + k for integers k}, or { : ±, ± 5, ± 7, ±,...}, so we have 7 5 5 7 Unlike the previous eample, we have two different families of points to consider, and we present two was of dealing with this kind of situation. One wa is to generalize what we did in the previous eample and use the formulas we found in our domain work to describe the intervals. To that end, we let a k = 6k+ + k = and b k = 5 6k+5 + k = for integers k. The goal now is to write the domain in terms of the a s an b s. We find a 0 =, a = 7, a = 5, a =, a =, b 0 = 5, b =, b =, b = 7 and b = 7. Hence, in terms of the a s and b s, our domain is... a, b b, a a, b b, a 0 a 0, b 0 b 0, a a, b... If we group these intervals in pairs, a, b b, a, a, b b, a 0, a 0, b 0 b 0, a and so forth, we see a pattern emerge of the form a k, b k b k, a k + for integers k so that our domain can be written as k= a k, b k b k, a k + = k= 6k +, 6k + 5 6k + 5, 6k + 7 A second approach to the problem eploits the periodic nature of f. Since cos and sin have period, it s not too difficult to show the function f repeats itself ever units. 5 This means if we can find a formula for the domain on an interval of length, we can epress the entire domain b translating our answer left and right on the -ais b adding integer multiples of. One such interval that arises from our domain work is [, 7 ]. The portion of the domain here is, 5 5, 7. Adding integer multiples of, we get the famil of intervals + k, 5 + k 5 + k, 7 + k for integers k. We leave it to the reader to show that getting common denominators leads to our previous answer. 5 This doesn t necessaril mean the period of f is. The tangent function is comprised of cos and sin, but its period is half theirs. The reader is invited to investigate the period of f.

870 Foundations of Trigonometr. To find the domain of f = cot, we first note that, due to the presence of the cot term, k for integers k. Net, we recall that for the square root to be defined, we need cot 0. Unlike the inequalities we solved in Eample 0.7., we are not restricted here to a given interval. Our strateg is to solve this inequalit over 0, the same interval which generates a fundamental ccle of cotangent and then add integer multiples of the period, in this case,. We let g = cot and set about making a sign diagram for g over the interval 0, to find where g 0. We note that g is undefined for = k for integers k, in particular, at the endpoints of our interval = 0 and =. Net, we look for the zeros of g. Solving g = 0, we get cot = or = + k for integers k and onl one of these, =, lies in 0,. Choosing the test values = 6 and =, we get g 6 =, and g =. 0 0 + We find g 0 on [,. Adding multiples of the period we get our solution to consist of the intervals [ [ + k, + k = k+, k +. Using etended interval notation, we epress our final answer as k= [ k +, k + We close this section with an eample which demonstrates how to solve equations and inequalities involving the inverse trigonometric functions. Eample 0.7.5. Solve the following equations and inequalities analticall. Check our answers using a graphing utilit.. arcsin =. arccos = 0. arcsec + =. arctan arctan = 0 5. arccos < 0 6. arccot > Solution.. To solve arcsin =, we first note that is in the range of the arcsine function so a solution eists! Net, we eploit the inverse propert of sine and arcsine from Theorem 0.6

0.7 Trigonometric Equations and Inequalities 87 arcsin = sin arcsin = sin = = Since sinarcsinu = u Graphing = arcsin and the horizontal line =, we see the intersect at 0.0.. Our first step in solving arccos = 0 is to isolate the arccosine. Doing so, we get arccos =. Since is in the range of arccosine, we ma appl Theorem 0.6 arccos = cos arccos = cos = Since cosarccosu = u The calculator confirms = arccos crosses = 0 the -ais at 0.707. = arcsin and = = arccos. From arcsec + =, we get arcsec =. As we saw in Section 0.6, there are two possible ranges for the arcsecant function. Fortunatel, both ranges contain. Appling Theorem 0.8 / 0.9, we get arcsec = secarcsec = sec = Since secarcsecu = u = To check using our calculator, we need to graph = arcsec +. To do so, we make use of the identit arcsecu = arccos u from Theorems 0.8 and 0.9. 6 We see the graph of = arccos + and the horizontal line = intersect at =.5. 6 Since we are checking for solutions where arcsecant is positive, we know u =, and so the identit applies in both cases.

87 Foundations of Trigonometr. With the presence of both arctan = arctan and arctan, we substitute u = arctan. The equation arctan arctan = 0 becomes u u = 0. Factoring, 7 we get u + u = 0, so u = arctan = or u = arctan =. Since is in the range of arctangent, but is not, we onl get solutions from the first equation. Using Theorem 0.7, we get arctan = tanarctan = tan = Since tanarctanu = u. The calculator verifies our result. = arcsec + and = = arctan arctan 5. Since the inverse trigonometric functions are continuous on their domains, we can solve inequalities featuring these functions using sign diagrams. Since all of the nonzero terms of arccos < 0 are on one side of the inequalit, we let f = arccos and note the domain of f is limited b the arccos to [, ]. Net, we find the zeros of f b setting f = arccos = 0. We get arccos = ±, and since the range of arccosine is [0, ], we focus our attention on arccos =. Using Theorem 0.6, we get = cos = 0 as our onl zero. Hence, we have two test intervals, [, 0 and 0, ]. Choosing test values = ±, we get f = < 0 and f = > 0. Since we are looking for where f = arccos < 0, our answer is [, 0. The calculator confirms that for these values of, the graph of = arccos is below = 0 the -ais. 7 It s not as bad as it looks... don t let the throw ou!

0.7 Trigonometric Equations and Inequalities 87 0 + 0 = arccos 6. To begin, we rewrite arccot > as arccot > 0. We let f = arccot, and note the domain of f is all real numbers,,. To find the zeros of f, we set f = arccot = 0 and solve. We get arccot =, and since is in the range of arccotangent, we ma appl Theorem 0.7 and solve arccot = cotarccot = cot = Since cotarccotu = u. = Net, we make a sign diagram for f. Since the domain of f is all real numbers, and there is onl one zero of f, =, we have two test intervals,, and,. Ideall, we wish to find test values in these intervals so that arccot corresponds to one of our oft-used common angles. After a bit of computation, 8 we choose = 0 for < and for >, we choose =. We find f0 = > 0 and f = < 0. Since we are looking for where f = arccot > 0, we get our answer,. To check graphicall, we use the technique in number c of Eample 0.6.5 in Section 0.6 to graph = arccot and we see it is above the horizontal line = on, =, 0.. + 0 = arccot and = 8 Set equal to the cotangents of the common angles and choose accordingl.

87 Foundations of Trigonometr 0.7. Eercises In Eercises - 8, find all of the eact solutions of the equation and then list those solutions which are in the interval [0,.. sin 5 = 0. cos =. sin =. tan 6 = 5. csc = 6. sec = 7. cot = 8. cos 9 = 9 9. sin = 0. cos + 5 = 0. sin =. cos + 7 = 6. csc = 0. tan = 5. tan = 6. sec = 7. cos = 8. sin = In Eercises 9 -, solve the equation, giving the eact solutions which lie in [0, 9. sin = cos 0. sin = sin. sin = cos. cos = sin. cos = cos. cos = 5 cos 5. cos + cos + = 0 6. cos = 5 sin 7. cos = sin + 8. sec = tan 9. tan = sec 0. cot = csc. sec = csc. cos csc cot = 6 cot. sin = tan. cot = csc 7 5. cos + csc = 0 6. tan = tan 7. tan = sec 8. cos = cos 9. tan cos = 0 0. csc + csc = csc +. tan = tan. tan = sec

0.7 Trigonometric Equations and Inequalities 875 In Eercises - 58, solve the equation, giving the eact solutions which lie in [0,. sin6 cos = cos6 sin. sin cos = cos sin 5. cos cos + sin sin = 6. cos5 cos sin5 sin = 7. sin + cos = 8. sin + cos = 9. cos sin = 50. sin + cos = 5. cos sin = 5. sin cos = 5. cos = cos5 5. cos = cos 55. sin5 = sin 56. cos5 = cos 57. sin6 + sin = 0 58. tan = cos In Eercises 59-68, solve the equation. 59. arccos = 60. arcsin = 6. arctan = 0 6. 6 arccot 5 = 0 6. arcsec = 6. arccsc = 65. 9 arcsin = 0 66. 9 arccos = 0 67. 8 arccot + = 0 arccot 68. 6 arctan = arctan + In Eercises 69-80, solve the inequalit. Epress the eact answer in interval notation, restricting our attention to 0. 69. sin 0 70. tan 7. sec 7. cos > 7. cos 0 7. sin + > 75. cot 76. cos 77. sin5 5 78. cos 79. sec 80. cot

876 Foundations of Trigonometr In Eercises 8-86, solve the inequalit. Epress the eact answer in interval notation, restricting our attention to. 8. cos > 8. sin > 8. sec 8. sin < 85. cot 86. cos sin In Eercises 87-9, solve the inequalit. Epress the eact answer in interval notation, restricting our attention to. 87. csc > 88. cos 5 89. cot 5 90. tan 9. sin sin 9. cos sin In Eercises 9-98, solve the given inequalit. 9. arcsin > 0 9. arccos 95. 6 arccot7 96. > arctan 97. arcsin > arcsin 98. arccos + > arccos In Eercises 99-07, epress the domain of the function using the etended interval notation. See page 756 in Section 0.. for details. 99. f = cos 00. f = cos sin + 0. f = tan 0. f = sec 0. f = csc 0. f = sin + cos 05. f = csc + sec 06. f = ln cos 07. f = arcsintan 08. With the help of our classmates, determine the number of solutions to sin = in [0,. Then find the number of solutions to sin =, sin = and sin = in [0,. A pattern should emerge. Eplain how this pattern would help ou solve equations like sin =. Now consider sin =, sin = and sin 5 =. What do ou find? Replace with and repeat the whole eploration.

0.7 Trigonometric Equations and Inequalities 877 0.7. Answers. = k 5 ; = 0, 5, 5, 5, 5,, 6 5, 7 5, 8 5, 9 5. = 9 + k or = 5 9 + k ; = 9, 5 9, 7 9, 9, 9, 7 9. = + k or = 5 6 + k; =, 5 6, 5, 6. = + k 6 ; =, 5, 8,, 7, 7 8, 5, 9, 8, 7,, 5 8 5. = 8 + k ; = 8, 7 8, 8, 5 8 6. = + k or = 7 + k ; =, 7,, 5, 7, 7. = + k ; =, 5 6,, 6 8. No solution 9. = + 6k or = 9 + 6k; = 0. = + k; =, 5. = + k or = + k; =,,, 7. = 9 + k or = + k; =, 5. No solution. = 5 8 + k ; = 8, 5 8, 9 8, 8 5. = + k or = + k; =,,, 5 6. = 6 + k or = 5 6 + k; = 6, 5 6, 7 6, 6 7. = + k ; =,, 5, 7 8. = + k or = + k; =,,, 5

878 Foundations of Trigonometr 9. =, 5 0. = 0,,, 5. = 6,, 5 6,. = 6, 5 6,. = 0,, 5. =,, arccos 7. = 7 6, 6, arcsin, arccos, arcsin. =, 5 6. = 6, 5 6 8. =, 7, arctan, + arctan 9. = 0,, 0. = 6, 5 6,. = arctan, + arctan. = 6, 7 6, 5 6, 6. = 0,,,, 5, 7. = 6,,, 5 6, 7 6, 5, 7, 6 5. =, 6. = 0,,,,, 5 7. =, 5 8. =, 9. = 6,, 5 6,. = 8, 5 8, 9 8, 8 0. = 6, 5 6, 7 6,, 6. No solution. = 0, 7, 7, 7, 7, 5 7, 6 7,, 8 7, 9 7, 0 7, 7, 7, 7. = 0,,, 5. = 0 6. = 8, 8, 8, 8, 5 8, 5 8, 7 8, 7 8, 9 8, 59 8, 6 8, 7 8, 7 8, 8 8, 85 8, 95 8 7. = 0, 8. =, 6 9. =, 7 5. = 7,,, 7 50. = 0,,, 5. = 6, 5 8, 5 6, 7 8,, 9 8

0.7 Trigonometric Equations and Inequalities 879 5. = 0,,,,, 5,, 7 5. = 0,,,,, 5 55. = 0, 8, 8, 5 8, 7 8,, 9 8, 8, 8, 5 8 56. = 7,, 7, 5 7,, 9 7, 7, 5, 7 57. = 0, 7, 7, 6 7, 8 7, 0 7, 58. = arcsin 7, 5, 5,, 7 5, 9 5 + 5 0.666, arcsin + 5.75 59. = 60. = 6. = 6. = 6. = 6. = 6 65. = ± 66. = 67. =, 0 68. = [ 69. [, ] 70., [, [ 7. 0, ] [, ] [ ] 5 [, 7. 0,, 5 [ 7., ] [ 5, 7 ] [ 7. 0, ] 6, 75. 0, ] [,, ] [ 5 [, 76. 0, ] [ ] 5, 7 ], 77. No solution 78. [0, ] [ 79. 0, ], [ ] 7, 80. [arccot, [ + arccot, 8. 6, 8. arcsin, arcsin 6 [ 8., [, ] ], 8.,,

880 Foundations of Trigonometr 85., ] 0, ] 86. [, ] 87.,, 0,, 88. [, ] 89., arccot5 ], arccot5 ] 0, arccot5], + arccot5] 90. 9. [ 7,, 5 ] [,, ] [,, ] [ 5,, 7 ] [, 5 [ 9. 6, 7 6 9. 0, ] ] [, ] [ 0, ] [, 5 } ] [ 6, 5 6 ] {,, ] 9. [, ] 95., 96., 97. [, 0 98. [, 99. 0. 0. 0. 05. 07. k= k= k= k= k= k= k, k + 00. {[ k +, {[ 6k, k k +, k k +, [ k, k + k +, ] 6k + k +, ] k + k= k + k, ]} } k + 0., 06. k= k, ] 7 ], k + k +

Chapter Applications of Trigonometr. Applications of Sinusoids In the same wa eponential functions can be used to model a wide variet of phenomena in nature, the cosine and sine functions can be used to model their fair share of natural behaviors. In section 0.5, we introduced the concept of a sinusoid as a function which can be written either in the form C = A cosω+φ+b for ω > 0 or equivalentl, in the form S = A sinω+φ+b for ω > 0. At the time, we remained undecided as to which form we preferred, but the time for such indecision is over. For clarit of eposition we focus on the sine function in this section and switch to the independent variable t, since the applications in this section are time-dependent. We reintroduce and summarize all of the important facts and definitions about this form of the sinusoid below. ˆ The amplitude is A Properties of the Sinusoid St = A sinωt + φ + B ˆ The angular frequenc is ω and the ordinar frequenc is f = ω ˆ The period is T = f = ω ˆ The phase is φ and the phase shift is φ ω ˆ The vertical shift or baseline is B Along with knowing these formulas, it is helpful to remember what these quantities mean in contet. The amplitude measures the maimum displacement of the sine wave from its baseline determined b the vertical shift, the period is the length of time it takes to complete one ccle of the sinusoid, the angular frequenc tells how man ccles are completed over an interval of length, and the ordinar frequenc measures how man ccles occur per unit of time. The phase indicates what See Section 6.5. Sine haters can use the co-function identit cos θ = sinθ to turn all of the sines into cosines.

88 Applications of Trigonometr angle φ corresponds to t = 0, and the phase shift represents how much of a head start the sinusoid has over the un-shifted sine function. The figure below is repeated from Section 0.5. amplitude baseline period In Section 0.., we introduced the concept of circular motion and in Section 0.., we developed formulas for circular motion. Our first fora into sinusoidal motion puts these notions to good use. Eample... Recall from Eercise 55 in Section 0. that The Giant Wheel at Cedar Point is a circle with diameter 8 feet which sits on an 8 foot tall platform making its overall height 6 feet. It completes two revolutions in minutes and 7 seconds. Assuming that the riders are at the edge of the circle, find a sinusoid which describes the height of the passengers above the ground t seconds after the pass the point on the wheel closest to the ground. Solution. We sketch the problem situation below and assume a counter-clockwise rotation. θ Q h P O Otherwise, we could just observe the motion of the wheel from the other side.

. Applications of Sinusoids 88 We know from the equations given on page 7 in Section 0.. that the -coordinate for counterclockwise motion on a circle of radius r centered at the origin with constant angular velocit frequenc ω is given b = r sinωt. Here, t = 0 corresponds to the point r, 0 so that θ, the angle measuring the amount of rotation, is in standard position. In our case, the diameter of the wheel is 8 feet, so the radius is r = 6 feet. Since the wheel completes two revolutions in minutes and 7 seconds which is 7 seconds the period T = 7 7 = seconds. Hence, the angular frequenc is ω = = radians per second. Putting these two pieces of information T 7 together, we have that = 6 sin 7 t describes the -coordinate on the Giant Wheel after t seconds, assuming it is centered at 0, 0 with t = 0 corresponding to the point Q. In order to find an epression for h, we take the point O in the figure as the origin. Since the base of the Giant Wheel ride is 8 feet above the ground and the Giant Wheel itself has a radius of 6 feet, its center is 7 feet above the ground. To account for this vertical shift upward, we add 7 to our formula for to obtain the new formula h = + 7 = 6 sin 7 t + 7. Net, we need to adjust things so that t = 0 corresponds to the point P instead of the point Q. This is where the phase comes into pla. Geometricall, we need to shift the angle θ in the figure back radians. From Section 0.., we know θ = ωt = 7t, so we temporaril write the height in terms of θ as h = 6 sin θ + 7. Subtracting from θ gives the final answer ht = 6 sin θ + 7 = 6 sin 7 t + 7. We can check the reasonableness of our answer b graphing = ht over the interval [ 0, 7 ]. 6 7 8 A few remarks about Eample.. are in order. First, note that the amplitude of 6 in our answer corresponds to the radius of the Giant Wheel. This means that passengers on the Giant Wheel never stra more than 6 feet verticall from the center of the Wheel, which makes sense. / Second, the phase shift of our answer works out to be /7 = 7 8 = 5.875. This represents the time dela in seconds we introduce b starting the motion at the point P as opposed to the point Q. Said differentl, passengers which start at P take 5.875 seconds to catch up to the point Q. Our net eample revisits the dalight data first introduced in Section.5, Eercise 6b. We are readjusting our baseline from = 0 to = 7. 7 t

88 Applications of Trigonometr Eample... According to the U.S. Naval Observator website, the number of hours H of dalight that Fairbanks, Alaska received on the st da of the nth month of 009 is given below. Here t = represents Januar, 009, t = represents Februar, 009, and so on. Month Number 5 6 7 8 9 0 Hours of Dalight 5.8 9.. 5.9 9..8 9. 5.6. 9. 5.6.. Find a sinusoid which models these data and use a graphing utilit to graph our answer along with the data.. Compare our answer to part to one obtained using the regression feature of a calculator. Solution.. To get a feel for the data, we plot it below. 0 8 6 0 8 6 H 5 6 7 8 9 0 The data certainl appear sinusoidal, 5 but when it comes down to it, fitting a sinusoid to data manuall is not an eact science. We do our best to find the constants A, ω, φ and B so that the function Ht = A sinωt + φ + B closel matches the data. We first go after the vertical shift B whose value determines the baseline. In a tpical sinusoid, the value of B is the average of the maimum and minimum values. So here we take B =.+.8 =.55. Net is the amplitude A which is the displacement from the baseline to the maimum and minimum values. We find A =.8.55 =.55. = 9.5. At this point, we have Ht = 9.5 sinωt + φ +.55. Net, we go after the angular frequenc ω. Since the data collected is over the span of a ear months, we take the period T = months. 6 This 5 Oka, it appears to be the shape we saw in some of the graphs in Section.. Just humor us. 6 Even though the data collected lies in the interval [, ], which has a length of, we need to think of the data point at t = as a representative sample of the amount of dalight for ever da in Januar. That is, it represents Ht over the interval [0, ]. Similarl, t = is a sample of Ht over [, ], and so forth. t

. Applications of Sinusoids 885 means ω = T = = 6. The last quantit to find is the phase φ. Unlike the previous eample, it is easier in this case to find the phase shift φ ω. Since we picked A > 0, the phase shift corresponds to the first value of t with Ht =.55 the baseline value. 7 Here, we choose t =, since its corresponding H value of. is closer to.55 than the net value, 5.9, which corresponds to t =. Hence, φ ω =, so φ = ω = 6 =. We have Ht = 9.5 sin 6 t +.55. Below is a graph of our data with the curve = Ht.. Using the SinReg command, we graph the calculator s regression below. While both models seem to be reasonable fits to the data, the calculator model is possibl the better fit. The calculator does not give us an r value like it did for linear regressions in Section.5, nor does it give us an R value like it did for quadratic, cubic and quartic regressions as in Section.. The reason for this, much like the reason for the absence of R for the logistic model in Section 6.5, is beond the scope of this course. We ll just have to use our own good judgment when choosing the best sinusoid model... Harmonic Motion One of the major applications of sinusoids in Science and Engineering is the stud of harmonic motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the motion of an object on a spring, to the response of an electronic circuit. In this subsection, we restrict our attention to modeling a simple spring sstem. Before we jump into the Mathematics, there are some Phsics terms and concepts we need to discuss. In Phsics, mass is defined as a measure of an object s resistance to straight-line motion whereas weight is the amount of force pull gravit eerts on an object. An object s mass cannot change, 8 while its weight could change. 7 See the figure on page 88. 8 Well, assuming the object isn t subjected to relativistic speeds...

886 Applications of Trigonometr An object which weighs 6 pounds on the surface of the Earth would weigh pound on the surface of the Moon, but its mass is the same in both places. In the English sstem of units, pounds lbs. is a measure of force weight, and the corresponding unit of mass is the slug. In the SI sstem, the unit of force is Newtons N and the associated unit of mass is the kilogram kg. We convert between mass and weight using the formula 9 w = mg. Here, w is the weight of the object, m is the mass and g is the acceleration due to gravit. In the English sstem, g = feet, and in the SI second sstem, g = 9.8 meters. Hence, on Earth a mass of slug weighs lbs. and a mass of kg weighs second 9.8 N. 0 Suppose we attach an object with mass m to a spring as depicted below. The weight of the object will stretch the spring. The sstem is said to be in equilibrium when the weight of the object is perfectl balanced with the restorative force of the spring. How far the spring stretches to reach equilibrium depends on the spring s spring constant. Usuall denoted b the letter k, the spring constant relates the force F applied to the spring to the amount d the spring stretches in accordance with Hooke s Law F = kd. If the object is released above or below the equilibrium position, or if the object is released with an upward or downward velocit, the object will bounce up and down on the end of the spring until some eternal force stops it. If we let t denote the object s displacement from the equilibrium position at time t, then t = 0 means the object is at the equilibrium position, t < 0 means the object is above the equilibrium position, and t > 0 means the object is below the equilibrium position. The function t is called the equation of motion of the object. t = 0 at the t < 0 above the t > 0 below the equilibrium position equilibrium position equilibrium position If we ignore all other influences on the sstem ecept gravit and the spring force, then Phsics tells us that gravit and the spring force will battle each other forever and the object will oscillate indefinitel. In this case, we describe the motion as free meaning there is no eternal force causing the motion and undamped meaning we ignore friction caused b surrounding medium, which in our case is air. The following theorem, which comes from Differential Equations, gives t as a function of the mass m of the object, the spring constant k, the initial displacement 0 of the 9 This is a consequence of Newton s Second Law of Motion F = ma where F is force, m is mass and a is acceleration. In our present setting, the force involved is weight which is caused b the acceleration due to gravit. 0 slug foot kg meter Note that pound = and Newton =. second second Look familiar? We saw Hooke s Law in Section... To keep units compatible, if we are using the English sstem, we use feet ft. to measure displacement. If we are in the SI sstem, we measure displacement in meters m. Time is alwas measured in seconds s.

. Applications of Sinusoids 887 object and initial velocit v 0 of the object. As with t, 0 = 0 means the object is released from the equilibrium position, 0 < 0 means the object is released above the equilibrium position and 0 > 0 means the object is released below the equilibrium position. As far as the initial velocit v 0 is concerned, v 0 = 0 means the object is released from rest, v 0 < 0 means the object is heading upwards and v 0 > 0 means the object is heading downwards. Theorem.. Equation for Free Undamped Harmonic Motion: Suppose an object of mass m is suspended from a spring with spring constant k. If the initial displacement from the equilibrium position is 0 and the initial velocit of the object is v 0, then the displacement from the equilibrium position at time t is given b t = A sinωt + φ where ˆ ω = k m and A = 0 + v0 ˆ A sinφ = 0 and Aω cosφ = v 0. ω It is a great eercise in dimensional analsis to verif that the formulas given in Theorem. work out so that ω has units s and A has units ft. or m, depending on which sstem we choose. Eample... Suppose an object weighing 6 pounds stretches a spring 8 feet.. If the object is attached to the spring and released feet below the equilibrium position from rest, find the equation of motion of the object, t. When does the object first pass through the equilibrium position? Is the object heading upwards or downwards at this instant?. If the object is attached to the spring and released feet below the equilibrium position with an upward velocit of 8 feet per second, find the equation of motion of the object, t. What is the longest distance the object travels above the equilibrium position? When does this first happen? Confirm our result using a graphing utilit. Solution. In order to use the formulas in Theorem., we first need to determine the spring constant k and the mass of the object m. To find k, we use Hooke s Law F = kd. We know the object weighs 6 lbs. and stretches the spring 8 ft.. Using F = 6 and d = 8, we get 6 = k 8, or k = 8 lbs. ft. ft.. To find m, we use w = mg with w = 6 lbs. and g =. We get m = slugs. We can s now proceed to appl Theorem.. k. With k = 8 and m =, we get ω = m = 8 =. We are told that the object is released feet below the equilibrium position from rest. This means 0 = and v 0 = 0. Therefore, A = 0 + v 0 ω = + 0 =. To determine the phase φ, we have A sinφ = 0, which in this case gives sinφ = so sinφ =. Onl φ = and angles coterminal to it The sign conventions here are carried over from Phsics. If not for the spring, the object would fall towards the ground, which is the natural or positive direction. Since the spring force acts in direct opposition to gravit, an movement upwards is considered negative.

888 Applications of Trigonometr satisf this condition, so we pick the phase to be φ =. Hence, the equation of motion is t = sin t +. To find when the object passes through the equilibrium position we solve t = sin t + = 0. Going through the usual analsis we find t = + k for integers k. Since we are interested in the first time the object passes through the equilibrium position, we look for the smallest positive t value which in this case is t = 0.78 seconds after the start of the motion. Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it first passes through it. To check this answer, we graph one ccle of t. Since our applied domain in this situation is t 0, and the period of t is T = ω = =, we graph t over the interval [0, ]. Remembering that t > 0 means the object is below the equilibrium position and t < 0 means the object is above the equilibrium position, the fact our graph is crossing through the t-ais from positive to negative at t = confirms our answer.. The onl difference between this problem and the previous problem is that we now release the object with an upward velocit of 8 ft s. We still have ω = and 0 =, but now we have v 0 = 8, the negative indicating the velocit is directed upwards. Here, we get A = 0 + v 0 ω = + = 5. From A sinφ = 0, we get 5 sinφ = which gives sinφ = 5. From Aω cosφ = v 0, we get 0 cosφ = 8, or cosφ = 5. This means that φ is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. Since t is epressed in terms of sine, we choose to epress φ = arcsin 5. Hence, t = 5 sin t + [ arcsin 5]. Since the amplitude of t is 5, the object will travel at most 5 feet above the equilibrium position. To find when this happens, we solve the equation t = 5 sin t + [ arcsin 5] = 5, the negative once again signifing that the object is above the equilibrium position. Going through the usual machinations, we get t = arcsin 5 + + k for integers k. The smallest of these values occurs when k = 0, that is, t = arcsin 5 +.07 seconds after the start of the motion. To check our answer using the calculator, we graph = 5 sin + [ arcsin 5] on a graphing utilit and confirm the coordinates of the first relative minimum to be approimatel.07, 5. t = sin t + t = 5 sin + [ arcsin ] 5 It is possible, though beond the scope of this course, to model the effects of friction and other eternal forces acting on the sstem. 5 While we ma not have the Phsics and Calculus background For confirmation, we note that Aω cosφ = v 0, which in this case reduces to 6 cosφ = 0. 5 Take a good Differential Equations class to see this!

. Applications of Sinusoids 889 to derive equations of motion for these scenarios, we can certainl analze them. We eamine three cases in the following eample. Eample.... Write t = 5e t/5 cost + 5e t/5 sint in the form t = At sinωt + φ. Graph t using a graphing utilit.. Write t = t + cost + t + sint in the form t = At sinωt + φ. Graph t using a graphing utilit.. Find the period of t = 5 sin6t 5 sin 8t. Graph t using a graphing utilit. Solution.. We start rewriting t = 5e t/5 cost + 5e t/5 sint b factoring out 5e t/5 from both terms to get t = 5e t/5 cost + sint. We convert what s left in parentheses to the required form using the formulas introduced in Eercise 6 from Section 0.5. We find cost + sint = sin t + so that t = 0e t/5 sin t +. Graphing this on the calculator as = 0e /5 sin + reveals some interesting behavior. The sinusoidal nature continues indefinitel, but it is being attenuated. In the sinusoid A sinω + φ, the coefficient A of the sine function is the amplitude. In the case of = 0e /5 sin +, we can think of the function A = 0e /5 as the amplitude. As, 0e /5 0 which means the amplitude continues to shrink towards zero. Indeed, if we graph = ±0e /5 along with = 0e /5 sin +, we see this attenuation taking place. This equation corresponds to the motion of an object on a spring where there is a slight force which acts to damp, or slow the motion. An eample of this kind of force would be the friction of the object against the air. In this model, the object oscillates forever, but with smaller and smaller amplitude. = 0e /5 sin + = 0e /5 sin +, = ±0e /5. Proceeding as in the first eample, we factor out t + from each term in the function t = t + cost + t + sint to get t = t + cost + sint. We find cost + sint = sin t +, so t = t + sin t +. Graphing this on the calculator as = + sin +, we find the sinusoid s amplitude growing. Since our amplitude function here is A = + = + 6, which continues to grow without bound

890 Applications of Trigonometr as, this is hardl surprising. The phenomenon illustrated here is forced motion. That is, we imagine that the entire apparatus on which the spring is attached is oscillating as well. In this case, we are witnessing a resonance effect the frequenc of the eternal oscillation matches the frequenc of the motion of the object on the spring. 6 = + sin + = + sin + = ± +. Last, but not least, we come to t = 5 sin6t 5 sin8t. To find the period of this function, we need to determine the length of the smallest interval on which both ft = 5 sin6t and gt = 5 sin8t complete a whole number of ccles. To do this, we take the ratio of their frequencies and reduce to lowest terms: 6 8 =. This tells us that for ever ccles f makes, g makes. In other words, the period of t is three times the period of ft which is four times the period of gt, or. We graph = 5 sin6 5 sin8 over [0, ] on the calculator to check this. This equation of motion also results from forced motion, but here the frequenc of the eternal oscillation is different than that of the object on the spring. Since the sinusoids here have different frequencies, the are out of snc and do not amplif each other as in the previous eample. Taking things a step further, we can use a sum to product identit to rewrite t = 5 sin6t 5 sin8t as t = 0 sint cos7t. The lower frequenc factor in this epression, 0 sint, plas an interesting role in the graph of t. Below we graph = 5 sin6 5 sin8 and = ±0 sin over [0, ]. This is an eample of the beat phenomena, and the curious reader is invited to eplore this concept as well. 7 = 5 sin6 5 sin8 over [0, ] = 5 sin6 5 sin8 and = ±0 sin over [0, ] 6 The reader is invited to investigate the destructive implications of resonance. 7 A good place to start is this article on beats.

. Applications of Sinusoids 89.. Eercises. The sounds we hear are made up of mechanical waves. The note A above the note middle C is a sound wave with ordinar frequenc f = 0 Hertz = 0 ccles second. Find a sinusoid which models this note, assuming that the amplitude is and the phase shift is 0.. The voltage V in an alternating current source has amplitude 0 and ordinar frequenc f = 60 Hertz. Find a sinusoid which models this voltage. Assume that the phase is 0.. The London Ee is a popular tourist attraction in London, England and is one of the largest Ferris Wheels in the world. It has a diameter of 5 meters and makes one revolution counterclockwise ever 0 minutes. It is constructed so that the lowest part of the Ee reaches ground level, enabling passengers to simpl walk on to, and off of, the ride. Find a sinsuoid which models the height h of the passenger above the ground in meters t minutes after the board the Ee at ground level.. On page 7 in Section 0.., we found the -coordinate of counter-clockwise motion on a circle of radius r with angular frequenc ω to be = r cosωt, where t = 0 corresponds to the point r, 0. Suppose we are in the situation of Eercise above. Find a sinsusoid which models the horizontal displacement of the passenger from the center of the Ee in meters t minutes after the board the Ee. Here we take t > 0 to mean the passenger is to the right of the center, while t < 0 means the passenger is to the left of the center. 5. In Eercise 5 in Section 0., we introduced the o-o trick Around the World in which a o-o is thrown so it sweeps out a vertical circle. As in that eercise, suppose the o-o string is 8 inches and it completes one revolution in seconds. If the closest the o-o ever gets to the ground is inches, find a sinsuoid which models the height h of the o-o above the ground in inches t seconds after it leaves its lowest point. 6. Suppose an object weighing 0 pounds is suspended from the ceiling b a spring which stretches feet to its equilibrium position when the object is attached. a Find the spring constant k in lbs. ft. and the mass of the object in slugs. b Find the equation of motion of the object if it is released from foot below the equilibrium position from rest. When is the first time the object passes through the equilibrium position? In which direction is it heading? c Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocit of feet per second. Find when the object passes through the equilibrium position heading downwards for the third time.

89 Applications of Trigonometr 7. Consider the pendulum below. Ignoring air resistance, the angular displacement of the pendulum from the vertical position, θ, can be modeled as a sinusoid. 8 θ The amplitude of the sinusoid is the same as the initial angular displacement, θ 0, of the pendulum and the period of the motion is given b l T = g where l is the length of the pendulum and g is the acceleration due to gravit. a Find a sinusoid which gives the angular displacement θ as a function of time, t. Arrange things so θ0 = θ 0. b In Eercise 0 section 5., ou found the length of the pendulum needed in Jeff s antique Seth-Thomas clock to ensure the period of the pendulum is of a second. Assuming the initial displacement of the pendulum is 5, find a sinusoid which models the displacement of the pendulum θ as a function of time, t, in seconds. 8. The table below lists the average temperature of Lake Erie as measured in Cleveland, Ohio on the first of the month for each month during the ears 97 000. 9 For eample, t = represents the average of the temperatures recorded for Lake Erie on ever March for the ears 97 through 000. Month Number, t 5 6 7 8 9 0 Temperature F, T 6 8 7 57 67 7 7 67 56 6 a Using the techniques discussed in Eample.., fit a sinusoid to these data. b Using a graphing utilit, graph our model along with the data set to judge the reasonableness of the fit. 8 Provided θ is kept small. Carl remembers the Rule of Thumb as being 0 or less. Check with our friendl neighborhood phsicist to make sure. 9 See this website: http://www.erh.noaa.gov/cle/climate/cle/normals/laketempcle.html.

. Applications of Sinusoids 89 c Use the model ou found in part 8a to predict the average temperature recorded for Lake Erie on April 5 th and September 5 th during the ears 97 000. 0 d Compare our results to those obtained using a graphing utilit. 9. The fraction of the moon illuminated at midnight Eastern Standard Time on the t th da of June, 009 is given in the table below. Da of June, t 6 9 5 8 7 0 Fraction Illuminated, F 0.8 0.98 0.98 0.8 0.57 0.7 0.0 0.0 0.6 0.58 a Using the techniques discussed in Eample.., fit a sinusoid to these data. b Using a graphing utilit, graph our model along with the data set to judge the reasonableness of the fit. c Use the model ou found in part 9a to predict the fraction of the moon illuminated on June, 009. d Compare our results to those obtained using a graphing utilit. 0. With the help of our classmates, research the phenomena mentioned in Eample.., namel resonance and beats.. With the help of our classmates, research Amplitude Modulation and Frequenc Modulation.. What other things in the world might be roughl sinusoidal? Look to see what models ou can find for them and share our results with our class. 0 The computed average is F for April 5 th and 7 F for September 5 th. See this website: http://www.usno.nav.mil/usno/astronomical-applications/data-services/frac-moon-ill. You ma want to plot the data before ou find the phase shift. The listed fraction is 0.6.

89 Applications of Trigonometr.. Answers. St = sin 880t. V t = 0 sin 0t. ht = 67.5 sin 5 t + 67.5. t = 67.5 cos 5 t = 67.5 sin 5 t 5. ht = 8 sin t + 0 6. a k = 5 lbs. ft. and m = 5 6 slugs b t = sin t +. The object first passes through the equilibrium point when t = 8 0.9 seconds after the motion starts. At this time, the object is heading upwards. c t = sin t + 7. The object passes through the equilibrium point heading downwards for the third time when t = 7 6. seconds. 7. a θt = θ 0 sin g l t + b θt = sin t + 8. a T t = 0.5 sin 6 t + 5.5 b Our function and the data set are graphed below. The sinusoid seems to be shifted to the right of our data. c The average temperature on April 5 th is approimatel T.5 9.00 F and the average temperature on September 5 th is approimatel T 9.5 7.8 F. d Using a graphing calculator, we get the following This model predicts the average temperature for April 5 th to be approimatel. F and the average temperature on September 5 th to be approimatel 70.05 F. This model appears to be more accurate.

. Applications of Sinusoids 895 9. a Based on the shape of the data, we either choose A < 0 or we find the second value of t which closel approimates the baseline value, F = 0.505. We choose the latter to obtain F t = 0.75 sin 5 t + 0.505 = 0.75 sin 5 t + 0.505 b Our function and the data set are graphed below. It s a prett good fit. c The fraction of the moon illuminated on June st, 009 is approimatel F 0.60 d Using a graphing calculator, we get the following. This model predicts that the fraction of the moon illuminated on June st, 009 is approimatel 0.59. This appears to be a better fit to the data than our first model.

896 Applications of Trigonometr. The Law of Sines Trigonometr literall means measuring triangles and with Chapter 0 under our belts, we are more than prepared to do just that. The main goal of this section and the net is to develop theorems which allow us to solve triangles that is, find the length of each side of a triangle and the measure of each of its angles. In Sections 0., 0. and 0.6, we ve had some eperience solving right triangles. The following eample reviews what we know. Eample... Given a right triangle with a hpotenuse of length 7 units and one leg of length units, find the length of the remaining side and the measures of the remaining angles. Epress the angles in decimal degrees, rounded to the nearest hundreth of a degree. Solution. For definitiveness, we label the triangle below. c = 7 α β a b = To find the length of the missing side a, we use the Pthagorean Theorem to get a + = 7 which then ields a = units. Now that all three sides of the triangle are known, there are several was we can find α using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths and 7 were given, so we want to relate these to α. According to Theorem 0., cosα = 7. Since α is an acute angle, α = arccos 7 radians. Converting to degrees, we find α 55.5. Now that we have the measure of angle α, we could find the measure of angle β using the fact that α and β are complements so α + β = 90. Once again, we opt to use the data given to us in the problem. According to Theorem 0., we have that sinβ = 7 so β = arcsin 7 radians and we have β.85. A few remarks about Eample.. are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle and the corresponding lowercase English letter represents the side opposite that angle. Thus, a is the side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs α, a, β, b and γ, c are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not alwas the easiest or fastest wa to proceed, it as well as the measure of said angle as well as the length of said side

. The Law of Sines 897 minimizes the chances of propagated error. Third, since man of the applications which require solving triangles in the wild rel on degree measure, we shall adopt this convention for the time being. The Pthagorean Theorem along with Theorems 0. and 0.0 allow us to easil handle an given right triangle problem, but what if the triangle isn t a right triangle? In certain cases, we can use the Law of Sines to help. Theorem.. The Law of Sines: Given a triangle with angle-side opposite pairs α, a, β, b and γ, c, the following ratios hold or, equivalentl, sinα a = sinβ b = sinγ c a sinα = b sinβ = c sinγ The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle ABC below, all of whose angles are acute, with angle-side opposite pairs α, a, β, b and γ, c. If we drop an altitude from verte B, we divide the triangle into two right triangles: ABQ and BCQ. If we call the length of the altitude h for height, we get from Theorem 0. that sinα = h c and sinγ = h a so that h = c sinα = a sinγ. After some rearrangement of the last equation, we get sinα a using the triangles ABQ and ACQ to get sinβ α c B β γ a = sinγ c. If we drop an altitude from verte A, we can proceed as above = sinγ, completing the proof for this case. α c b A C A C b Q A C b For our net case consider the triangle ABC below with obtuse angle α. Etending an altitude from verte A gives two right triangles, as in the previous case: ABQ and ACQ. Proceeding as before, we get h = b sinγ and h = c sinβ so that sinβ b = sinγ c. B h γ c a c β h B Q γ B c β α a γ B c β Q h a γ A b C A b C Your Science teachers should thank us for this. Don t worr! Radians will be back before ou know it!

898 Applications of Trigonometr Dropping an altitude from verte B also generates two right triangles, ABQ and BCQ. We know that sinα = h c so that h = c sinα. Since α = 80 α, sinα = sinα, so in fact, we have h = c sinα. Proceeding to BCQ, we get sinγ = h a so h = a sinγ. Putting this together with the previous equation, we get sinγ c B = sinα a, and we are finished with this case. β a h c α α γ Q A b C The remaining case is when ABC is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 0. and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The net eample showcases some of the power, and the pitfalls, of the Law of Sines. Eample... Solve the following triangles. Give eact answers and decimal approimations rounded to hundredths and sketch the triangle.. α = 0, a = 7 units, β = 5. α = 85, β = 0, c = 5.5 units. α = 0, a = units, c = units. α = 0, a = units, c = units 5. α = 0, a = units, c = units 6. α = 0, a = units, c = units Solution.. Knowing an angle-side opposite pair, namel α and a, we ma proceed in using the Law of Sines. Since β = 5 b, we use sin5 = 7 sin0 so b = 7 sin5 sin0 = 7 6 5.7 units. Now that we have two angle-side pairs, it is time to find the third. To find γ, we use the fact that the sum of the measures of the angles in a triangle is 80. Hence, γ = 80 0 5 = 5. To find c, we have no choice but to used the derived value γ = 5, et we can minimize the propagation of error here b using the given angle-side opposite pair α, a. The Law of Sines c gives us sin5 = 7 sin0 so that c = 7 sin5 sin0.09 units.5. In this eample, we are not immediatel given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since γ = 80 85 0 = 65. As in the previous eample, we are forced to use a derived value in our computations since the onl 5 The eact value of sin5 could be found using the difference identit for sine or a half-angle formula, but that becomes unnecessaril mess for the discussion at hand. Thus eact here means 7 sin5 sin0.

. The Law of Sines 899 a angle-side pair available is γ, c. The Law of Sines gives sin85 = 5.5 sin65. After the usual rearrangement, we get a = 5.5 sin85 sin65 5.77 units. To find b we use the angle-side pair γ, c b which ields sin0 = 5.5 sin65 hence b = 5.5 sin0 sin65.90 units. β = 0 β = 5 a = 7 c = 5.5 a 5.77 c.09 α = 0 γ = 5 α = 85 γ = 65 b 5.7 b.90 Triangle for number Triangle for number. Since we are given α, a and c, we use the Law of Sines to find the measure of γ. We start with sinγ = sin0 and get sinγ = sin 0 =. Since the range of the sine function is [, ], there is no real number with sinγ =. Geometricall, we see that side a is just too short to make a triangle. The net three eamples keep the same values for the measure of α and the length of c while varing the length of a. We will discuss this case in more detail after we see what happens in those eamples.. In this case, we have the measure of α = 0, a = and c =. Using the Law of Sines, we get sinγ = sin0 so sinγ = sin 0 =. Now γ is an angle in a triangle which also contains α = 0. This means that γ must measure between 0 and 50 in order to fit inside the triangle with α. The onl angle that satisfies this requirement and has sinγ = is γ = 90. In other words, we have a right triangle. We find the measure of β to be β = 80 0 90 = 60 and then determine b using the Law of Sines. We find b = sin60 sin0 =.6 units. In this case, the side a is precisel long enough to form a unique right triangle. c = a = c = β = 60 a = α = 0 α = 0 b.6 Diagram for number Triangle for number 5. Proceeding as we have in the previous two eamples, we use the Law of Sines to find γ. In this case, we have sinγ = sin0 or sinγ = sin0 =. Since γ lies in a triangle with α = 0,

900 Applications of Trigonometr we must have that 0 < γ < 50. There are two angles γ that fall in this range and have sinγ = : γ = arcsin radians.8 and γ = arcsin radians 8.9. At this point, we pause to see if it makes sense that we actuall have two viable cases to consider. As we have discussed, both candidates for γ are compatible with the given angle-side pair α, a = 0, in that both choices for γ can fit in a triangle with α and both have a sine of. The onl other given piece of information is that c = units. Since c > a, it must be true that γ, which is opposite c, has greater measure than α which is opposite a. In both cases, γ > α, so both candidates for γ are compatible with this last piece of given information as well. Thus have two triangles on our hands. In the case γ = arcsin radians.8, we find 6 β 80 0.8 = 08.9. Using the Law of Sines with the angle-side opposite pair α, a and β, we find b sin08.9 sin0 5.70 units. In the case γ = arcsin radians 8.9, we repeat the eact same steps and find β.8 and b. units. 7 triangles are drawn below. Both β.8 c = α = 0 β 08.9 a = γ.8 b 5.70 c = a = α = 0 γ 8.9 b. 6. For this last problem, we repeat the usual Law of Sines routine to find that sinγ = sin0 so that sinγ =. Since γ must inhabit a triangle with α = 0, we must have 0 < γ < 50. Since the measure of γ must be strictl less than 50, there is just one angle which satisfies both required conditions, namel γ = 0. So β = 80 0 0 = 0 and, using the Law of Sines one last time, b = sin0 sin0 = 6.9 units. c = β = 0 a = α = 0 γ = 0 b 6.9 Some remarks about Eample.. are in order. We first note that if we are given the measures of two of the angles in a triangle, sa α and β, the measure of the third angle γ is uniquel 6 To find an eact epression for β, we convert everthing back to radians: α = 0 = radians, γ = arcsin 6 radians and 80 = radians. Hence, β = arcsin 6 = 5 arcsin 6 radians 08.9. 7 An eact answer for β in this case is β = arcsin radians 6.8.

. The Law of Sines 90 determined using the equation γ = 80 α β. Knowing the measures of all three angles of a triangle completel determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to find the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers and above. In number, the given side is adjacent to just one of the angles this is called the Angle-Angle-Side AAS case. 8 In number, the given side is adjacent to both angles which means we are in the so-called Angle-Side-Angle ASA case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, onl one of which is adjacent to the given angle, we are in the Angle-Side-Side ASS case. 9 In number, the length of the one given side a was too short to even form a triangle; in number, the length of a was just long enough to form a right triangle; in 5, a was long enough, but not too long, so that two triangles were possible; and in number 6, side a was long enough to form a triangle but too long to swing back and form two. These four cases eemplif all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem. Theorem.. Suppose α, a and γ, c are intended to be angle-side pairs in a triangle where α, a and c are given. Let h = c sinα ˆ ˆ ˆ ˆ If a < h, then no triangle eists which satisfies the given criteria. If a = h, then γ = 90 so eactl one right triangle eists which satisfies the criteria. If h < a < c, then two distinct triangles eist which satisf the given criteria. If a c, then γ is acute and eactl one triangle eists which satisfies the given criteria Theorem. is proved on a case-b-case basis. If a < h, then a < c sinα. If a triangle were to eist, the Law of Sines would have sinγ c = sinα a so that sinγ = c sinα a > a a =, which is impossible. In the figure below, we see geometricall wh this is the case. c a c h = c sinα a = h = c sinα α α a < h, no triangle a = h, γ = 90 Simpl put, if a < h the side a is too short to connect to form a triangle. This means if a h, we are alwas guaranteed to have at least one triangle, and the remaining parts of the theorem 8 If this sounds familiar, it should. From high school Geometr, we know there are four congruence conditions for triangles: Angle-Angle-Side AAS, Angle-Side-Angle ASA, Side-Angle-Side SAS and Side-Side-Side SSS. If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that eactl one triangle eists which satisfies the given criteria. 9 In more reputable books, this is called the Side-Side-Angle or SSA case.

90 Applications of Trigonometr tell us what kind and how man triangles to epect in each case. If a = h, then a = c sinα and the Law of Sines gives sinα a = sinγ c so that sinγ = c sinα a = a a =. Here, γ = 90 as required. Moving along, now suppose h < a < c. As before, the Law of Sines 0 gives sinγ = c sinα a. Since h < a, c sinα < a or c sinα a < which means there are two solutions to sinγ = c sinα a : an acute angle which we ll call γ 0, and its supplement, 80 γ 0. We need to argue that each of these angles fit into a triangle with α. Since α, a and γ 0, c are angle-side opposite pairs, the assumption c > a in this case gives us γ 0 > α. Since γ 0 is acute, we must have that α is acute as well. This means one triangle can contain both α and γ 0, giving us one of the triangles promised in the theorem. If we manipulate the inequalit γ 0 > α a bit, we have 80 γ 0 < 80 α which gives 80 γ 0 + α < 80. This proves a triangle can contain both of the angles α and 80 γ 0, giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume a c. Then α γ, which forces γ to be an acute angle. Hence, we get onl one triangle in this case, completing the proof. c a α γ 0 γ 0 h h < a < c, two triangles a c a h α γ a c, one triangle One last comment before we use the Law of Sines to solve an application problem. In the Angle- Side-Side case, if ou are given an obtuse angle to begin with then it is impossible to have the two triangle case. Think about this before reading further. Eample... Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is 0 and at the second point the angle is 5. Assuming a straight coastline, find the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point? Solution. We sketch the problem below with the first observation point labeled as P and the second as Q. In order to use the Law of Sines to find the distance d from Q to the island, we first need to find the measure of β which is the angle opposite the side of length 5 miles. To that end, we note that the angles γ and 5 are supplemental, so that γ = 80 5 = 5. We can now find β = 80 0 γ = 80 0 5 = 5. B the Law of Sines, we have d sin0 = 5 sin5 which gives d = 5 sin0 sin5 9.66 miles. Net, to find the point on the coast closest to the island, which we ve labeled as C, we need to find the perpendicular distance from the island to the coast. 0 Remember, we have alread argued that a triangle eists in this case! Do ou see wh C must lie to the right of Q?

. The Law of Sines 90 Let denote the distance from the second observation point Q to the point C and let denote the distance from C to the island. Using Theorem 0., we get sin 5 = d. After some rearranging, we find = d sin 5 9.66 6.8 miles. Hence, the island is approimatel 6.8 miles from the coast. To find the distance from Q to C, we note that β = 80 90 5 = 5 so b smmetr, we get = 6.8 miles. Hence, the point on the shore closest to the island is approimatel 6.8 miles down the coast from the second observation point. Sasquatch Island Sasquatch Island β β d 9.66 miles d 9.66 miles miles P 0 γ Q 5 Shoreline Q 5 C 5 miles miles We close this section with a new formula to compute the area enclosed b a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an eercise. Theorem.. Suppose α, a, β, b and γ, c are the angle-side opposite pairs of a triangle. Then the area A enclosed b the triangle is given b A = bc sinα = ac sinβ = ab sinγ Eample... Find the area of the triangle in Eample.. number. Solution. From our work in Eample.. number, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose A = ac sinβ from Theorem. because it uses the most pieces of given information. We are given a = 7 and β = 5, and we calculated c = 7 sin5 sin0. Using these values, we find A = 7 7 sin5 sin0 sin 5 = 5.8 square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem.. Or b Theorem 0. again...

90 Applications of Trigonometr.. Eercises In Eercises - 0, solve for the remaining sides and angles if possible. As in the tet, α, a, β, b and γ, c are angle-side opposite pairs.. α =, β = 7, a = 5. α = 7., β = 5., a = 7. α = 95, β = 85, a =.. α = 95, β = 6, a =. 5. α = 7, a = 5, b = 6. α = 7, a = 5, b = 7. α = 68.7, a = 88, b = 9 8. α =, a = 7, b =.5 9. α = 68.7, a = 70, b = 90 0. α = 0, a = 7, b =. α =, a = 9, b =.5. γ = 5, α = 5, c = 8.0. α = 6, a = 57, b = 00. γ = 7.6, c =, a =.05 5. β = 0, b = 6.75, c = 6. β = 0, b = 6.75, c = 8 7. β = 0, γ = 5, b = 6.75 8. β = 9., γ = 8.95, b =.5 9. γ = 0, β = 6, c = 0. α = 50, a = 5, b =.5. Find the area of the triangles given in Eercises, and 0 above. Another Classic Application: Grade of a Road The grade of a road is much like the pitch of a roof See Eample 0.6.6 in that it epresses the ratio of rise/run. In the case of a road, this ratio is alwas positive because it is measured going uphill and it is usuall given as a percentage. For eample, a road which rises 7 feet for ever 00 feet of horizontal forward progress is said to have a 7% grade. However, if we want to appl an Trigonometr to a stor problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Eercises -, we first have ou change road grades into angles and then use the Law of Sines in an application.. Using a right triangle with a horizontal leg of length 00 and vertical leg with length 7, show that a 7% grade means that the road hpotenuse makes about a angle with the horizontal. It will not be eactl, but it s prett close.. What grade is given b a 9.65 angle made b the road and the horizontal? I have friends who live in Pacifica, CA and their road is actuall this steep. It s not a nice road to drive.

. The Law of Sines 905. Along a long, straight stretch of mountain road with a 7% grade, ou see a tall tree standing perfectl plumb alongside the road. From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is 6. Use the Law of Sines to find the height of the tree. Hint: First show that the tree makes a 9 angle with the road. Another Classic Application: Bearings In the net several eercises we introduce and work with the navigation tool known as bearings. Simpl put, a bearing is the direction ou are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation to the east or to the west awa from the north-south up and down line of a compass rose. For eample, N0 E read 0 east of north is a bearing which is rotated clockwise 0 from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of θ = 50. Similarl, S50 W would point into Quadrant III along the terminal side of θ = 0 because we started out pointing due south along θ = 70 and rotated clockwise 50 back to 0. Counter-clockwise rotations would be found in the bearings N60 W which is on the terminal side of θ = 50 and S7 E which lies along the terminal side of θ = 97. These four bearings are drawn in the plane below. N60 W 60 N 0 N0 E W E S50 W 50 S 7 S7 E The cardinal directions north, south, east and west are usuall not given as bearings in the fashion described above, but rather, one just refers to them as due north, due south, due east and due west, respectivel, and it is assumed that ou know which quadrantal angle goes with each cardinal direction. Hint: Look at the diagram above. 5. Find the angle θ in standard position with 0 θ < 60 which corresponds to each of the bearings given below. a due west b S8 E c N5.5 E d due south The word plumb here means that the tree is perpendicular to the horizontal.

906 Applications of Trigonometr e N.5 W f S7 W 5 g N5 E h S5 W 6. The Colonel spots a campfire at a of bearing N E from his current position. Sarge, who is positioned 000 feet due east of the Colonel, reckons the bearing to the fire to be N0 W from his current position. Determine the distance from the campfire to each man, rounded to the nearest foot. 7. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N5 W which brings her to the Muffin Ridge Observator. From there, she knows a bearing of S65 E will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muffin Ridge Observator to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 8. The captain of the SS Bigfoot sees a signal flare at a bearing of N5 E from her current location. From his position, the captain of the HMS Sasquatch finds the signal flare to be at a bearing of N75 W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50 E, find the distances from the flare to each vessel, rounded to the nearest tenth of a mile. 9. Carl spies a potential Sasquatch nest at a bearing of N0 E and radios Jeff, who is at a bearing of N50 E from Carl s position. From Jeff s position, the nest is at a bearing of S70 W. If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round our answer to the nearest foot. 0. A hiker determines the bearing to a lodge from her current position is S0 W. She proceeds to hike miles at a bearing of S0 E at which point she determines the bearing to the lodge is S75 W. How far is she from the lodge at this point? Round our answer to the nearest hundredth of a mile.. A watchtower spots a ship off shore at a bearing of N70 E. A second tower, which is 50 miles from the first at a bearing of S80 E from the first tower, determines the bearing to the ship to be N5 W. How far is the boat from the second tower? Round our answer to the nearest tenth of a mile.. Skipp and Sall decide to hunt UFOs. One night, the position themselves miles apart on an abandoned stretch of desert runwa. An hour into their investigation, Skipp spies a UFO hovering over a spot on the runwa directl between him and Sall. He records the angle of inclination from the ground to the craft to be 75 and radios Sall immediatel to find the angle of inclination from her position to the craft is 50. How high off the ground is the UFO at this point? Round our answer to the nearest foot. Recall: mile is 580 feet. 5 See Eample 0.. in Section 0. for a review of the DMS sstem.

. The Law of Sines 907. The angle of depression from an observer in an apartment comple to a gargole on the building net door is 55. From a point five stories below the original observer, the angle of inclination to the gargole is 0. Find the distance from each observer to the gargole and the distance from the gargole to the apartment comple. Round our answers to the nearest foot. Use the rule of thumb that one stor of a building is 9 feet.. Prove that the Law of Sines holds when ABC is a right triangle. 5. Discuss with our classmates wh knowing onl the three angles of a triangle is not enough to determine an of the sides. 6. Discuss with our classmates wh the Law of Sines cannot be used to find the angles in the triangle when onl the three sides are given. Also discuss what happens if onl two sides and the angle between them are given. Said another wa, eplain wh the Law of Sines cannot be used in the SSS and SAS cases. 7. Given α = 0 and b = 0, choose four different values for a so that a the information ields no triangle b the information ields eactl one right triangle c the information ields two distinct triangles d the information ields eactl one obtuse triangle Eplain wh ou cannot choose a in such a wa as to have α = 0, b = 0 and our choice of a ield onl one triangle where that unique triangle has three acute angles. 8. Use the cases and diagrams in the proof of the Law of Sines Theorem. to prove the area formulas given in Theorem.. Wh do those formulas ield square units when four quantities are being multiplied together?

908 Applications of Trigonometr.. Answers. α = β = 7 γ = 50 a = 5 b 6.50 c.. α = 7. β = 5. γ = 5.7 a = 7 b 99.00 c 97.. Information does not produce a triangle. α = 95 β = 6 γ = a =. b 9.5 c.07 5. Information does not produce a triangle 6. α = 7 β 56. γ 6.7 a = 5 b = c 5.89 7. α = 68.7 β 76.9 γ. a = 88 b = 9 c 5.6 8. α = β 67.66 γ 70. a = 7 b =.5 c.9 α = 68.7 β 0. γ 8. a = 88 b = 9 c.7 α = β. γ 5.66 a = 7 b =.5 c.00 9. Information does not produce a triangle 0. α = 0 β = 90 γ = 60 a = 7 b = c = 7. α = β.78 γ. a = 9 b =.5 c 5.5. α = 5 β = 7 γ = 5 a = 8.0 b.7 c = 8.0. α = 6 β 69. γ.57 a = 57 b = 00 c.5. α 78.59 β 6.8 γ = 7.6 a =.05 b.0 c = α = 6 β 0.57 γ 6. a = 57 b = 00 c 55.5 α 0. β.99 γ = 7.6 a =.05 b 0.7 c = 5. α 8.6 β = 0 γ 9.9 a 8.0 b = 6.75 c = 6. Information does not produce a triangle 7. α = β = 0 γ = 5 a.68 b = 6.75 c 9.8 8. α = 66.9 β = 9. γ = 8.95 a 59.69 b =.5 c 6.75 9. Information does not produce a triangle 0. α = 50 β.5 γ 07.8 a = 5 b =.5 c.. The area of the triangle from Eercise is about 8. square units. The area of the triangle from Eercise is about 77. square units. The area of the triangle from Eercise 0 is about 9 square units.. arctan 7 00 0.699 radians, which is equivalent to.00. About 7%. About 5 feet

. The Law of Sines 909 5. a θ = 80 b θ = 5 c θ = 8.5 d θ = 70 e θ =.5 f θ = 97 8 8 g θ = 5 h θ = 5 6. The Colonel is about 9 feet from the campfire. Sarge is about 55 feet to the campfire. 7. The distance from the Muffin Ridge Observator to Sasquach Point is about 7. miles. The distance from Sasquatch Point to the Chupacabra Trailhead is about.6 miles. 8. The SS Bigfoot is about. miles from the flare. The HMS Sasquatch is about.9 miles from the flare. 9. Jeff is about 7 feet from the nest. 0. She is about.0 miles from the lodge. The boat is about 5. miles from the second tower.. The UFO is hovering about 959 feet above the ground.. The gargole is about feet from the observer on the upper floor. The gargole is about 7 feet from the observer on the lower floor. The gargole is about 5 feet from the other building.

90 Applications of Trigonometr. The Law of Cosines In Section., we developed the Law of Sines Theorem. to enable us to solve triangles in the Angle-Angle-Side AAS, the Angle-Side-Angle ASA and the ambiguous Angle-Side-Side ASS cases. In this section, we develop the Law of Cosines which handles solving triangles in the Side-Angle-Side SAS and Side-Side-Side SSS cases. We state and prove the theorem below. Theorem.5. Law of Cosines: Given a triangle with angle-side opposite pairs α, a, β, b and γ, c, the following equations hold a = b + c bc cosα b = a + c ac cosβ c = a + b ab cosγ or, solving for the cosine in each equation, we have cosα = b + c a bc cosβ = a + c b ac cosγ = a + b c ab To prove the theorem, we consider a generic triangle with the verte of angle α at the origin with side b positioned along the positive -ais. B = c cosα, c sinα c a α A = 0, 0 b C = b, 0 From this set-up, we immediatel find that the coordinates of A and C are A0, 0 and Cb, 0. From Theorem 0., we know that since the point B, lies on a circle of radius c, the coordinates Here, Side-Angle-Side means that we are given two sides and the included angle - that is, the given angle is adjacent to both of the given sides.

. The Law of Cosines 9 of B are B, = Bc cosα, c sinα. This would be true even if α were an obtuse or right angle so although we have drawn the case when α is acute, the following computations hold for an angle α drawn in standard position where 0 < α < 80. We note that the distance between the points B and C is none other than the length of side a. Using the distance formula, Equation., we get a = c cosα b + c sinα 0 a = c cosα b + c sin α a = c cosα b + c sin α a = c cos α bc cosα + b + c sin α a = c cos α + sin α + b bc cosα a = c + b bc cosα Since cos α + sin α = a = c + b bc cosα The remaining formulas given in Theorem.5 can be shown b simpl reorienting the triangle to place a different verte at the origin. We leave these details to the reader. What s important about a and α in the above proof is that α, a is an angle-side opposite pair and b and c are the sides adjacent to α the same can be said of an other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula which has its roots in the Pthagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the Pthagorean Theorem. If we have a triangle in which γ = 90, then cosγ = cos 90 = 0 so we get the familiar relationship c = a +b. What this means is that in the larger mathematical sense, the Law of Cosines and the Pthagorean Theorem amount to prett much the same thing. Eample... Solve the following triangles. Give eact answers and decimal approimations rounded to hundredths and sketch the triangle.. β = 50, a = 7 units, c = units. a = units, b = 7 units, c = 5 units Solution.. We are given the lengths of two sides, a = 7 and c =, and the measure of the included angle, β = 50. With no angle-side opposite pair to use, we appl the Law of Cosines. We get b = 7 + 7 cos 50 which ields b = 5 8 cos 50 5.9 units. In order to determine the measures of the remaining angles α and γ, we are forced to used the derived value for b. There are two was to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair β, b we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not This shouldn t come as too much of a shock. All of the theorems in Trigonometr can ultimatel be traced back to the definition of the circular functions along with the distance formula and hence, the Pthagorean Theorem.

9 Applications of Trigonometr enough to determine if the angle in question is acute or obtuse. Since both authors of the tetbook prefer the Law of Cosines, we proceed with this method first. When using the Law of Cosines, it s alwas best to find the measure of the largest unknown angle first, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to find α first. To that end, we use the formula cosα = b +c a bc and substitute a = 7, b = 5 8 cos 50 and c =. We get cosα = 7 cos 50 5 8 cos 50 Since α is an angle in a triangle, we know the radian measure of α must lie between 0 and radians. This matches the range of the arccosine function, so we have 7 cos 50 α = arccos radians.99 5 8 cos 50 At this point, we could find γ using γ = 80 α β 80.99 50 = 5.0, that is if we trust our approimation for α. To minimize propagation of error, however, we could use the Law of Cosines again, in this case using cosγ = a +b c ab. Plugging in a = 7, b = 5 8 cos 50 7 cos50 and c =, we get γ = arccos radians 5.0. We sketch the triangle below. 5 8 cos50 β = 50 a = 7 c = α.99 γ 5.0 b 5.9 As we mentioned earlier, once we ve determined b it is possible to use the Law of Sines to find the remaining angles. Here, however, we must proceed with caution as we are in the ambiguous ASS case. It is advisable to first find the smallest of the unknown angles, since we are guaranteed it will be acute. 5 In this case, we would find γ since the side opposite γ is smaller than the side opposite the other unknown angle, α. Using the angle-side opposite pair β, b, we get sinγ sin50. The usual calculations produces γ 5 8 cos50 5.0 and = α = 80 β γ 80 50 5.0 =.99.. Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we find β first, since it is opposite the longest side, b. We get cosβ = a +c b ac = 5, so we get β = arccos 5 radians 0.5. As in after simplifing... Your instructor will let ou know which procedure to use. It all boils down to how much ou trust our calculator. 5 There can onl be one obtuse angle in the triangle, and if there is one, it must be the largest.

. The Law of Cosines 9 the previous problem, now that we have obtained an angle-side opposite pair β, b, we could proceed using the Law of Sines. The Law of Cosines, however, offers us a rare opportunit to find the remaining angles using onl the data given to us in the statement of the problem. Using this, we get γ = arccos 5 7 radians. and α = arccos 9 5 radians.05. β 0.5 c = 5 a = α.05 γ. We note that, depending on how man decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approimate answers ou obtain ma differ slightl from those the authors obtain in the Eamples and the Eercises. A great eample of this is number in Eample.., where the approimate values we record for the measures of the angles sum to 80.0, which is geometricall impossible. Net, we have an application of the Law of Cosines. Eample... A researcher wishes to determine the width of a vernal pond as drawn below. From a point P, he finds the distance to the eastern-most point of the pond to be 950 feet, while the distance to the western-most point of the pond from P is 000 feet. If the angle between the two lines of sight is 60, find the width of the pond. b = 7 000 feet 60 950 feet P Solution. We are given the lengths of two sides and the measure of an included angle, so we ma appl the Law of Cosines to find the length of the missing side opposite the given angle. Calling this length w for width, we get w = 950 + 000 950000 cos 60 = 95500 from which we get w = 95500 976 feet.

9 Applications of Trigonometr In Section., we used the proof of the Law of Sines to develop Theorem. as an alternate formula for the area enclosed b a triangle. In this section, we use the Law of Cosines to derive another such formula - Heron s Formula. Theorem.6. Heron s Formula: Suppose a, b and c denote the lengths of the three sides of a triangle. Let s be the semiperimeter of the triangle, that is, let s = a + b + c. Then the area A enclosed b the triangle is given b A = ss as bs c We prove Theorem.6 using Theorem.. Using the convention that the angle γ is opposite the side c, we have A = ab sinγ from Theorem.. In order to simplif computations, we start b manipulating the epression for A. A = ab sinγ = a b sin γ = a b cos γ since sin γ = cos γ. The Law of Cosines tells us cosγ = a +b c ab, so substituting this into our equation for A gives A = a b cos γ [ = a b a + b c ] ab [ = a b a + b c ] a b [ = a b a b a + b c ] a b = a b a + b c 6 = ab a + b c 6 [ ab a + b c ] ab + [ a + b c ] = = 6 c a + ab b a + ab + b c 6 difference of squares.

. The Law of Cosines 95 A = = = = = c [ a ab + b ] [ a + ab + b ] c 6 c a b a + b c 6 c a bc + a ba + b ca + b + c 6 b + c aa + c ba + b ca + b + c 6 b + c a a + c b a + b c a + b + c perfect square trinomials. difference of squares. At this stage, we recognize the last factor as the semiperimeter, s = a+b+c a + b + c =. To complete the proof, we note that s a = a + b + c a = a + b + c a = b + c a Similarl, we find s b = a+c b and s c = a+b c. Hence, we get A = b + c a a + c b = s as bs cs a + b c a + b + c so that A = ss as bs c as required. We close with an eample of Heron s Formula. Eample... Find the area enclosed of the triangle in Eample.. number. Solution. We are given a =, b = 7 and c = 5. Using these values, we find s = + 7 + 5 = 8, s a = 8 =, s b = 8 7 = and s c = 8 5 =. Using Heron s Formula, we get A = ss as bs c = 8 = 96 = 6 9.80 square units.

96 Applications of Trigonometr.. Eercises In Eercises - 0, use the Law of Cosines to find the remaining sides and angles if possible.. a = 7, b =, γ = 59.. α = 0, b = 5, c = 7. a = 5, β = 8., c = 5. a =, b =, γ = 90 5. α = 0, b =, c = 6. a = 7, b = 0, c = 7. a =, b =, c = 5 8. a = 00, b = 0, c = 8 9. a = 5, b = 5, c = 5 0. a = 5, b =, ; c = In Eercises - 6, solve for the remaining sides and angles, if possible, using an appropriate technique.. a = 8, α = 6, b = 0. a = 7, b = 5, c = 6. a = 6, α = 6, b = 0. a =, α = 6, b = 0 5. α =, b = 7, c = 88 6. β = 7, γ = 70, c = 98.6 7. Find the area of the triangles given in Eercises 6, 8 and 0 above. 8. The hour hand on m antique Seth Thomas schoolhouse clock in inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o clock. Round our answer to the nearest hundredth of an inch. 9. A geologist wants to measure the diameter of a crater. From her camp, it is miles to the northern-most point of the crater and miles to the southern-most point. If the angle between the two lines of sight is 7, what is the diameter of the crater? Round our answer to the nearest hundredth of a mile. 0. From the Pedimaus International Airport a tour helicopter can fl to Cliffs of Insanit Point b following a bearing of N8. E for 9 miles and it can fl to Bigfoot Falls b following a bearing of S68.5 E for 07 miles. 6 Find the distance between Cliffs of Insanit Point and Bigfoot Falls. Round our answer to the nearest mile.. Cliffs of Insanit Point and Bigfoot Falls from Eericse 0 above both lie on a straight stretch of the Great Sasquatch Canon. What bearing would the tour helicopter need to follow to go directl from Bigfoot Falls to Cliffs of Insanit Point? Round our angle to the nearest tenth of a degree. 6 Please refer to Page 905 in Section. for an introduction to bearings.

. The Law of Cosines 97. A naturalist sets off on a hike from a lodge on a bearing of S80 W. After.5 miles, she changes her bearing to S7 W and continues hiking for miles. Find her distance from the lodge at this point. Round our answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round our angle to the nearest degree.. The HMS Sasquatch leaves port on a bearing of N E and travels for 5 miles. It then changes course and follows a heading of S E for miles. How far is it from port? Round our answer to the nearest hundredth of a mile. What is its bearing to port? Round our angle to the nearest degree.. The SS Bigfoot leaves a harbor bound for Nessie Island which is 00 miles awa at a bearing of N E. A storm moves in and after 00 miles, the captain of the Bigfoot finds he has drifted off course. If his bearing to the harbor is now S70 W, how far is the SS Bigfoot from Nessie Island? Round our answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round our angle to the nearest tenth of a degree. 5. From a point 00 feet above level ground in a firetower, a ranger spots two fires in the Yeti National Forest. The angle of depression 7 made b the line of sight from the ranger to the first fire is.5 and the angle of depression made b line of sight from the ranger to the second fire is.. The angle formed b the two lines of sight is 7. Find the distance between the two fires. Round our answer to the nearest foot. Hint: In order to use the 7 angle between the lines of sight, ou will first need to use right angle Trigonometr to find the lengths of the lines of sight. This will give ou a Side-Angle-Side case in which to appl the Law of Cosines. fire 7 fire firetower 6. If ou appl the Law of Cosines to the ambiguous Angle-Side-Side ASS case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not ield a triangle. If the equation has onl one positive real zero then eactl one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Appl the Law of Cosines to Eercises, and above in order to demonstrate this result. 7. Discuss with our classmates wh Heron s Formula ields an area in square units even though four lengths are being multiplied together. 7 See Eercise 78 in Section 0. for the definition of this angle.

98 Applications of Trigonometr.. Answers. α 5.5 β 85.6 γ = 59. a = 7 b = c 0.6. α = 0 β 9.0 γ 6.60 a 9. b = 5 c = 7. α 85.90 β = 8. γ 85.90 a = 5 b.88 c = 5. α 6.87 β 5. γ = 90 a = b = c = 5 5. α = 0 β 5.8 γ.7 a = 7 b = c = 6. α. β 9.58 γ 98. a = 7 b = 0 c = 7. Information does not produce a triangle 8. α 8.05 β 87.8 γ 9. a = 00 b = 0 c = 8 9. α = 60 β = 60 γ = 60 a = 5 b = 5 c = 5 0. α.6 β 67.8 γ = 90 a = 5 b = c =. α = 6 β 98. γ 8.89 a = 8 b = 0 c 6.5. α 55.0 β 89.0 γ 5.0 a = 7 b = 5 c = 6 α = 6 β 8.89 γ 5. a = 8 b = 0 c.6. Information does not produce a triangle. α = 6 β 5. γ 6.9 a = b = 0 c.98 5. α = β 89. γ 8.77 a 78.0 b = 7 c = 88 6. α β = 7 γ = 70 a 9.7 b 69. c = 98.6 7. The area of the triangle given in Eercise 6 is 00 = 0.6 square units. The area of the triangle given in Eercise 8 is 57675 79.75 square units. The area of the triangle given in Eercise 0 is eactl 0 square units. 8. The distance between the ends of the hands at four o clock is about 8.6 inches. 9. The diameter of the crater is about 5. miles. 0. About miles. N.8 W. She is about.9 miles from the lodge and her bearing to the lodge is N7 E.. It is about.50 miles from port and its heading to port is S7 W.. It is about 9.6 miles from the island and the captain should set a course of N6. E to reach the island. 5. The fires are about 756 feet apart. Tr to avoid rounding errors.

. Polar Coordinates 99. Polar Coordinates In Section., we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane. We defined the Cartesian coordinate plane using two number lines one horizontal and one vertical which intersect at right angles at a point we called the origin. To plot a point, sa P,, we start at the origin, travel horizontall to the left units, then up units. Alternativel, we could start at the origin, travel up units, then to the left units and arrive at the same location. For the most part, the motions of the Cartesian sstem over and up describe a rectangle, and most points can be thought of as the corner diagonall across the rectangle from the origin. For this reason, the Cartesian coordinates of a point are often called rectangular coordinates. In this section, we introduce a new sstem for assigning coordinates to points in the plane polar coordinates. We start with an origin point, called the pole, and a ra called the polar ais. We then locate a point P using two coordinates, r, θ, where r represents a directed distance from the pole and θ is a measure of rotation from the polar ais. Roughl speaking, the polar coordinates r, θ of a point measure how far out the point is from the pole that s r, and how far to rotate from the polar ais, that s θ. P, P r, θ r θ Pole r Polar Ais For eample, if we wished to plot the point P with polar coordinates, 5 6, we d start at the pole, move out along the polar ais units, then rotate 5 6 radians counter-clockwise. P, 5 6 r = θ = 5 6 Pole Pole Pole We ma also visualize this process b thinking of the rotation first. To plot P, 5 6 this wa, we rotate 5 6 counter-clockwise from the polar ais, then move outwards from the pole units. Ecluding, of course, the points in which one or both coordinates are 0. We will eplain more about this momentaril. As with anthing in Mathematics, the more was ou have to look at something, the better. The authors encourage the reader to take time to think about both approaches to plotting points given in polar coordinates.

90 Applications of Trigonometr Essentiall we are locating a point on the terminal side of 5 6 which is units awa from the pole. P, 5 6 θ = 5 6 θ = 5 6 Pole Pole Pole If r < 0, we begin b moving in the opposite direction on the polar ais from the pole. For eample, to plot Q.5, we have r =.5 Pole θ = Pole Pole Q.5, If we interpret the angle first, we rotate radians, then move back through the pole.5 units. Here we are locating a point.5 units awa from the pole on the terminal side of 5, not. θ = θ = Pole Pole Pole Q.5, As ou ma have guessed, θ < 0 means the rotation awa from the polar ais is clockwise instead of counter-clockwise. Hence, to plot R.5, we have the following. r =.5 Pole Pole Pole θ = R.5, From an angles first approach, we rotate R is the point on the terminal side of θ = then move out.5 units from the pole. We see that which is.5 units from the pole. Pole Pole Pole θ = θ = R.5,

. Polar Coordinates 9 The points Q and R above are, in fact, the same point despite the fact that their polar coordinate representations are different. Unlike Cartesian coordinates where a, b and c, d represent the same point if and onl if a = c and b = d, a point can be represented b infinitel man polar coordinate pairs. We eplore this notion more in the following eample. Eample... For each point in polar coordinates given below plot the point and then give two additional epressions for the point, one of which has r > 0 and the other with r < 0.. P, 0. P, 7 6. P 7, 5. P, Solution.. Whether we move units along the polar ais and then rotate 0 or rotate 0 then move out units from the pole, we plot P, 0 below. θ = 0 Pole Pole P, 0 We now set about finding alternate descriptions r, θ for the point P. Since P is units from the pole, r = ±. Net, we choose angles θ for each of the r values. The given representation for P is, 0 so the angle θ we choose for the r = case must be coterminal with 0. Can ou see wh? One such angle is θ = 0 so one answer for this case is, 0. For the case r =, we visualize our rotation starting units to the left of the pole. From this position, we need onl to rotate θ = 60 to arrive at location coterminal with 0. Hence, our answer here is, 60. We check our answers b plotting them. Pole θ = 0 θ = 60 Pole P, 0 P, 60. We plot, 7 6 b first moving units to the left of the pole and then rotating 7 6 radians. Since r = < 0, we find our point lies units from the pole on the terminal side of 6. P, 7 6 Pole Pole θ = 7 6

9 Applications of Trigonometr To find alternate descriptions for P, we note that the distance from P to the pole is units, so an representation r, θ for P must have r = ±. As we noted above, P lies on the terminal side of 6, so this, coupled with r =, gives us, 6 as one of our answers. To find a different representation for P with r =, we ma choose an angle coterminal with the angle in the original representation of P, 7 6. We pick 5 6 and get, 5 6 as our second answer. P, 6 P, 5 6 θ = 5 6 θ = 6 Pole Pole. To plot P 7, 5, we move along the polar ais 7 units from the pole and rotate clockwise 5 radians as illustrated below. Pole Pole θ = 5 P 7, 5 Since P is 7 units from the pole, an representation r, θ for P satisfies r = ±7. For the r = 7 case, we can take θ to be an angle coterminal with 5. In this case, we choose θ =, and get 7, as one answer. For the r = 7 case, we visualize moving left 7 units from the pole and then rotating through an angle θ to reach P. We find that θ = satisfies this requirement, so our second answer is 7,. Pole Pole θ = θ = P 7, P 7,

. Polar Coordinates 9. We move three units to the left of the pole and follow up with a clockwise rotation of radians to plot P,. We see that P lies on the terminal side of. P, θ = Pole Pole Since P lies on the terminal side of, one alternative representation for P is,. To find a different representation for P with r =, we ma choose an angle coterminal with. We choose θ = 7 for our final answer, 7. P, P, 7 θ = θ = 7 Pole Pole Now that we have had some practice with plotting points in polar coordinates, it should come as no surprise that an given point epressed in polar coordinates has infinitel man other representations in polar coordinates. The following result characterizes when two sets of polar coordinates determine the same point in the plane. It could be considered as a definition or a theorem, depending on our point of view. We state it as a propert of the polar coordinate sstem. Equivalent Representations of Points in Polar Coordinates Suppose r, θ and r, θ are polar coordinates where r 0, r 0 and the angles are measured in radians. Then r, θ and r, θ determine the same point P if and onl if one of the following is true: ˆ ˆ r = r and θ = θ + k for some integer k r = r and θ = θ + k + for some integer k All polar coordinates of the form 0, θ represent the pole regardless of the value of θ. The ke to understanding this result, and indeed the whole polar coordinate sstem, is to keep in mind that r, θ means directed distance from pole, angle of rotation. If r = 0, then no matter how much rotation is performed, the point never leaves the pole. Thus 0, θ is the pole for all

9 Applications of Trigonometr values of θ. Now let s assume that neither r nor r is zero. If r, θ and r, θ determine the same point P then the non-zero distance from P to the pole in each case must be the same. Since this distance is controlled b the first coordinate, we have that either r = r or r = r. If r = r, then when plotting r, θ and r, θ, the angles θ and θ have the same initial side. Hence, if r, θ and r, θ determine the same point, we must have that θ is coterminal with θ. We know that this means θ = θ + k for some integer k, as required. If, on the other hand, r = r, then when plotting r, θ and r, θ, the initial side of θ is rotated radians awa from the initial side of θ. In this case, θ must be coterminal with + θ. Hence, θ = + θ + k which we rewrite as θ = θ +k + for some integer k. Conversel, if r = r and θ = θ +k for some integer k, then the points P r, θ and P r, θ lie the same directed distance from the pole on the terminal sides of coterminal angles, and hence are the same point. Now suppose r = r and θ = θ + k + for some integer k. To plot P, we first move a directed distance r from the pole; to plot P, our first step is to move the same distance from the pole as P, but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated eactl radians apart. Since θ = θ + k + = θ + + k for some integer k, we see that θ is coterminal to θ + and it is this etra radians of rotation which aligns the points P and P. Net, we marr the polar coordinate sstem with the Cartesian rectangular coordinate sstem. To do so, we identif the pole and polar ais in the polar sstem to the origin and positive -ais, respectivel, in the rectangular sstem. We get the following result. Theorem.7. Conversion Between Rectangular and Polar Coordinates: Suppose P is represented in rectangular coordinates as, and in polar coordinates as r, θ. Then ˆ = r cosθ and = r sinθ ˆ + = r and tanθ = provided 0 In the case r > 0, Theorem.7 is an immediate consequence of Theorem 0. along with the quotient identit tanθ = sinθ cosθ. If r < 0, then we know an alternate representation for r, θ is r, θ +. Since cosθ + = cosθ and sinθ + = sinθ, appling the theorem to r, θ + gives = r cosθ + = r cosθ = r cosθ and = r sinθ + = r sinθ = r sinθ. Moreover, + = r = r, and = tanθ + = tanθ, so the theorem is true in this case, too. The remaining case is r = 0, in which case r, θ = 0, θ is the pole. Since the pole is identified with the origin 0, 0 in rectangular coordinates, the theorem in this case amounts to checking 0 = 0. The following eample puts Theorem.7 to good use. Eample... Convert each point in rectangular coordinates given below into polar coordinates with r 0 and 0 θ <. Use eact values if possible and round an approimate values to two decimal places. Check our answer b converting them back to rectangular coordinates.. P,. Q,. R0,. S,

. Polar Coordinates 95 Solution.. Even though we are not eplicitl told to do so, we can avoid man common mistakes b taking the time to plot the points before we do an calculations. Plotting P, shows that it lies in Quadrant IV. With = and =, we get r = + = + = + = 6 so r = ±. Since we are asked for r 0, we choose r =. To find θ, we have that tanθ = = =. This tells us θ has a reference angle of, and since P lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have 0 θ <, so we choose θ = 5. Hence, our answer is, 5. To check, we convert r, θ =, 5 back to rectangular coordinates and we find = r cosθ = cos 5 = = and = r sinθ = sin 5 = =, as required.. The point Q, lies in Quadrant III. Using = =, we get r = + = 8 so r = ± 8 = ±. Since we are asked for r 0, we choose r =. We find tanθ = =, which means θ has a reference angle of. Since Q lies in Quadrant III, we choose θ = 5, which satisfies the requirement that 0 θ <. Our final answer is r, θ =, 5. To check, we find = r cosθ = cos 5 = = and = r sinθ = sin 5 = =, so we are done. θ = 5 θ = 5 P Q P has rectangular coordinates, Q has rectangular coordinates, P has polar coordinates, 5 Q has polar coordinates, 5. The point R0, lies along the negative -ais. While we could go through the usual computations to find the polar form of R, in this case we can find the polar coordinates of R using the definition. Since the pole is identified with the origin, we can easil tell the point R is units from the pole, which means in the polar representation r, θ of R we know r = ±. Since we require r 0, we choose r =. Concerning θ, the angle θ = satisfies 0 θ < Since = 0, we would have to determine θ geometricall.

96 Applications of Trigonometr with its terminal side along the negative -ais, so our answer is,. To check, we note = r cosθ = cos = 0 = 0 and = r sinθ = sin = =.. The point S, lies in Quadrant II. With = and =, we get r = + = 5 so r = ±5. As usual, we choose r = 5 0 and proceed to determine θ. We have tanθ = = =, and since this isn t the tangent of one the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisf 0 θ <, we choose θ = arctan radians. Hence, our answer is r, θ = 5, arctan 5,.. To check our answers requires a bit of tenacit since we need to simplif epressions of the form: cos arctan and sin arctan. These are good review eercises and are hence left to the reader. We find cos arctan = 5 and sin arctan = 5, so that = r cosθ = 5 5 = and = r sinθ = 5 5 = which confirms our answer. S θ = θ = arctan R R has rectangular coordinates 0, S has rectangular coordinates, R has polar coordinates, S has polar coordinates 5, arctan Now that we ve had practice converting representations of points between the rectangular and polar coordinate sstems, we now set about converting equations from one sstem to another. Just as we ve used equations in and to represent relations in rectangular coordinates, equations in the variables r and θ represent relations in polar coordinates. We convert equations between the two sstems using Theorem.7 as the net eample illustrates. Eample.... Convert each equation in rectangular coordinates into an equation in polar coordinates. a + = 9 b = c =. Convert each equation in polar coordinates into an equation in rectangular coordinates. a r = b θ = c r = cosθ

. Polar Coordinates 97 Solution.. One strateg to convert an equation from rectangular to polar coordinates is to replace ever occurrence of with r cosθ and ever occurrence of with r sinθ and use identities to simplif. This is the technique we emplo below. a We start b substituting = r cosθ and = sinθ into + = 9 and simplifing. With no real direction in which to proceed, we follow our mathematical instincts and see where the take us. 5 r cosθ + r sinθ = 9 r cos θ 6r cosθ + 9 + r sin θ = 9 r cos θ + sin θ 6r cosθ = 0 Subtract 9 from both sides. r 6r cosθ = 0 Since cos θ + sin θ = rr 6 cosθ = 0 Factor. We get r = 0 or r = 6 cosθ. From Section 7. we know the equation + = 9 describes a circle, and since r = 0 describes just a point namel the pole/origin, we choose r = 6 cosθ for our final answer. 6 b Substituting = r cosθ and = r sinθ into = gives r sinθ = r cosθ. Rearranging, we get r cosθ + r sinθ = 0 or rcosθ + sinθ = 0. This gives r = 0 or cosθ + sinθ = 0. Solving the latter equation for θ, we get θ = + k for integers k. As we did in the previous eample, we take a step back and think geometricall. We know = describes a line through the origin. As before, r = 0 describes the origin, but nothing else. Consider the equation θ =. In this equation, the variable r is free, 7 meaning it can assume an and all values including r = 0. If we imagine plotting points r, for all conceivable values of r positive, negative and zero, we are essentiall drawing the line containing the terminal side of θ = which is none other than =. Hence, we can take as our final answer θ = here.8 c We substitute = r cosθ and = r sinθ into = and get r sinθ = r cosθ, or r cos θ r sinθ = 0. Factoring, we get rr cos θ sinθ = 0 so that either r = 0 or r cos θ = sinθ. We can solve the latter equation for r b dividing both sides of the equation b cos θ, but as a general rule, we never divide through b a quantit that ma be 0. In this particular case, we are safe since if cos θ = 0, then cosθ = 0, and for the equation r cos θ = sinθ to hold, then sinθ would also have to be 0. Since there are no angles with both cosθ = 0 and sinθ = 0, we are not losing an 5 Eperience is the mother of all instinct, and necessit is the mother of invention. Stud this eample and see what techniques are emploed, then tr our best to get our answers in the homework to match Jeff s. 6 Note that when we substitute θ = into r = 6 cosθ, we recover the point r = 0, so we aren t losing anthing b disregarding r = 0. 7 See Section 8.. 8 We could take it to be an of θ = + k for integers k.

98 Applications of Trigonometr information b dividing both sides of r cos θ = sinθ b cos θ. Doing so, we get r = sinθ, or r = secθ tanθ. As before, the r = 0 case is recovered in the solution cos θ r = secθ tanθ let θ = 0, so we state the latter as our final answer.. As a general rule, converting equations from polar to rectangular coordinates isn t as straight forward as the reverse process. We could solve r = + for r to get r = ± + and solving tanθ = requires the arctangent function to get θ = arctan + k for integers k. Neither of these epressions for r and θ are especiall user-friendl, so we opt for a second strateg rearrange the given polar equation so that the epressions r = +, r cosθ =, r sinθ = and/or tanθ = present themselves. a Starting with r =, we can square both sides to get r = or r = 9. We ma now substitute r = + to get the equation + = 9. As we have seen, 9 squaring an equation does not, in general, produce an equivalent equation. The concern here is that the equation r = 9 might be satisfied b more points than r =. On the surface, this appears to be the case since r = 9 is equivalent to r = ±, not just r =. However, an point with polar coordinates, θ can be represented as, θ +, which means an point r, θ whose polar coordinates satisf the relation r = ± has an equivalent 0 representation which satisfies r =. to get tanθ = tan =. b We take the tangent of both sides the equation θ = Since tanθ =, we get = or =. Of course, we pause a moment to wonder if, geometricall, the equations θ = and = generate the same set of points. The same argument presented in number b applies equall well here so we are done. c Once again, we need to manipulate r = cosθ a bit before using the conversion formulas given in Theorem.7. We could square both sides of this equation like we did in part a above to obtain an r on the left hand side, but that does nothing helpful for the right hand side. Instead, we multipl both sides b r to obtain r = r r cosθ. We now have an r and an r cosθ in the equation, which we can easil handle, but we also have another r to deal with. Rewriting the equation as r = r + r cosθ and squaring both sides ields r = r + r cosθ. Substituting r = + and r cosθ = gives + = + +. Once again, we have performed some 9 Eercise 5.. in Section 5., for instance... 0 Here, equivalent means the represent the same point in the plane. As ordered pairs,, 0 and, are different, but when interpreted as polar coordinates, the correspond to the same point in the plane. Mathematicall speaking, relations are sets of ordered pairs, so the equations r = 9 and r = represent different relations since the correspond to different sets of ordered pairs. Since polar coordinates were defined geometricall to describe the location of points in the plane, however, we concern ourselves onl with ensuring that the sets of points in the plane generated b two equations are the same. This was not an issue, b the wa, when we first defined relations as sets of points in the plane in Section.. Back then, a point in the plane was identified with a unique ordered pair given b its Cartesian coordinates. In addition to taking the tangent of both sides of an equation There are infinitel man solutions to tanθ =, and θ = is onl one of them!, we also went from =, in which cannot be 0, to = in which we assume can be 0.

. Polar Coordinates 99 algebraic maneuvers which ma have altered the set of points described b the original equation. First, we multiplied both sides b r. This means that now r = 0 is a viable solution to the equation. In the original equation, r = cosθ, we see that θ = 0 gives r = 0, so the multiplication b r doesn t introduce an new points. The squaring of both sides of this equation is also a reason to pause. Are there points with coordinates r, θ which satisf r = r + r cosθ but do not satisf r = r + r cosθ? Suppose r, θ satisfies r = r + r cosθ. Then r = ± r + r cosθ. If we have that r = r +r cosθ, we are done. What if r = r + r cosθ = r r cosθ? We claim that the coordinates r, θ +, which determine the same point as r, θ, satisf r = r + r cosθ. We substitute r = r and θ = θ + into r = r + r cosθ to see if we get a true statement. r? = r + r cosθ + r r cosθ? = r r cosθ + Since r = r r cosθ r + r cosθ r + r cosθ? = r r cosθ Since cosθ + = cosθ = r + r cosθ Since both sides worked out to be equal, r, θ + satisfies r = r + r cosθ which means that an point r, θ which satisfies r = r + r cosθ has a representation which satisfies r = r + r cosθ, and we are done. In practice, much of the pedantic verification of the equivalence of equations in Eample.. is left unsaid. Indeed, in most tetbooks, squaring equations like r = to arrive at r = 9 happens without a second thought. Your instructor will ultimatel decide how much, if an, justification is warranted. If ou take anthing awa from Eample.., it should be that relativel nice things in rectangular coordinates, such as =, can turn ugl in polar coordinates, and vice-versa. In the net section, we devote our attention to graphing equations like the ones given in Eample.. number on the Cartesian coordinate plane without converting back to rectangular coordinates. If nothing else, number c above shows the price we pa if we insist on alwas converting to back to the more familiar rectangular coordinate sstem.

90 Applications of Trigonometr.. Eercises In Eercises - 6, plot the point given in polar coordinates and then give three different epressions for the point such that a r < 0 and 0 θ, b r > 0 and θ 0 c r > 0 and θ.,. 5, 7., 5., 5 6 5., 7 6., 5 7., 7 8. 6, 6 9. 0, 0., 5.,.,.,..5, 5. 5, 6., 6 In Eercises 7-6, convert the point from polar coordinates into rectangular coordinates. 7. 5, 7 8., 9., 7 0. 0, 6. 5,., 5. 9, 7. 5, 9 6 5., 6. 7, 7 7. 6, arctan 8. 0, arctan 6 9., arctan 0. 5, arctan., arctan., arctan 5., + arctan., + arctan 5., arctan 6., arctan 5 In Eercises 7-56, convert the point from rectangular coordinates into polar coordinates with r 0 and 0 θ <. 7. 0, 5 8., 9. 7, 7 0.,., 0.,.,.,

. Polar Coordinates 9 5. 0, 6. 5, 5 7. 6, 8 8. 5, 5 0 9. 8, 50. 0, 6 5 5 0 5. 5, 5. 5, 5 6 65 5., 7 5., 9 55., 56. 5, 65 5 In Eercises 57-76, convert the equation from rectangular coordinates into polar coordinates. Solve for r in all but #60 through #6. In Eercises 60-6, ou need to solve for θ 57. = 6 58. = 59. = 7 60. = 0 6. = 6. = 6. = 6. + = 5 65. + = 7 66. = 9 67. = + 68. = 69. = 70. + = 0 7. + = 0 7. + = 7. = 7 7. + + = 75. + = 9 76. + = In Eercises 77-96, convert the equation from polar coordinates into rectangular coordinates. 77. r = 7 78. r = 79. r = 80. θ = 8. θ = 8. θ = 8. θ = 8. r = cosθ 85. 5r = cosθ 86. r = sinθ 87. r = sinθ 88. r = 7 secθ 89. r = cscθ 90. r = secθ 9. r = 5 cscθ 9. r = secθ tanθ 9. r = cscθ cotθ 9. r = sinθ 95. r = cosθ 96. r = + sinθ 97. Convert the origin 0, 0 into polar coordinates in four different was. 98. With the help of our classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates.

9 Applications of Trigonometr.. Answers.,,,, 5,, 7. 5, 7, 5, 5,, 5, 5.,,,,,, 7. 5, 5, 5 6, 6 5 5, 7, 6, 7 6

. Polar Coordinates 9 5., 7,, 6 6, 9,, 7 6 6 6 9 6 6., 5,, 7,,, 7.,,, 0,,, 8. 7,, 7 6, 5 6 7 7,, 6, 6

9 Applications of Trigonometr 9. 0,, 0, 0,, 0, 0 0 0 0 0., 5,,, 7,, 9.,,, 8,,,.,,, 5,,, 7

. Polar Coordinates 95.,,, 6 6, 5,, 9 6 6..5,,.5, 7.5, 5,.5, 5. 5,, 5, 5, 5,, 6.,,,,,,

96 Applications of Trigonometr 5 7., 5. 0, 5 8., 9.,.,. 0, 9. 5., 6. 7, 0 7. 9.. 7. 9 5, 5 5, 5 5, 0.,.. 8..,. 5. 5, 6. 65, 9. arctan 8 5., + arctan 5 7 5. 5, arctan 55., 9, 9, 6, 0, 5 5. 9.. 7. 6 5 5, 5 5 5 5, 5 5, + + 7, 7 8, 0, arctan 50. 0, arctan 5. 5., + arctan 5, arctan 0. 0, 0 5, 5 8. 0, 0. 6 6 5, 5 5 6. 5, 0.. 56., arctan, 7 6, 6 8. 5, arctan 57. r = 6 secθ 58. r = secθ 59. r = 7 cscθ 60. θ = 0 6. θ = 6. θ = 6. θ = arctan 6. r = 5 65. r = 7 66. r = 9 cosθ sinθ 67. = cosθ sinθ 68. r = secθ tanθ 69. r = cscθ cotθ 70. r = sinθ 7. r = cosθ 7. r = cosθ

. Polar Coordinates 97 7. r = 7 sinθ 7. r = cosθ 75. r = 6 sinθ 76. r = sinθ 77. + = 9 78. + = 9 79. + = 80. = 8. = 8. = 0 8. = 0 8. + = or + = 85. 5 + 5 = or + = 0 00 87. + = or + + = 88. = 7 86. + = or + = 9 89. = 90. = 9. = 5 9. = 9. = 9. + = 95. + + = + 96. + + = + 97. An point of the form 0, θ will work, e.g. 0,, 0, 7, 0, and 0, 0.

98 Applications of Trigonometr.5 Graphs of Polar Equations In this section, we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. Since an given point in the plane has infinitel man different representations in polar coordinates, our Fundamental Graphing Principle in this section is not as clean as it was for graphs of rectangular equations on page. We state it below for completeness. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points which satisf the equation. That is, a point P r, θ is on the graph of an equation if and onl if there is a representation of P, sa r, θ, such that r and θ satisf the equation. Our first eample focuses on some of the more structurall simple polar equations. Eample.5.. Graph the following polar equations.. r =. r =. θ = 5. θ = Solution. In each of these equations, onl one of the variables r and θ is present making the other variable free. This makes these graphs easier to visualize than others.. In the equation r =, θ is free. The graph of this equation is, therefore, all points which have a polar coordinate representation, θ, for an choice of θ. Graphicall this translates into tracing out all of the points units awa from the origin. This is eactl the definition of circle, centered at the origin, with a radius of. θ > 0 θ < 0 In r =, θ is free The graph of r =. Once again we have θ being free in the equation r =. Plotting all of the points of the form, θ gives us a circle of radius centered at the origin. See the discussion in Eample.. number a.

.5 Graphs of Polar Equations 99 θ < 0 θ > 0 In r =, θ is free The graph of r =. In the equation θ = 5, r is free, so we plot all of the points with polar representation r, 5. What we find is that we are tracing out the line which contains the terminal side of θ = 5 when plotted in standard position. r < 0 θ = 5 r = 0 r > 0 In θ = 5, r is free The graph of θ = 5. As in the previous eample, the variable r is free in the equation θ =. Plotting r, for various values of r shows us that we are tracing out the -ais.

90 Applications of Trigonometr r > 0 r = 0 θ = r < 0 In θ =, r is free The graph of θ = Hopefull, our eperience in Eample.5. makes the following result clear. Theorem.8. Graphs of Constant r and θ: Suppose a and α are constants, a 0. ˆ ˆ The graph of the polar equation r = a on the Cartesian plane is a circle centered at the origin of radius a. The graph of the polar equation θ = α on the Cartesian plane is the line containing the terminal side of α when plotted in standard position. Suppose we wish to graph r = 6 cosθ. A reasonable wa to start is to treat θ as the independent variable, r as the dependent variable, evaluate r = fθ at some friendl values of θ and plot the resulting points. We generate the table below. θ r = 6 cosθ r, θ 0 6 6, 0, 0 0,, 6 6, 5, 5 0 0, 7, 7 6 6, 6 For a review of these concepts and this process, see Sections. and.6.

.5 Graphs of Polar Equations 9 Despite having nine ordered pairs, we get onl four distinct points on the graph. For this reason, we emplo a slightl different strateg. We graph one ccle of r = 6 cosθ on the θr-plane and use it to help graph the equation on the -plane. We see that as θ ranges from 0 to, r ranges from 6 to 0. In the -plane, this means that the curve starts 6 units from the origin on the positive -ais θ = 0 and graduall returns to the origin b the time the curve reaches the -ais θ =. The arrows drawn in the figure below are meant to help ou visualize this process. In the θr-plane, the arrows are drawn from the θ-ais to the curve r = 6 cosθ. In the -plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise wa to generate the graph than computing the actual function values, but it is markedl faster. 6 r θ runs from 0 to θ 6 Net, we repeat the process as θ ranges from to. Here, the r values are all negative. This means that in the -plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative -ais. 6 r θ runs from to θ 6 r < 0 so we plot here As θ ranges from to, the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the r for these values of θ match the r values for θ in The graph looks eactl like = 6 cos in the -plane, and for good reason. At this stage, we are just graphing the relationship between r and θ before we interpret them as polar coordinates r, θ on the -plane.

9 Applications of Trigonometr [ ] 0,, we have that the curve begins to retrace itself at this point. Proceeding further, we find that when θ, we retrace the portion of the curve in Quadrant IV that we first traced out as θ. The reader is invited to verif that plotting an range of θ outside the interval [0, ] results in retracting some portion of the curve. We present the final graph below. 6 r θ 6 6 r = 6 cosθ in the θr-plane r = 6 cosθ in the -plane Eample.5.. Graph the following polar equations.. r = sinθ. r = + cosθ. r = 5 sinθ. r = 6 cosθ Solution.. We first plot the fundamental ccle of r = sinθ on the θr-aes. To help us visualize what is going on graphicall, we divide up [0, ] into the usual four subintervals [ 0, ] [, [ ] [, ],, and, ], and proceed as we did above. As θ ranges from 0 to, r decreases from to. This means that the curve in the -plane starts units from the origin on the positive -ais and graduall pulls in towards the origin as it moves towards the positive -ais. r 6 θ runs from 0 to θ The graph of r = 6 cosθ looks suspiciousl like a circle, for good reason. See number a in Eample...

.5 Graphs of Polar Equations 9 Net, as θ runs from to, we see that r increases from to. Picking up where we left off, we graduall pull the graph awa from the origin until we reach the negative -ais. 6 r θ runs from to θ Over the interval [, ], we see that r increases from to 6. On the -plane, the curve sweeps out awa from the origin as it travels from the negative -ais to the negative -ais. r 6 θ θ runs from to Finall, as θ takes on values from to, r decreases from 6 back to. The graph on the -plane pulls in from the negative -ais to finish where we started. r 6 θ θ runs from to We leave it to the reader to verif that plotting points corresponding to values of θ outside the interval [0, ] results in retracing portions of the curve, so we are finished.

9 Applications of Trigonometr r 6 r = sinθ in the θr-plane θ 6 r = sinθ in the -plane.. The first thing to note when graphing r = + cosθ on the θr-plane over the interval [0, ] is that the graph crosses through the θ-ais. This corresponds to the graph of the curve passing through the origin in the -plane, and our first task is to determine when this happens. Setting r = 0 we get + cosθ = 0, or cosθ =. Solving for θ in [0, ] gives θ = and θ =. Since these values of θ are important geometricall, we break the interval [0, ] into si subintervals: [ 0, ] [,, ] [,, ], [, ] [,, ] [ and, ]. As θ ranges from 0 to, r decreases from 6 to. Plotting this on the -plane, we start 6 units out from the origin on the positive -ais and slowl pull in towards the positive -ais. 6 r θ runs from 0 to θ On the interval [, ], r decreases from to 0, which means the graph is heading into and will eventuall cross through the origin. Not onl do we reach the origin when θ =, a theorem from Calculus 5 states that the curve hugs the line θ = as it approaches the origin. 5 The tangents at the pole theorem from second semester Calculus.

.5 Graphs of Polar Equations 95 6 r θ = θ On the interval [, ], r ranges from 0 to. Since r 0, the curve passes through the origin in the -plane, following the line θ = and continues upwards through Quadrant IV towards the positive -ais. 6 Since r is increasing from 0 to, the curve pulls awa from the origin to finish at a point on the positive -ais. 6 r θ = θ Net, as θ progresses from to, r ranges from to 0. Since r 0, we continue our graph in the first quadrant, heading into the origin along the line θ =. 6 Recall that one wa to visualize plotting polar coordinates r, θ with r < 0 is to start the rotation from the left side of the pole - in this case, the negative -ais. Rotating between and radians from the negative -ais in this case determines the region between the line θ = and the -ais in Quadrant IV.

96 Applications of Trigonometr 6 r θ = θ On the interval [, ], r returns to positive values and increases from 0 to. We hug the line θ = as we move through the origin and head towards the negative -ais. 6 r θ = θ As we round out the interval, we find that as θ runs through to, r increases from out to 6, and we end up back where we started, 6 units from the origin on the positive -ais. 6 r θ θ runs from to

.5 Graphs of Polar Equations 97 Again, we invite the reader to show that plotting the curve for values of θ outside [0, ] results in retracing a portion of the curve alread traced. Our final graph is below. 6 r θ = θ θ = 6 r = + cosθ in the θr-plane r = + cosθ in the -plane. As usual, we start b graphing a fundamental ccle of r = 5 sinθ in the θr-plane, which in this case, occurs as θ ranges from 0 to. We partition our interval into subintervals to help us with the graphing, namel [ 0, ] [,, ] [,, ] [ and, ]. As θ ranges from 0 to, r increases from 0 to 5. This means that the graph of r = 5 sinθ in the -plane starts at the origin and graduall sweeps out so it is 5 units awa from the origin on the line θ =. 5 r θ 5 Net, we see that r decreases from 5 to 0 as θ runs through [, ], and furthermore, r is heading negative as θ crosses. Hence, we draw the curve hugging the line θ = the -ais as the curve heads to the origin.

98 Applications of Trigonometr 5 r θ 5 As θ runs from to, r becomes negative and ranges from 0 to 5. Since r 0, the curve pulls awa from the negative -ais into Quadrant IV. 5 r θ 5 For θ, r increases from 5 to 0, so the curve pulls back to the origin. 5 r θ 5

.5 Graphs of Polar Equations 99 Even though we have finished with one complete ccle of r = 5 sinθ, if we continue plotting beond θ =, we find that the curve continues into the third quadrant! Below we present a graph of a second ccle of r = 5 sinθ which continues on from the first. The boed labels on the θ-ais correspond to the portions with matching labels on the curve in the -plane. 5 r 5 7 θ 5 We have the final graph below. 5 r 5 5 7 θ 5 5 5 5 r = 5 sinθ in the θr-plane r = 5 sinθ in the -plane. Graphing r = 6 cosθ is complicated b the r, so we solve to get r = ± 6 cosθ = ± cosθ. How do we sketch such a curve? First off, we sketch a fundamental period of r = cosθ which we have dotted in the figure below. When cosθ < 0, cosθ is undefined, so we don t have an values on the interval,. On the intervals which remain, cosθ ranges from 0 to, inclusive. Hence, cosθ ranges from 0 to as well. 7 From this, we know r = ± cosθ ranges continuousl from 0 to ±, respectivel. Below we graph both r = cosθ and r = cosθ on the θr plane and use them to sketch the corresponding pieces of the curve r = 6 cosθ in the -plane. As we have seen in earlier 7 Owing to the relationship between = and = over [0, ], we also know cosθ cosθ wherever the former is defined.

950 Applications of Trigonometr eamples, the lines θ = and θ =, which are the zeros of the functions r = ± cosθ, serve as guides for us to draw the curve as is passes through the origin. r θ = θ = θ r = cosθ and r = cosθ As we plot points corresponding to values of θ outside of the interval [0, ], we find ourselves retracing parts of the curve, 8 so our final answer is below. r θ = θ = θ r = ± cosθ in the θr-plane r = 6 cosθ in the -plane A few remarks are in order. First, there is no relation, in general, between the period of the function fθ and the length of the interval required to sketch the complete graph of r = fθ in the plane. As we saw on page 9, despite the fact that the period of fθ = 6 cosθ is, we sketched the complete graph of r = 6 cosθ in the -plane just using the values of θ as θ ranged from 0 to. In Eample.5., number, the period of fθ = 5 sinθ is, but in order to obtain the complete graph of r = 5 sinθ, we needed to run θ from 0 to. While man of the common polar graphs can be grouped into families, 9 the authors trul feel that taking the time to work through each graph in the manner presented here is the best wa to not onl understand the polar 8 In this case, we could have generated the entire graph b using just the plot r = cosθ, but graphed over the interval [0, ] in the θr-plane. We leave the details to the reader. 9 Numbers and in Eample.5. are eamples of limaçons, number is an eample of a polar rose, and number is the famous Lemniscate of Bernoulli.

.5 Graphs of Polar Equations 95 coordinate sstem, but also prepare ou for what is needed in Calculus. Second, the smmetr seen in the eamples is also a common occurrence when graphing polar equations. In addition to the usual kinds of smmetr discussed up to this point in the tet smmetr about each ais and the origin, it is possible to talk about rotational smmetr. We leave the discussion of smmetr to the Eercises. In our net eample, we are given the task of finding the intersection points of polar curves. According to the Fundamental Graphing Principle for Polar Equations on page 98, in order for a point P to be on the graph of a polar equation, it must have a representation P r, θ which satisfies the equation. What complicates matters in polar coordinates is that an given point has infinitel man representations. As a result, if a point P is on the graph of two different polar equations, it is entirel possible that the representation P r, θ which satisfies one of the equations does not satisf the other equation. Here, more than ever, we need to rel on the Geometr as much as the Algebra to find our solutions. Eample.5.. Find the points of intersection of the graphs of the following polar equations.. r = sinθ and r = sinθ. r = and r = cosθ. r = and r = 6 cosθ. r = sin θ and r = cos θ Solution.. Following the procedure in Eample.5., we graph r = sinθ and find it to be a circle centered at the point with rectangular coordinates 0, with a radius of. The graph of r = sinθ is a special kind of limaçon called a cardioid. 0 r = sinθ and r = sinθ It appears as if there are three intersection points: one in the first quadrant, one in the second quadrant, and the origin. Our net task is to find polar representations of these points. In 0 Presumabl, the name is derived from its resemblance to a stlized human heart.

95 Applications of Trigonometr order for a point P to be on the graph of r = sinθ, it must have a representation P r, θ which satisfies r = sinθ. If P is also on the graph of r = sinθ, then P has a possibl different representation P r, θ which satisfies r = sinθ. We first tr to see if we can find an points which have a single representation P r, θ that satisfies both r = sinθ and r = sinθ. Assuming such a pair r, θ eists, then equating the epressions for r gives sinθ = sinθ or sinθ =. From this, we get θ = 6 + k or θ = 5 6 + k for integers k. Plugging θ = 6 into r = sinθ, we get r = sin 6 = =, which is also the value we obtain when we substitute it into r = sinθ. Hence,, 6 is one representation for the point of intersection in the first quadrant. For the point of intersection in the second quadrant, we tr θ = 5 6. Both equations give us the point, 5 6, so this is our answer here. What about the origin? We know from Section. that the pole ma be represented as 0, θ for an angle θ. On the graph of r = sinθ, we start at the origin when θ = 0 and return to it at θ =, and as the reader can verif, we are at the origin eactl when θ = k for integers k. On the curve r = sinθ, however, we reach the origin when θ =, and more generall, when θ = + k for integers k. There is no integer value of k for which k = + k which means while the origin is on both graphs, the point is never reached simultaneousl. In an case, we have determined the three points of intersection to be, 6,, 5 6 and the origin.. As before, we make a quick sketch of r = and r = cosθ to get feel for the number and location of the intersection points. The graph of r = is a circle, centered at the origin, with a radius of. The graph of r = cosθ is also a circle - but this one is centered at the point with rectangular coordinates, 0 and has a radius of. r = and r = cosθ We have two intersection points to find, one in Quadrant I and one in Quadrant IV. Proceeding as above, we first determine if an of the intersection points P have a representation r, θ which satisfies both r = and r = cosθ. Equating these two epressions for r, we get cosθ =. To solve this equation, we need the arccosine function. We get We are reall using the technique of substitution to solve the sstem of equations { r = sinθ r = sinθ

.5 Graphs of Polar Equations 95 θ = arccos + k or θ = arccos + k for integers k. From these solutions, we get, arccos as one representation for our answer in Quadrant I, and, arccos as one representation for our answer in Quadrant IV. The reader is encouraged to check these results algebraicall and geometricall.. Proceeding as above, we first graph r = and r = 6 cosθ to get an idea of how man intersection points to epect and where the lie. The graph of r = is a circle centered at the origin with a radius of and the graph of r = 6 cosθ is another four-leafed rose. 6 6 6 6 r = and r = 6 cosθ It appears as if there are eight points of intersection - two in each quadrant. We first look to see if there an points P r, θ with a representation that satisfies both r = and r = 6 cosθ. For these points, 6 cosθ = or cosθ =. Solving, we get θ = 6 + k or θ = 5 6 + k for integers k. Out of all of these solutions, we obtain just four distinct points represented b, 6,, 5 6,, 7 6 and, 6. To determine the coordinates of the remaining four points, we have to consider how the representations of the points of intersection can differ. We know from Section. that if r, θ and r, θ represent the same point and r 0, then either r = r or r = r. If r = r, then θ = θ+k, so one possibilit is that an intersection point P has a representation r, θ which satisfies r = and another representation r, θ+k for some integer, k which satisfies r = 6 cosθ. At this point, if we replace ever occurrence of θ in the equation r = 6 cosθ with θ+k and then see if, b equating the resulting epressions for r, we get an more solutions for θ. Since cosθ + k = cosθ + k = cosθ for ever integer k, however, the equation r = 6 cosθ + k reduces to the same equation we had before, r = 6 cosθ, which means we get no additional solutions. Moving on to the case where r = r, we have that θ = θ + k + for integers k. We look to see if we can find points P which have a representation r, θ that satisfies r = and another, See Eample.5. number. The authors have chosen to replace θ with θ +k in the equation r = 6 cosθ for illustration purposes onl. We could have just as easil chosen to do this substitution in the equation r =. Since there is no θ in r =, however, this case would reduce to the previous case instantl. The reader is encouraged to follow this latter procedure in the interests of efficienc.

95 Applications of Trigonometr r, θ + k +, that satisfies r = 6 cosθ. To do this, we substitute r for r and θ + k + for θ in the equation r = 6 cosθ and get r = 6 cosθ + k +. Since cosθ + k + = cosθ + k + = cosθ for all integers k, the equation r = 6 cosθ + k + reduces to r = 6 cosθ, or r = 6 cosθ. Coupling this equation with r = gives 6 cosθ = or cosθ =. We get θ = +k or θ = +k. From these solutions, we obtain 5 the remaining four intersection points with representations,,,,, and, 5, which we can readil check graphicall.. As usual, we begin b graphing r = sin θ and r = cos θ. Using the techniques presented in Eample.5., we find that we need to plot both functions as θ ranges from 0 to to obtain the complete graph. To our surprise and/or delight, it appears as if these two equations describe the same curve! r = sin θ and r = cos θ appear to determine the same curve in the -plane To verif this incredible claim, 6 we need to show that, in fact, the graphs of these two equations intersect at all points on the plane. Suppose P has a representation r, θ which satisfies both r = sin θ and r = cos θ. Equating these two epressions for r gives the equation sin θ = cos θ. While normall we discourage dividing b a variable epression in case it could be 0, we note here that if cos θ = 0, then for our equation to hold, sin θ = 0 as well. Since no angles have both cosine and sine equal to zero, we are safe to divide both sides of the equation sin θ = cos θ b cos θ to get tan θ = which gives θ = + k for integers k. From these solutions, however, we Again, we could have easil chosen to substitute these into r = which would give r =, or r =. 5 We obtain these representations b substituting the values for θ into r = 6 cosθ, once again, for illustration purposes. Again, in the interests of efficienc, we could plug these values for θ into r = where there is no θ and,,,, and, 5. While it is not true that, represents the same point get the list of points:, as,, we still get the same set of solutions. 6 A quick sketch of r = sin θ and r = cos θ in the θr-plane will convince ou that, viewed as functions of r, these are two different animals.

.5 Graphs of Polar Equations 955 get onl one intersection point which can be represented b,. We now investigate other representations for the intersection points. Suppose P is an intersection point with a representation r, θ which satisfies r = sin θ and the same point P has a different representation r, θ + k for some integer k which satisfies r = cos θ. Substituting into the latter, we get r = cos [θ + k] = cos θ + k. Using the sum formula for cosine, we epand cos θ + k = cos θ cosk sin θ sin k = ± cos θ, since sink = 0 for all integers k, and cos k = ± for all integers k. If k is an even integer, we get the same equation r = cos θ as before. If k is odd, we get r = cos θ. This latter epression for r leads to the equation sin θ = cos θ, or tan θ =. Solving, we get θ = + k for integers k, which gives the intersection point,. Net, we assume P has a representation r, θ which satisfies r = sin θ and a representation r, θ + k + which satisfies r = cos θ for some integer k. Substituting r for r and θ + k + in for θ into r = cos θ gives r = cos [θ + k + ]. Once again, we use the sum formula for cosine to get cos [θ + k + ] = cos θ + k+ = cos θ cos k+ = ± sin θ where the last equalit is true since cos k+ = 0 and sin Hence, r = cos [θ + k + ] can be rewritten as r = ± sin θ then sin k+ sin θ sin k+ k+ = ± for integers k.. If we choose k = 0, = sin =, and the equation r = cos [θ + k + ] in this case reduces to r = sin θ, or r = sin θ which is the other equation under consideration! What this means is that if a polar representation r, θ for the point P satisfies r = sin θ, then the representation r, θ + for P automaticall satisfies r = cos θ. Hence the equations r = sin θ and r = cos θ determine the same set of points in the plane. Our work in Eample.5. justifies the following. Guidelines for Finding Points of Intersection of Graphs of Polar Equations To find the points of intersection of the graphs of two polar equations E and E : ˆ Sketch the graphs of E and E. Check to see if the curves intersect at the origin pole. ˆ Solve for pairs r, θ which satisf both E and E. ˆ Substitute θ + k for θ in either one of E or E but not both and solve for pairs r, θ which satisf both equations. Keep in mind that k is an integer. ˆ Substitute r for r and θ + k + for θ in either one of E or E but not both and solve for pairs r, θ which satisf both equations. Keep in mind that k is an integer.

956 Applications of Trigonometr Our last eample ties together graphing and points of intersection to describe regions in the plane. Eample.5.. Sketch the region in the -plane described b the following sets.. { r, θ 0 r 5 sinθ, 0 θ }. { r, θ r 6 cosθ, 0 θ } 6. { r, θ + cosθ r 0, θ. { r, θ 0 r sinθ, 0 θ } { 6 r, θ 0 r sinθ, 6 θ } } Solution. Our first step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this eample are found in either Eample.5. or Eample.5., most of the work is done for us.. We know from Eample.5. number that the graph of r = 5 sinθ is a rose. Moreover, we know from our work there that as 0 θ, we are tracing out the leaf of the rose which lies in the first quadrant. The inequalit 0 r 5 sinθ means we want to capture all [ of ] the points between the origin r = 0 and the curve r = 5 sinθ as θ runs through 0,. Hence, the region we seek is the leaf itself. 5 r θ 5 { } r, θ 0 r 5 sinθ, 0 θ. We know from Eample.5. number that r = and r = 6 cosθ intersect at θ = 6, so the region that is being described here is the set of points whose directed distance r from the origin is at least but no more than 6 cosθ as θ runs from 0 to 6. In other words, we are looking at the points outside or on the circle since r but inside or on the rose since r 6 cosθ. We shade the region below. θ = 6 r = and r = 6 cosθ { } r, θ r 6 cosθ, 0 θ 6

.5 Graphs of Polar Equations 957. From Eample.5. number, we know that the graph of r = + cosθ is a limaçon whose inner loop is traced out as θ runs through the given values to. Since the values r takes on in this interval are non-positive, the inequalit + cosθ r 0 makes sense, and we are looking for all of the points between the pole r = 0 and the limaçon as θ ranges over the interval [ ]. In other words, we shade in the inner loop of the limaçon. r 6, θ = θ θ = { r, θ + cosθ r 0, θ }. We have two regions described here connected with the union smbol. We shade each in turn and find our final answer b combining the two. In Eample.5., number, we found that the curves r = sinθ and r = sinθ intersect when θ = 6. Hence, for the first region, { r, θ 0 r sinθ, 0 θ } 6, we are shading the region between the origin r = 0 out to the circle r = sinθ as θ ranges from 0 to 6, which is the angle of intersection of the two curves. For the second region, { r, θ 0 r sinθ, 6 θ }, θ picks up where it left off at 6 and continues to. In this case, however, we are shading from the origin r = 0 out to the cardioid r = sinθ which pulls into the origin at θ =. Putting these two regions together gives us our final answer. θ = 6 r = sinθ and r = sinθ { } r, θ 0 r sinθ, 0 θ { 6 r, θ 0 r sinθ, 6 θ }

958 Applications of Trigonometr.5. Eercises In Eercises - 0, plot the graph of the polar equation b hand. Carefull label our graphs.. Circle: r = 6 sinθ. Circle: r = cosθ. Rose: r = sinθ. Rose: r = cosθ 5. Rose: r = 5 sinθ 6. Rose: r = cos5θ 7. Rose: r = sinθ 8. Rose: r = cosθ 9. Cardioid: r = cosθ 0. Cardioid: r = 5 + 5 sinθ. Cardioid: r = + cosθ. Cardioid: r = sinθ. Limaçon: r = cosθ. Limaçon: r = sinθ 5. Limaçon: r = + cosθ 6. Limaçon: r = 5 cosθ 7. Limaçon: r = 5 sinθ 8. Limaçon: r = + 7 sinθ 9. Lemniscate: r = sinθ 0. Lemniscate: r = cosθ In Eercises - 0, find the eact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole origin.. r = cosθ and r = + cosθ. r = + sinθ and r = cosθ. r = sinθ and r =. r = cosθ and r = 5. r = cosθ and r = sinθ 6. r = cosθ and r = sinθ 7. r = cosθ and r = 8. r = sinθ and r = 9. r = cosθ and r = 0. r = sinθ and r = In Eercises - 0, sketch the region in the -plane described b the given set.. {r, θ 0 r, 0 θ }. {r, θ 0 r sinθ, 0 θ }. { r, θ 0 r cosθ, θ }. { r, θ 0 r sinθ, 0 θ }

.5 Graphs of Polar Equations 959 5. { r, θ 0 r cosθ, θ } 6. { r, θ r cosθ, θ } 7. { r, θ + cosθ r cosθ, θ } { 8. r, θ r } sinθ, θ 7 9. { r, θ 0 r sinθ, 0 θ } { 6 r, θ 0 r cosθ, 6 θ } 0. { r, θ 0 r sinθ, 0 θ } { r, θ 0 r, θ } In Eercises - 50, use set-builder notation to describe the polar region. Assume that the region contains its bounding curves.. The region inside the circle r = 5.. The region inside the circle r = 5 which lies in Quadrant III.. The region inside the left half of the circle r = 6 sinθ.. The region inside the circle r = cosθ which lies in Quadrant IV. 5. The region inside the top half of the cardioid r = cosθ 6. The region inside the cardioid r = sinθ which lies in Quadrants I and IV. 7. The inside of the petal of the rose r = cosθ which lies on the positive -ais 8. The region inside the circle r = 5 but outside the circle r =. 9. The region which lies inside of the circle r = cosθ but outside of the circle r = sinθ 50. The region in Quadrant I which lies inside both the circle r = as well as the rose r = 6 sinθ While the authors trul believe that graphing polar curves b hand is fundamental to our understanding of the polar coordinate sstem, we would be derelict in our duties if we totall ignored the graphing calculator. Indeed, there are some important polar curves which are simpl too difficult to graph b hand and that makes the calculator an important tool for our further studies in Mathematics, Science and Engineering. We now give a brief demonstration of how to use the graphing calculator to plot polar curves. The first thing ou must do is switch the MODE of our calculator to POL, which stands for polar.

960 Applications of Trigonometr This changes the Y= menu as seen above in the middle. Let s plot the polar rose given b r = cosθ from Eercise 8 above. We tpe the function into the r= menu as seen above on the right. We need to set the viewing window so that the curve displas properl, but when we look at the WINDOW menu, we find three etra lines. In order for the calculator to be able to plot r = cosθ in the -plane, we need to tell it not onl the dimensions which and will assume, but we also what values of θ to use. From our previous work, we know that we need 0 θ, so we enter the data ou see above. I ll sa more about the θ-step in just a moment. Hitting GRAPH ields the curve below on the left which doesn t look quite right. The issue here is that the calculator screen is 96 piels wide but onl 6 piels tall. To get a true geometric perspective, we need to hit ZOOM SQUARE seen below in the middle to produce a more accurate graph which we present below on the right. In function mode, the calculator automaticall divided the interval [Xmin, Xma] into 96 equal subintervals. In polar mode, however, we must specif how to split up the interval [θmin, θma] using the θstep. For most graphs, a θstep of 0. is fine. If ou make it too small then the calculator takes a long time to graph. It ou make it too big, ou get chunk garbage like this. You will need to eperiment with the settings in order to get a nice graph. Eercises 5-60 give ou some curves to graph using our calculator. Notice that some of them have eplicit bounds on θ and others do not.

.5 Graphs of Polar Equations 96 5. r = θ, 0 θ 5. r = lnθ, θ 5. r = e.θ, 0 θ 5. r = θ θ,. θ. 55. r = sin5θ cosθ 56. r = sin θ + cos θ 57. r = arctanθ, θ 58. r = 59. r = cosθ 60. r = cosθ cosθ 6. How man petals does the polar rose r = sinθ have? What about r = sinθ, r = sinθ and r = sin5θ? With the help of our classmates, make a conjecture as to how man petals the polar rose r = sinnθ has for an natural number n. Replace sine with cosine and repeat the investigation. How man petals does r = cosnθ have for each natural number n? Looking back through the graphs in the section, it s clear that man polar curves enjo various forms of smmetr. However, classifing smmetr for polar curves is not as straight-forward as it was for equations back on page 6. In Eercises 6-6, we have ou and our classmates eplore some of the more basic forms of smmetr seen in common polar curves. 6. Show that if f is even 7 then the graph of r = fθ is smmetric about the -ais. a Show that fθ = + cosθ is even and verif that the graph of r = + cosθ is indeed smmetric about the -ais. See Eample.5. number. b Show that fθ = sin θ is not even, et the graph of r = sin θ is smmetric about the -ais. See Eample.5. number. 6. Show that if f is odd 8 then the graph of r = fθ is smmetric about the origin. a Show that fθ = 5 sinθ is odd and verif that the graph of r = 5 sinθ is indeed smmetric about the origin. See Eample.5. number. b Show that fθ = cos θ is not odd, et the graph of r = cos θ is smmetric about the origin. See Eample.5. number. 6. Show that if f θ = fθ for all θ in the domain of f then the graph of r = fθ is smmetric about the -ais. a For fθ = sinθ, show that f θ = fθ and the graph of r = sinθ is smmetric about the -ais, as required. See Eample.5. number. 7 Recall that this means f θ = fθ for θ in the domain of f. 8 Recall that this means f θ = fθ for θ in the domain of f.

96 Applications of Trigonometr b For fθ = 5 sinθ, show that f f, et the graph of r = 5 sinθ is smmetric about the -ais. See Eample.5. number. In Section.7, we discussed transformations of graphs. classmates eplore transformations of polar graphs. In Eercise 65 we have ou and our 65. For Eercises 65a and 65b below, let fθ = cosθ and gθ = sinθ. a Using our graphing calculator, compare the graph of r = fθ to each of the graphs of r = f θ +, r = f θ +, r = f θ and r = f θ. Repeat this process for gθ. In general, how do ou think the graph of r = fθ + α compares with the graph of r = fθ? b Using our graphing calculator, compare the graph of r = fθ to each of the graphs of r = f θ, r = f θ, r = f θ and r = fθ. Repeat this process for gθ. In general, how do ou think the graph of r = k fθ compares with the graph of r = fθ? Does it matter if k > 0 or k < 0? 66. In light of Eercises 6-6, how would the graph of r = f θ compare with the graph of r = fθ for a generic function f? What about the graphs of r = fθ and r = fθ? What about r = fθ and r = f θ? Test out our conjectures using a variet of polar functions found in this section with the help of a graphing utilit. 67. With the help of our classmates, research cardioid microphones. 68. Back in Section., in the paragraph before Eercise 5, we gave ou this link to a fascinating list of curves. Some of these curves have polar representations which we invite ou and our classmates to research.

.5 Graphs of Polar Equations 96.5. Answers. Circle: r = 6 sinθ. Circle: r = cosθ 6 6 6 6. Rose: r = sinθ. Rose: r = cosθ θ = θ = 5. Rose: r = 5 sinθ 6. Rose: r = cos5θ θ = 5 θ = θ = 7 0 θ = 0 θ = 9 0 θ = 0 5 5 5

96 Applications of Trigonometr 7. Rose: r = sinθ θ = θ = 8. Rose: r = cosθ θ = 5 θ = 8 8 θ = 7 8 θ = 8 9. Cardioid: r = cosθ 0. Cardioid: r = 5 + 5 sinθ 6 0 5 6 6 0 5 5 0 5 6 0. Cardioid: r = + cosθ. Cardioid: r = sinθ

.5 Graphs of Polar Equations 965. Limaçon: r = cosθ θ =. Limaçon: r = sinθ θ = 5 6 θ = 6 θ = 5 5. Limaçon: r = + cosθ + θ = 5 6 + θ = 7 6 6. Limaçon: r = 5 cosθ 8 8 8 8 θ = arccos 5 θ = arccos 5 7. Limaçon: r = 5 sinθ θ = arcsin 5 8 θ = arcsin 5 8. Limaçon: r = + 7 sinθ 9 5 8 8 9 9 θ = + arcsin 7 θ = arcsin 7 8 9

966 Applications of Trigonometr 9. Lemniscate: r = sinθ 0. Lemniscate: r = cosθ θ = θ =. r = cosθ and r = + cosθ,,, 5, pole. r = + sinθ and r = cosθ +,,, 7, pole

.5 Graphs of Polar Equations 967. r = sinθ and r =, 7 6,, 6. r = cosθ and r =,,,,, 0 5. r = cosθ and r = sinθ,, pole 6

968 Applications of Trigonometr 6. r = cosθ and r = sinθ 0 0, arctan, pole 7. r = cosθ and r =,, 6, 5, 6, 7, 6, 6 8. r = sinθ and r =,,, 5,,,, 7

.5 Graphs of Polar Equations 969 9. r = cosθ and r =,,, 5 6 6, 6,,, 7 6,,,,,,, 5, 0. r = sinθ and r =,,, 5, 7,, 9, 7,,,,,,,,

970 Applications of Trigonometr. {r, θ 0 r, 0 θ }. {r, θ 0 r sinθ, 0 θ }. { r, θ 0 r cosθ, θ }. { r, θ 0 r sinθ, 0 θ } 5. { r, θ 0 r cosθ, θ } 6. { r, θ r cosθ, θ }

.5 Graphs of Polar Equations 97 7. { r, θ + cosθ r cosθ, θ } 8. { r, θ r } sinθ, θ 7 9. { r, θ 0 r sinθ, 0 θ } { 6 r, θ 0 r cosθ, 6 θ }

97 Applications of Trigonometr 0. { r, θ 0 r sinθ, 0 θ } { r, θ 0 r, θ }. {r, θ 0 r 5, 0 θ }. { r, θ 0 r 5, θ }. { r, θ 0 r 6 sinθ, θ }. { r, θ cosθ r 0, θ } 5. {r, θ 0 r cosθ, 0 θ } 6. { r, θ 0 r sinθ, 0 θ } { r, θ 0 r sinθ, θ } or { r, θ 0 r sinθ, θ 5 } 7. { r, θ 0 r cosθ, 0 θ } { 8 r, θ 0 r cosθ, 5 8 θ } or { r, θ 0 r cosθ, 8 θ } 8 8. {r, θ r 5, 0 θ } 9. { r, θ 0 r cosθ, θ 0} {r, θ sinθ r cosθ, 0 θ arctan} 50. { r, θ 0 r 6 sinθ, 0 θ } { r, θ 0 r, θ 5 } { r, θ 0 r 6 sinθ, 5 θ }

.6 Hooked on Conics Again 97.6 Hooked on Conics Again In this section, we revisit our friends the Conic Sections which we began studing in Chapter 7. Our first task is to formalize the notion of rotating aes so this subsection is actuall a follow-up to Eample 8.. in Section 8.. In that eample, we saw that the graph of = is actuall a hperbola. More specificall, it is the hperbola obtained b rotating the graph of = counter-clockwise through a 5 angle. Armed with polar coordinates, we can generalize the process of rotating aes as shown below..6. Rotation of Aes Consider the - and -aes below along with the dashed - and -aes obtained b rotating the - and -aes counter-clockwise through an angle θ and consider the point P,. The coordinates, are rectangular coordinates and are based on the - and -aes. Suppose we wished to find rectangular coordinates based on the - and -aes. That is, we wish to determine P,. While this seems like a formidable challenge, it is nearl trivial if we use polar coordinates. Consider the angle φ whose initial side is the positive -ais and whose terminal side contains the point P. P, = P, θ φ θ We relate P, and P, b converting them to polar coordinates. Converting P, to polar coordinates with r > 0 ields = r cosθ + φ and = r sinθ + φ. To convert the point P, into polar coordinates, we first match the polar ais with the positive -ais, choose the same r > 0 since the origin is the same in both sstems and get = r cosφ and = r sinφ. Using the sum formulas for sine and cosine, we have = r cosθ + φ = r cosθ cosφ r sinθ sinφ Sum formula for cosine = r cosφ cosθ r sinφ sinθ = cosθ sinθ Since = r cosφ and = r sinφ

97 Applications of Trigonometr Similarl, using the sum formula for sine we get = sinθ + cosθ. These equations enable us to easil convert points with -coordinates back into -coordinates. The also enable us to easil convert equations in the variables and into equations in the variables in terms of and. If we want equations which enable us to convert points with -coordinates into -coordinates, we need to solve the sstem { cosθ sinθ = sinθ + cosθ = for and. Perhaps the cleanest wa to solve this sstem is to write it as a matri equation. Using the machiner developed in Section 8., we write the above sstem as the matri equation AX = X where [ cosθ sinθ A = sinθ cosθ ] [, X = ] [, X = Since deta = cosθcosθ sinθsinθ = cos θ + sin θ =, the determinant of A is not zero so A is invertible and X = A X. Using the formula given in Equation 8. with deta =, we find [ ] A cosθ sinθ = sinθ cosθ so that X = A X [ ] [ ] [ ] cosθ sinθ = sinθ cosθ [ ] [ ] cosθ + sinθ = sinθ + cosθ From which we get = cosθ + sinθ and = sinθ + cosθ. To summarize, Theorem.9. Rotation of Aes: Suppose the positive and aes are rotated counterclockwise through an angle θ to produce the aes and, respectivel. Then the coordinates P, and P, are related b the following sstems of equations ] { = cosθ sinθ = sinθ + cosθ and { = cosθ + sinθ = sinθ + cosθ We put the formulas in Theorem.9 to good use in the following eample. Sound familiar? In Section., the equations = r cosθ and = r sinθ make it eas to convert points from polar coordinates into rectangular coordinates, and the make it eas to convert equations from rectangular coordinates into polar coordinates. We could, of course, interchange the roles of and, and and replace φ with φ to get and in terms of and, but that seems like cheating. The matri A introduced here is revisited in the Eercises.

.6 Hooked on Conics Again 975 Eample.6.. Suppose the - and - aes are both rotated counter-clockwise through the angle θ = to produce the - and - aes, respectivel.. Let P, =, and find P,. Check our answer algebraicall and graphicall.. Convert the equation + 0 + = to an equation in and and graph. Solution.. If P, =, then = and =. Using these values for and along with θ =, Theorem.9 gives = cosθ + sinθ = cos + sin which simplifies to =. Similarl, = sinθ + cosθ = sin + cos which gives = =. Hence P, =,. To check our answer algebraicall, we use the formulas in Theorem.9 to convert P, =, back into and coordinates. We get = cosθ sinθ = cos sin = = Similarl, using = sinθ + cosθ, we obtain = as required. To check our answer graphicall, we sketch in the -ais and -ais to see if the new coordinates P, =,.6,.7 seem reasonable. Our graph is below. P, =, P,.6,.7. To convert the equation +0 + = to an equation in the variables and, we substitute = cos sin = and = sin + cos = +

976 Applications of Trigonometr and simplif. While this is b no means a trivial task, it is nothing more than a heft dose of Beginning Algebra. We will not go through the entire computation, but rather, the reader should take the time to do it. Start b verifing that = +, =, = + + To our surprise and delight, the equation + 0 + = in -coordinates reduces to 6 + 6 =, or + 9 = in -coordinates. The latter is an ellipse centered at 0, 0 with vertices along the -ais with -coordinates 0, ± and whose minor ais has endpoints with -coordinates ±, 0. We graph it below. + 0 + = The elimination of the troublesome term from the equation + 0 + = in Eample.6. number allowed us to graph the equation b hand using what we learned in Chapter 7. It is natural to wonder if we can alwas do this. That is, given an equation of the form A +B+C +D+E+F = 0, with B 0, is there an angle θ so that if we rotate the and - aes counter-clockwise through that angle θ, the equation in the rotated variables and contains no term? To eplore this conjecture, we make the usual substitutions = cosθ sinθ and = sinθ + cosθ into the equation A + B + C + D + E + F = 0 and set the coefficient of the term equal to 0. Terms containing in this epression will come from the first three terms of the equation: A, B and C. We leave it to the reader to verif that = cos θ cosθ sinθ + sinθ = cosθ sinθ + cos θ sin θ cosθ sinθ = sin θ + cosθ sinθ + cos θ

.6 Hooked on Conics Again 977 The contribution to the -term from A is A cosθ sinθ, from B it is B cos θ sin θ, and from C it is C cosθ sinθ. Equating the -term to 0, we get A cosθ sinθ + B cos θ sin θ + C cosθ sinθ = 0 A sinθ + B cosθ + C sinθ = 0 Double Angle Identities From this, we get B cosθ = A C sinθ, and our goal is to solve for θ in terms of the coefficients A, B and C. Since we are assuming B 0, we can divide both sides of this equation b B. To solve for θ we would like to divide both sides of the equation b sinθ, provided of course that we have assurances that sinθ 0. If sinθ = 0, then we would have B cosθ = 0, and since B 0, this would force cosθ = 0. Since no angle θ can have both sinθ = 0 and cosθ = 0, we can safel assume sinθ 0. We get cosθ sinθ = A C A C B, or cotθ = B. We have just proved the following theorem. Theorem.0. The equation A + B + C + D + E + F = 0 with B 0 can be transformed into an equation in variables and without an terms b rotating the - and - aes counter-clockwise through an angle θ which satisfies cotθ = A C B. We put Theorem.0 to good use in the following eample. Eample.6.. Graph the following equations.. 5 + 6 + 5 6 + 6 0 = 0. 6 + + 9 + 5 0 = 0 Solution.. Since the equation 5 + 6 + 5 6 + 6 0 = 0 is alread given to us in the form required b Theorem.0, we identif A = 5, B = 6 and C = 5 so that cotθ = A C B = 5 5 6 = 0. This means cotθ = 0 which gives θ = + k for integers k. We choose θ = so that our rotation equations are = and = +. The reader should verif that = +, =, = + + Making the other substitutions, we get that 5 + 6 + 5 6 + 6 0 = 0 reduces to 8 8 + 0 = 0, or 9 =. The latter is the equation of a hperbola centered at the -coordinates 0, opening in the direction with vertices ±, in -coordinates and asmptotes = ± +. We graph it below. The reader is invited to think about the case sinθ = 0 geometricall. What happens to the aes in this case?

978 Applications of Trigonometr. From 6 + + 9 + 5 0 = 0, we get A = 6, B = and C = 9 so that cotθ = 7. Since this isn t one of the values of the common angles, we will need to use inverse functions. Ultimatel, we need to find cosθ and sinθ, which means we have two options. If we use the arccotangent function immediatel, after the usual calculations we get θ = arccot 7. To get cosθ and sinθ from this, we would need to use half angle identities. Alternativel, we can start with cotθ = 7, use a double angle identit, and then go after cosθ and sinθ. We adopt the second approach. From cotθ = 7, we have tanθ = tanθ 7. Using the double angle identit for tangent, we have tan θ = 7, which gives tan θ+ tanθ = 0. Factoring, we get tanθ+ tanθ = 0 which gives tanθ = or tanθ =. While either of these values of tanθ satisfies the equation cotθ = 7, we choose tanθ =, since this produces an acute angle, θ = arctan. To find the rotation equations, we need cosθ = cos arctan and sinθ = sin arctan. Using the techniques developed in Section 0.6 we get cosθ = 5 and sinθ = 5. Our rotation equations are = cosθ sinθ = 5 5 and = sinθ + cosθ = 5 + 5. As usual, we now substitute these quantities into 6 + + 9 + 5 0 = 0 and simplif. As a first step, the reader can verif = 6 5 5 +9 5, = + 7 5 5, = 9 5 5 + 5 +6 5 Once the dust settles, we get 5 5 = 0, or =, whose graph is a parabola opening along the positive -ais with verte 0, 0. We graph this equation below. θ = arctan θ = 5 + 6 + 5 6 + 6 0 = 0 6 + + 9 + 5 0 = 0 As usual, there are infinitel man solutions to tanθ =. We choose the acute angle θ = arctan. The reader is encouraged to think about wh there is alwas at least one acute answer to cotθ = A C and what this B means geometricall in terms of what we are tring to accomplish b rotating the aes. The reader is also encouraged to keep a sharp lookout for the angles which satisf tanθ = in our final graph. Hint: =.

.6 Hooked on Conics Again 979 We note that even though the coefficients of and were both positive numbers in parts and of Eample.6., the graph in part turned out to be a hperbola and the graph in part worked out to be a parabola. Whereas in Chapter 7, we could easil pick out which conic section we were dealing with based on the presence or absence of quadratic terms and their coefficients, Eample.6. demonstrates that all bets are off when it comes to conics with an term which require rotation of aes to put them into a more standard form. Nevertheless, it is possible to determine which conic section we have b looking at a special, familiar combination of the coefficients of the quadratic terms. We have the following theorem. Theorem.. Suppose the equation A + B + C + D + E + F = 0 describes a non-degenerate conic section. a ˆ ˆ ˆ If B AC > 0 then the graph of the equation is a hperbola. If B AC = 0 then the graph of the equation is a parabola. If B AC < 0 then the graph of the equation is an ellipse or circle. a Recall that this means its graph is either a circle, parabola, ellipse or hperbola. See page 97. As ou ma epect, the quantit B AC mentioned in Theorem. is called the discriminant of the conic section. While we will not attempt to eplain the deep Mathematics which produces this coincidence, we will at least work through the proof of Theorem. mechanicall to show that it is true. 5 First note that if the coefficient B = 0 in the equation A +B+C +D+E+F = 0, Theorem. reduces to the result presented in Eercise in Section 7.5, so we proceed here under the assumption that B 0. We rotate the -aes counter-clockwise through an angle θ which satisfies cotθ = A C B to produce an equation with no -term in accordance with Theorem.0: A + C + D + E + F = 0. In this form, we can invoke Eercise in Section 7.5 once more using the product A C. Our goal is to find the product A C in terms of the coefficients A, B and C in the original equation. To that end, we make the usual substitutions = cosθ sinθ = sinθ + cosθ into A + B + C + D + E + F = 0. We leave it to the reader to show that, after gathering like terms, the coefficient A on and the coefficient C on are A = A cos θ + B cosθ sinθ + C sin θ C = A sin θ B cosθ sinθ + C cos θ In order to make use of the condition cotθ = A C B, we rewrite our formulas for A and C using the power reduction formulas. After some regrouping, we get Net, we tr to make sense of the product A = [A + C + A C cosθ] + B sinθ C = [A + C A C cosθ] B sinθ A C = {[A + C + A C cosθ] + B sinθ} {[A + C A C cosθ] B sinθ} 5 We hope that someda ou get to see wh this works the wa it does.

980 Applications of Trigonometr We break this product into pieces. First, we use the difference of squares to multipl the first quantities in each factor to get [A + C + A C cosθ] [A + C A C cosθ] = A + C A C cos θ Net, we add the product of the outer and inner quantities in each factor to get B sinθ [A + C + A C cosθ] +B sinθ [A + C A C cosθ] = BA C cosθ sinθ The product of the last quantit in each factor is B sinθ B sinθ = B sin θ. Putting all of this together ields A C = A + C A C cos θ BA C cosθ sinθ B sin θ From cotθ = A C cosθ B, we get sinθ = A C B, or A C sinθ = B cosθ. We use this substitution twice along with the Pthagorean Identit cos θ = sin θ to get A C = A + C A C cos θ BA C cosθ sinθ B sin θ = A + C A C [ sin θ ] B cosθb cosθ B sin θ = A + C A C + A C sin θ B cos θ B sin θ = A + C A C + [A C sinθ] B cos θ B sin θ = A + C A C + [B cosθ] B cos θ B sin θ = A + C A C + B cos θ B cos θ B sin θ = A + C A C B cos θ B sin θ = A + C A C B [ cos θ + sin θ ] = A + C A C B = A + AC + C A AC + C B = AC B Hence, B AC = A C, so the quantit B AC has the opposite sign of A C. The result now follows b appling Eercise in Section 7.5. Eample.6.. Use Theorem. to classif the graphs of the following non-degenerate conics.. + 0 + =. 5 + 6 + 5 6 + 6 0 = 0. 6 + + 9 + 5 0 = 0 Solution. This is a straightforward application of Theorem..

.6 Hooked on Conics Again 98. We have A =, B = 0 and C = so B AC = 0 = 0 < 0. Theorem. predicts the graph is an ellipse, which checks with our work from Eample.6. number.. Here, A = 5, B = 6 and C = 5, so B AC = 6 55 = 576 > 0. Theorem. classifies the graph as a hperbola, which matches our answer to Eample.6. number.. Finall, we have A = 6, B = and C = 9 which gives 69 = 0. Theorem. tells us that the graph is a parabola, matching our result from Eample.6. number..6. The Polar Form of Conics In this subsection, we start from scratch to reintroduce the conic sections from a more unified perspective. We have our new definition below. Definition.. Given a fied line L, a point F not on L, and a positive number e, a conic section is the set of all points P such that the distance from P to F the distance from P to L = e The line L is called the directri of the conic section, the point F is called a focus of the conic section, and the constant e is called the eccentricit of the conic section. We have seen the notions of focus and directri before in the definition of a parabola, Definition 7.. There, a parabola is defined as the set of points equidistant from the focus and directri, giving an eccentricit e = according to Definition.. We have also seen the concept of eccentricit before. It was introduced for ellipses in Definition 7.5 in Section 7., and later etended to hperbolas in Eercise in Section 7.5. There, e was also defined as a ratio of distances, though in these cases the distances involved were measurements from the center to a focus and from the center to a verte. One wa to reconcile the old ideas of focus, directri and eccentricit with the new ones presented in Definition. is to derive equations for the conic sections using Definition. and compare these parameters with what we know from Chapter 7. We begin b assuming the conic section has eccentricit e, a focus F at the origin and that the directri is the vertical line = d as in the figure below. d r cosθ P r, θ r θ = d O = F

98 Applications of Trigonometr Using a polar coordinate representation P r, θ for a point on the conic with r > 0, we get e = the distance from P to F the distance from P to L = so that r = ed + r cosθ. Solving this equation for r, ields ed r = e cosθ r d + r cosθ At this point, we convert the equation r = ed + r cosθ back into a rectangular equation in the variables and. If e > 0, but e, the usual conversion process outlined in Section. gives 6 e e d e + e = e d e d We leave it to the reader to show if 0 < e <, this is the equation of an ellipse centered at e d, 0 e with major ais along the -ais. Using the notation from Section 7., we have a = e d and e b = e d, so the major ais has length ed and the minor ais has length ed e e e. Moreover, we find that one focus is 0, 0 and working through the formula given in Definition 7.5 gives the eccentricit to be e, as required. If e >, then the equation generates a hperbola with center e d, 0 whose e transverse ais lies along the -ais. Since such hperbolas have the form h a b =, we need to take the opposite reciprocal of the coefficient of to find b. We get 7 a = e d e = e d e and b = e d = e d ed, so the transverse ais has length and the conjugate ais has length e e e ed. e Additionall, we verif that one focus is at 0, 0, and the formula given in Eercise in Section ed 7.5 gives the eccentricit is e in this case as well. If e =, the equation r = reduces to e cosθ d r = cosθ which gives the rectangular equation = d + d. This is a parabola with verte d, 0 opening to the right. In the language of Section 7., p = d so p = d, the focus is 0, 0, the focal diameter is d and the directri is = d, as required. Hence, we have shown that in all cases, our new understanding of conic section, focus, eccentricit and directri as presented in Definition. correspond with the old definitions given in Chapter 7. Before we summarize our findings, we note that in order to arrive at our general equation of a conic ed r = e cosθ, we assumed that the directri was the line = d for d > 0. We could have just as easil chosen the directri to be = d, = d or = d. As the reader can verif, in these cases ed we obtain the forms r = +e cosθ, r = ed e sinθ and r = ed +e sinθ, respectivel. The ke thing to remember is that in an of these cases, the directri is alwas perpendicular to the major ais of an ellipse and it is alwas perpendicular to the transverse ais of the hperbola. For parabolas, knowing the focus is 0, 0 and the directri also tells us which wa the parabola opens. We have established the following theorem. 6 Turn r = ed + r cosθ into r = ed + and square both sides to get r = e d +. Replace r with +, epand d +, combine like terms, complete the square on and clean things up. 7 Since e > in this case, e < 0. Hence, we rewrite e = e to help simplif things later on.

.6 Hooked on Conics Again 98 Theorem.. Suppose e and d are positive numbers. Then ˆ the graph of r = ed e cosθ ˆ the graph of r = ed +e cosθ ˆ the graph of r = ed e sinθ ˆ the graph of r = ed +e sinθ is the graph of a conic section with directri = d. is the graph of a conic section with directri = d. is the graph of a conic section with directri = d. is the graph of a conic section with directri = d. In each case above, 0, 0 is a focus of the conic and the number e is the eccentricit of the conic. ˆ If 0 < e <, the graph is an ellipse whose major ais has length ed ais has length ed e e and whose minor ˆ If e =, the graph is a parabola whose focal diameter is d. ˆ If e >, the graph is a hperbola whose transverse ais has length ed conjugate ais has length ed e. e and whose We test out Theorem. in the net eample. Eample.6.. Sketch the graphs of the following equations.. r = Solution. sinθ. r = cosθ. r = 6 + sinθ. From r = sinθ, we first note e = which means we have a parabola on our hands. Since ed =, we have d = and considering the form of the equation, this puts the directri at =. Since the focus is at 0, 0, we know that the verte is located at the point in rectangular coordinates 0, and must open upwards. With d =, we have a focal diameter of d = 8, so the parabola contains the points ±, 0. We graph r = sinθ below.. We first rewrite r = cosθ in the form found in Theorem., namel r = / cosθ. Since e = satisfies 0 < e <, we know that the graph of this equation is an ellipse. Since ed =, we have d = and, based on the form of the equation, the directri is =. This means that the ellipse has its major ais along the -ais. We can find the vertices of the ellipse b finding the points of the ellipse which lie on the -ais. We find r0 = 6 and r = which correspond to the rectangular points, 0 and 6, 0, so these are our vertices. The center of the ellipse is the midpoint of the vertices, which in this case is, 0. 8 We know one focus is 0, 0, which is from the center, 0 and this allows us to find the 8 As a quick check, we have from Theorem. the major ais should have length ed e = / = 9.

98 Applications of Trigonometr other focus, 0, even though we are not asked to do so. Finall, we know from Theorem ed. that the length of the minor ais is = = 6 e / which means the endpoints of the minor ais are, ±. We now have everthing we need to graph r = cosθ. = 5 6 = r = sinθ r = cosθ 6. From r = + sinθ we get e = > so the graph is a hperbola. Since ed = 6, we get d =, and from the form of the equation, we know the directri is =. This means the transverse ais of the hperbola lies along the -ais, so we can find the vertices b looking where the hperbola intersects the -ais. We find r = and r = 6. These two points correspond to the rectangular points 0, and 0, 6 which puts the center of the hperbola at 0,. Since one focus is at 0, 0, which is units awa from the center, we know the other focus is at 0, 8. According to Theorem., the conjugate ais has a length ed of = 6 = e. Putting this together with the location of the vertices, we get that the asmptotes of the hperbola have slopes ± = ±. Since the center of the hperbola is 0,, the asmptotes are = ± +. We graph the hperbola below. 8 7 6 5 = 5 5 r = 6 + sinθ

.6 Hooked on Conics Again 985 In light of Section.6., the reader ma wonder what the rotated form of the conic sections would look like in polar form. We know from Eercise 65 in Section.5 that replacing θ with θ φ in an epression r = fθ rotates the graph of r = fθ counter-clockwise b an angle φ. For instance, to graph r = sinθ all we need to do is rotate the graph of r = sinθ, which we obtained in Eample.6. number, counter-clockwise b radians, as shown below. r = sinθ Using rotations, we can greatl simplif the form of the conic sections presented in Theorem., since an three of the forms given there can be obtained from the fourth b rotating through some multiple of. Since rotations do not affect lengths, all of the formulas for lengths Theorem. remain intact. In the theorem below, we also generalize our formula for conic sections to include circles centered at the origin b etending the concept of eccentricit to include e = 0. We conclude this section with the statement of the following theorem. Theorem.. Given constants l > 0, e 0 and φ, the graph of the equation r = l e cosθ φ is a conic section with eccentricit e and one focus at 0, 0. ˆ If e = 0, the graph is a circle centered at 0, 0 with radius l. ˆ If e 0, then the conic has a focus at 0, 0 and the directri contains the point with polar coordinates d, φ where d = l e. If 0 < e <, the graph is an ellipse whose major ais has length ed and whose minor ais has length ed e If e =, the graph is a parabola whose focal diameter is d. If e >, the graph is a hperbola whose transverse ais has length ed conjugate ais has length ed e. e e and whose

986 Applications of Trigonometr.6. Eercises Graph the following equations.. + + + 6 = 0. 7 + 5 = 0. 5 + 6 + 5 + = 0. + + + 6 = 0 5. + 7 6 = 0 6. + 8 = 0 7. + 5 5 = 0 8. 8 + + 7 0 = 0 Graph the following equations. 9. r =. r =. r = cosθ cosθ + cosθ 0. r =. r =. r = + sinθ + sinθ sinθ 6 5. r = + sinθ 6. r = cos θ + [ ] cosθ sinθ The matri Aθ = is called a rotation matri. We ve seen this matri most sinθ cosθ recentl in the proof of used in the proof of Theorem.9. 7. Show the matri from Eample 8.. in Section 8. is none other than A. 8. Discuss with our classmates how to use Aθ to rotate points in the plane. 9. Using the even / odd identities for cosine and sine, show Aθ = A θ. Interpret this geometricall.

.6 Hooked on Conics Again 987.6. Answers. + + + 6 = 0 becomes = after rotating counter-clockwise through θ =.. 7 + 5 = 0 becomes 9 + = after rotating counter-clockwise through θ = θ = θ = + + + 6 = 0. 5 + 6 + 5 + = 0 becomes + + = after rotating counter-clockwise through θ =. 7 + 5 = 0. + + + 6 = 0 becomes = + after rotating counter-clockwise through θ = θ = θ = 5 + 6 + 5 + = 0 + + + 6 = 0

988 Applications of Trigonometr 5. + 7 6 = 0 becomes 6 = after rotating counter-clockwise through θ = 6. θ = 6 6. + 8 = 0 becomes = after rotating counter-clockwise through θ = θ = + 7 6 = 0 7. + 5 5 = 0 becomes = after rotating counter-clockwise through θ = arctan. + 8 = 0 8. 8 + + 7 0 = 0 becomes + = after rotating counter-clockwise through θ = arctan θ = arctan θ = arctan + 5 5 = 0 8 + + 7 0 = 0

.6 Hooked on Conics Again 989 9. r = cosθ is a parabola directri =, verte, 0 focus 0, 0, focal diameter 0. r = +sinθ = + is an ellipse sinθ directri =, vertices 0,, 0, center 0,, foci 0, 0, 0, minor ais length. r = cosθ = is an ellipse cosθ directri =, vertices, 0,, 0 center, 0, foci 0, 0,, 0 minor ais length. r = +sinθ is a parabola directri =, verte 0, focus 0, 0, focal diameter

990 Applications of Trigonometr. r = + cosθ is a hperbola directri =, vertices, 0,, 0 center, 0, foci 0, 0,, 0 conjugate ais length. r = sinθ is a hperbola directri =, vertices 0,, 0, center 0,, foci 0, 0, 0, 8 conjugate ais length 5. r = +sinθ is the parabola r = +sinθ rotated through φ = 6. r = 6 cosθ+ is the ellipse r = 6 cosθ = cosθ rotated through φ = φ = φ =

.7 Polar Form of Comple Numbers 99.7 Polar Form of Comple Numbers In this section, we return to our stud of comple numbers which were first introduced in Section.. Recall that a comple number is a number of the form z = a + bi where a and b are real numbers and i is the imaginar unit defined b i =. The number a is called the real part of z, denoted Rez, while the real number b is called the imaginar part of z, denoted Imz. From Intermediate Algebra, we know that if z = a + bi = c + di where a, b, c and d are real numbers, then a = c and b = d, which means Rez and Imz are well-defined. To start off this section, we associate each comple number z = a + bi with the point a, b on the coordinate plane. In this case, the -ais is relabeled as the real ais, which corresponds to the real number line as usual, and the -ais is relabeled as the imaginar ais, which is demarcated in increments of the imaginar unit i. The plane determined b these two aes is called the comple plane. Imaginar Ais i i, z = + i i i, 0 z = 0 Real Ais i i i 0, z = i i The Comple Plane Since the ordered pair a, b gives the rectangular coordinates associated with the comple number z = a + bi, the epression z = a + bi is called the rectangular form of z. Of course, we could just as easil associate z with a pair of polar coordinates r, θ. Although it is not as straightforward as the definitions of Rez and Imz, we can still give r and θ special names in relation to z. Definition.. The Modulus and Argument of Comple Numbers: Let z = a + bi be a comple number with a = Rez and b = Imz. Let r, θ be a polar representation of the point with rectangular coordinates a, b where r 0. ˆ The modulus of z, denoted z, is defined b z = r. ˆ ˆ The angle θ is an argument of z. The set of all arguments of z is denoted argz. If z 0 and < θ, then θ is the principal argument of z, written θ = Argz. Well-defined means that no matter how we epress z, the number Rez is alwas the same, and the number Imz is alwas the same. In other words, Re and Im are functions of comple numbers.

99 Applications of Trigonometr Some remarks about Definition. are in order. We know from Section. that ever point in the plane has infinitel man polar coordinate representations r, θ which means it s worth our time to make sure the quantities modulus, argument and principal argument are well-defined. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the onl r-value which can be used here is r = 0. Hence for z = 0, z = 0 is well-defined. If z 0, then the point associated with z is not the origin, and there are two possibilities for r: one positive and one negative. However, we stipulated r 0 in our definition so this pins down the value of z to one and onl one number. Thus the modulus is well-defined in this case, too. Even with the requirement r 0, there are infinitel man angles θ which can be used in a polar representation of a point r, θ. If z 0 then the point in question is not the origin, so all of these angles θ are coterminal. Since coterminal angles are eactl radians apart, we are guaranteed that onl one of them lies in the interval, ], and this angle is what we call the principal argument of z, Argz. In fact, the set argz of all arguments of z can be described using set-builder notation as argz = {Argz + k k is an integer}. Note that since argz is a set, we will write θ argz to mean θ is in the set of arguments of z. If z = 0 then the point in question is the origin, which we know can be represented in polar coordinates as 0, θ for an angle θ. In this case, we have arg0 =, and since there is no one value of θ which lies, ], we leave Arg0 undefined. It is time for an eample. Eample.7.. For each of the following comple numbers find Rez, Imz, z, argz and Argz. Plot z in the comple plane.. z = i. z = + i. z = i. z = 7 Solution.. For z = i = + i, we have Rez = and Imz =. To find z, argz and Argz, we need to find a polar representation r, θ with r 0 for the point P, associated with z. We know r = + =, so r = ±. Since we require r 0, we choose r =, so z =. Net, we find a corresponding angle θ. Since r > 0 and P lies in Quadrant IV, θ is a Quadrant IV angle. We know tanθ = =, so θ = 6 + k for integers k. Hence, argz = { 6 + k k is an integer}. Of these values, onl θ = 6 satisfies the requirement that < θ, hence Argz = 6.. The comple number z = + i has Rez =, Imz =, and is associated with the point P,. Our net task is to find a polar representation r, θ for P where r 0. Running through the usual calculations gives r = 5, so z = 5. To find θ, we get tanθ =, and since r > 0 and P lies in Quadrant II, we know θ is a Quadrant II angle. We find θ = + arctan + k, or, more succinctl θ = arctan + k for integers k. Hence argz = { arctan + k k is an integer}. Onl θ = arctan satisfies the requirement < θ, so Argz = arctan. In case ou re wondering, the use of the absolute value notation z for modulus will be eplained shortl. Recall the smbol being used here,, is the mathematical smbol which denotes membership in a set. If we had Calculus, we would regard Arg0 as an indeterminate form. But we don t, so we won t.

.7 Polar Form of Comple Numbers 99. We rewrite z = i as z = 0 + i to find Rez = 0 and Imz =. The point in the plane which corresponds to z is 0, and while we could go through the usual calculations to find the required polar form of this point, we can almost see the answer. The point 0, lies units awa from the origin on the positive -ais. Hence, r = z = and θ = + k for integers k. We get argz = { + k k is an integer} and Argz =.. As in the previous problem, we write z = 7 = 7 + 0i so Rez = 7 and Imz = 0. The number z = 7 corresponds to the point 7, 0, and this is another instance where we can determine the polar form b ee. The point 7, 0 is 7 units awa from the origin along the negative -ais. Hence, r = z = 7 and θ = + = k + k for integers k. We have argz = {k + k is an integer}. Onl one of these values, θ =, just barel lies in the interval, ] which means and Argz =. We plot z along with the other numbers in this eample below. Imaginar Ais z = + i z = 7 i i i i z = i 7 i z = i Real Ais Now that we ve had some practice computing the modulus and argument of some comple numbers, it is time to eplore their properties. We have the following theorem. Theorem.. Properties of the Modulus: Let z and w be comple numbers. ˆ z is the distance from z to 0 in the comple plane ˆ z 0 and z = 0 if and onl if z = 0 ˆ z = Rez + Imz ˆ Product Rule: zw = z w ˆ ˆ Power Rule: z n = z n for all natural numbers, n Quotient Rule: z = z w w, provided w 0 To prove the first three properties in Theorem., suppose z = a + bi where a and b are real numbers. To determine z, we find a polar representation r, θ with r 0 for the point a, b. From Section., we know r = a + b so that r = ± a + b. Since we require r 0, then it must be that r = a + b, which means z = a + b. Using the distance formula, we find the distance

99 Applications of Trigonometr from 0, 0 to a, b is also a + b, establishing the first propert. 5 For the second propert, note that since z is a distance, z 0. Furthermore, z = 0 if and onl if the distance from z to 0 is 0, and the latter happens if and onl if z = 0, which is what we were asked to show. 6 For the third propert, we note that since a = Rez and b = Imz, z = a + b = Rez + Imz. To prove the product rule, suppose z = a + bi and w = c + di for real numbers a, b, c and d. Then zw = a + bic + di. After the usual arithmetic 7 we get zw = ac bd + ad + bci. Therefore, zw = ac bd + ad + bc = a c abcd + b d + a d + abcd + b c Epand = a c + a d + b c + b d Rearrange terms = a c + d + b c + d Factor = a + b c + d Factor = a + b c + d Product Rule for Radicals = z w Definition of z and w Hence zw = z w as required. Now that the Product Rule has been established, we use it and the Principle of Mathematical Induction 8 to prove the power rule. Let P n be the statement z n = z n. Then P is true since z = z = z. Net, assume P k is true. That is, assume z k = z k for some k. Our job is to show that P k + is true, namel z k+ = z k+. As is customar with induction proofs, we first tr to reduce the problem in such a wa as to use the Induction Hpothesis. z k+ = z k z Properties of Eponents = z k z Product Rule = z k z Induction Hpothesis = z k+ Properties of Eponents Hence, P k + is true, which means z n = z n is true for all natural numbers n. Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume w 0 so w 0 and we get z = w z w = z w Product Rule. 5 Since the absolute value of a real number can be viewed as the distance from to 0 on the number line, this first propert justifies the notation z for modulus. We leave it to the reader to show that if z is real, then the definition of modulus coincides with absolute value so the notation z is unambiguous. 6 This ma be considered b some to be a bit of a cheat, so we work through the underling Algebra to see this is true. We know z = 0 if and onl if a + b = 0 if and onl if a + b = 0, which is true if and onl if a = b = 0. The latter happens if and onl if z = a + bi = 0. There. 7 See Eample.. in Section. for a review of comple number arithmetic. 8 See Section 9. for a review of this technique.

.7 Polar Form of Comple Numbers 995 Hence, the proof reall boils down to showing w = w. This is left as an eercise. Net, we characterize the argument of a comple number in terms of its real and imaginar parts. Theorem.5. Properties of the Argument: Let z be a comple number. ˆ If Rez 0 and θ argz, then tanθ = Imz Rez. ˆ If Rez = 0 and Imz > 0, then argz = { + k k is an integer}. ˆ If Rez = 0 and Imz < 0, then argz = { + k k is an integer}. ˆ If Rez = Imz = 0, then z = 0 and argz =,. To prove Theorem.5, suppose z = a+bi for real numbers a and b. B definition, a = Rez and b = Imz, so the point associated with z is a, b = Rez, Imz. From Section., we know that if r, θ is a polar representation for Rez, Imz, then tanθ = Imz Rez, provided Rez 0. If Rez = 0 and Imz > 0, then z lies on the positive imaginar ais. Since we take r > 0, we have that θ is coterminal with, and the result follows. If Rez = 0 and Imz < 0, then z lies on the negative imaginar ais, and a similar argument shows θ is coterminal with. The last propert in the theorem was alread discussed in the remarks following Definition.. Our net goal is to completel marr the Geometr and the Algebra of the comple numbers. To that end, consider the figure below. Imaginar Ais a, b z = a + bi r, θ bi z = a + b = r 0 θ argz a Real Ais Polar coordinates, r, θ associated with z = a + bi with r 0. We know from Theorem.7 that a = r cosθ and b = r sinθ. Making these substitutions for a and b gives z = a + bi = r cosθ + r sinθi = r [cosθ + i sinθ]. The epression cosθ + i sinθ is abbreviated cisθ so we can write z = rcisθ. Since r = z and θ argz, we get Definition.. A Polar Form of a Comple Number: Suppose z is a comple number and θ argz. The epression: is called a polar form for z. z cisθ = z [cosθ + i sinθ]

996 Applications of Trigonometr Since there are infinitel man choices for θ argz, there infinitel man polar forms for z, so we used the indefinite article a in Definition.. It is time for an eample. Eample.7... Find the rectangular form of the following comple numbers. Find Rez and Imz. a z = cis b z = cis c z = cis0 d z = cis. Use the results from Eample.7. to find a polar form of the following comple numbers. a z = i b z = + i c z = i d z = 7 Solution.. The ke to this problem is to write out cisθ as cosθ + i sinθ. a B definition, z = cis [ = cos + i sin ]. After some simplifing, we get z = + i, so that Rez = and Imz =. = [ cos b Epanding, we get z = cis z = i, so Rez = = Imz. + i sin ]. From this, we find c We get z = cis0 = [cos0 + i sin0] =. Writing = + 0i, we get Rez = and Imz = 0, which makes sense seeing as is a real number. d Lastl, we have z = cis = cos + i sin = i. Since i = 0 + i, we get Rez = 0 and Imz =. Since i is called the imaginar unit, these answers make perfect sense.. To write a polar form of a comple number z, we need two pieces of information: the modulus z and an argument not necessaril the principal argument of z. We shamelessl mine our solution to Eample.7. to find what we need. a For z = i, z = and θ = 6, so z = cis 6. We can check our answer b converting it back to rectangular form to see that it simplifies to z = i. b For z = + i, z = 5 and θ = arctan. Hence, z = 5cis arctan. It is a good eercise to actuall show that this polar form reduces to z = + i. c For z = i, z = and θ =. In this case, z = cis. This can be checked geometricall. Head out units from 0 along the positive real ais. Rotating radians counter-clockwise lands ou eactl units above 0 on the imaginar ais at z = i. d Last but not least, for z = 7, z = 7 and θ =. We get z = 7cis. As with the previous problem, our answer is easil checked geometricall.

.7 Polar Form of Comple Numbers 997 The following theorem summarizes the advantages of working with comple numbers in polar form. Theorem.6. Products, Powers and Quotients Comple Numbers in Polar Form: Suppose z and w are comple numbers with polar forms z = z cisα and w = w cisβ. Then ˆ Product Rule: zw = z w cisα + β ˆ Power Rule DeMoivre s Theorem : z n = z n cisnθ for ever natural number n ˆ Quotient Rule: z w = z cisα β, provided w 0 w The proof of Theorem.6 requires a health mi of definition, arithmetic and identities. We first start with the product rule. zw = [ z cisα] [ w cisβ] = z w [cosα + i sinα] [cosβ + i sinβ] We now focus on the quantit in brackets on the right hand side of the equation. [cosα + i sinα] [cosβ + i sinβ] = cosα cosβ + i cosα sinβ + i sinα cosβ + i sinα sinβ = cosα cosβ + i sinα sinβ Rearranging terms + i sinα cosβ + i cosα sinβ = cosα cosβ sinα sinβ Since i = + i sinα cosβ + cosα sinβ Factor out i = cosα + β + i sinα + β Sum identities = cisα + β Definition of cis Putting this together with our earlier work, we get zw = z w cisα + β, as required. Moving right along, we net take aim at the Power Rule, better known as DeMoivre s Theorem. 9 We proceed b induction on n. Let P n be the sentence z n = z n cisnθ. Then P is true, since z = z = z cisθ = z cis θ. We now assume P k is true, that is, we assume z k = z k ciskθ for some k. Our goal is to show that P k + is true, or that z k+ = z k+ cisk + θ. We have z k+ = z k z Properties of Eponents = z k ciskθ z cisθ Induction Hpothesis = z k z ciskθ + θ Product Rule = z k+ cisk + θ 9 Compare this proof with the proof of the Power Rule in Theorem..

998 Applications of Trigonometr Hence, assuming P k is true, we have that P k + is true, so b the Principle of Mathematical Induction, z n = z n cisnθ for all natural numbers n. The last propert in Theorem.6 to prove is the quotient rule. Assuming w 0 we have z w = z cisα w cisβ z cosα + i sinα = w cosβ + i sinβ Net, we multipl both the numerator and denominator of the right hand side b cosβ i sinβ which is the comple conjugate of cosβ + i sinβ to get z w = z cosα + i sinα cosβ i sinβ w cosβ + i sinβ cosβ i sinβ If we let the numerator be N = [cosα + i sinα] [cosβ i sinβ] and simplif we get N = [cosα + i sinα] [cosβ i sinβ] = cosα cosβ i cosα sinβ + i sinα cosβ i sinα sinβ Epand = [cosα cosβ + sinα sinβ] + i [sinα cosβ cosα sinβ] Rearrange and Factor = cosα β + i sinα β Difference Identities = cisα β Definition of cis If we call the denominator D then we get D = [cosβ + i sinβ] [cosβ i sinβ] = cos β i cosβ sinβ + i cosβ sinβ i sin β Epand = cos β i sin β Simplif = cos β + sin β Again, i = = Pthagorean Identit Putting it all together, we get z w = = z cosα + i sinα w cosβ + i sinβ z cisα β w = z cisα β w cosβ i sinβ cosβ i sinβ and we are done. The net eample makes good use of Theorem.6.

.7 Polar Form of Comple Numbers 999 Eample.7.. Let z = + i and w = + i. Use Theorem.6 to find the following.. zw. w 5. Write our final answers in rectangular form. Solution. In order to use Theorem.6, we need to write z and w in polar form. For z = +i, we find z = + = 6 =. If θ argz, we know tanθ = Imz Rez = =. Since z lies in Quadrant I, we have θ = 6 + k for integers k. Hence, z = cis 6. For w = + i, we have w = + =. For an argument θ of w, we have tanθ = =. Since w lies in Quadrant II, θ = + k for integers k and w = cis. We can now proceed.. We get zw = cis 6 cis = 8cis 6 + = 8cis 5 [ 6 = 8 cos 5 6 + i sin 5 ] 6. After simplifing, we get zw = + i.. We use DeMoivre s Theorem which ields w 5 = [ cis ] 5 = 5 cis 5 = cis 0. Since 0 is coterminal with, we get w5 = [ cos + i sin ] = 6 6i.. Last, but not least, we have z w = cis 6 cis = cis 6 = cis. Since is a quadrantal angle, we can see the rectangular form b moving out units along the positive real ais, then rotating radians clockwise to arrive at the point units below 0 on the imaginar ais. The long and short of it is that z w = i. Some remarks are in order. First, the reader ma not be sold on using the polar form of comple numbers to multipl comple numbers especiall if the aren t given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers z and w in Eample.7. into polar form, compute their product, and convert back to rectangular form certainl more work than is required to multipl out zw = + i + i the old-fashioned wa. However, Theorem.6 pas huge dividends when computing powers of comple numbers. Consider how we computed w 5 above and compare that to using the Binomial Theorem, Theorem 9., to accomplish the same feat b epanding + i 5. Division is trick in the best of times, and we saved ourselves a lot of time and effort using Theorem.6 to find and simplif z w using their polar forms as opposed to starting with +i +i, rationalizing the denominator, and so forth. There is geometric reason for studing these polar forms and we would be derelict in our duties if we did not mention the Geometr hidden in Theorem.6. Take the product rule, for instance. If z = z cisα and w = w cisβ, the formula zw = z w cisα + β can be viewed geometricall as a two step process. The multiplication of z b w can be interpreted as magnifing 0 the distance z from z to 0, b the factor w. Adding the argument of w to the argument of z can be interpreted geometricall as a rotation of β radians counter-clockwise. Focusing on z and w from Eample 0 Assuming w >. Assuming β > 0. z w

000 Applications of Trigonometr.7., we can arrive at the product zw b plotting z, doubling its distance from 0 since w =, and rotating radians counter-clockwise. The sequence of diagrams below attempts to describe this process geometricall. Imaginar Ais Imaginar Ais 6i 5i i i i i z = cis 6 z w = 8cis 6 zw = 8cis 6 + 6i 5i i i i i z w = 8cis 6 0 5 6 7 Real Ais 7 6 5 0 5 6 7 Real Ais Multipling z b w =. Rotating counter-clockwise b Argw = radians. Visualizing zw for z = cis 6 and w = cis. We ma also visualize division similarl. Here, the formula z w = z w cisα β ma be interpreted as shrinking the distance from 0 to z b the factor w, followed up b a clockwise rotation of β radians. In the case of z and w from Eample.7., we arrive at z w b first halving the distance from 0 to z, then rotating clockwise radians. Imaginar Ais Imaginar Ais i i z = cis w 6 i i z = cis 6 z = cis w 6 i 0 Real Ais 0 Real Ais i zw = cis 6 Dividing z b w =. Rotating clockwise b Argw = radians. Visualizing z w for z = cis 6 and w = cis. Our last goal of the section is to reverse DeMoivre s Theorem to etract roots of comple numbers. Definition.. Let z and w be comple numbers. If there is a natural number n such that w n = z, then w is an n th root of z. Unlike Definition 5. in Section 5., we do not specif one particular prinicpal n th root, hence the use of the indefinite article an as in an n th root of z. Using this definition, both and are Again, assuming w >. Again, assuming β > 0.

.7 Polar Form of Comple Numbers 00 square roots of 6, while 6 means the principal square root of 6 as in 6 =. Suppose we wish to find all comple third cube roots of 8. Algebraicall, we are tring to solve w = 8. We know that there is onl one real solution to this equation, namel w = 8 =, but if we take the time to rewrite this equation as w 8 = 0 and factor, we get w w + w + = 0. The quadratic factor gives two more cube roots w = ± i, for a total of three cube roots of 8. In accordance with Theorem., since the degree of pw = w 8 is three, there are three comple zeros, counting multiplicit. Since we have found three distinct zeros, we know these are all of the zeros, so there are eactl three distinct cube roots of 8. Let us now solve this same problem using the machiner developed in this section. To do so, we epress z = 8 in polar form. Since z = 8 lies 8 units awa on the positive real ais, we get z = 8cis0. If we let w = w cisα be a polar form of w, the equation w = 8 becomes w = 8 w cisα = 8cis0 w cisα = 8cis0 DeMoivre s Theorem The comple number on the left hand side of the equation corresponds to the point with polar coordinates w, α, while the comple number on the right hand side corresponds to the point with polar coordinates 8, 0. Since w 0, so is w, which means w, α and 8, 0 are two polar representations corresponding to the same comple number, both with positive r values. From Section., we know w = 8 and α = 0 + k for integers k. Since w is a real number, we solve w = 8 b etracting the principal cube root to get w = 8 =. As for α, we get α = k for integers k. This produces three distinct points with polar coordinates corresponding to k = 0, and : specificall, 0,, and,. These correspond to the comple numbers w 0 = cis0, w = cis and w = cis, respectivel. Writing these out in rectangular form ields w 0 =, w = + i and w = i. While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for eample, all of the fifth roots of. Tr using Chapter techniques on that! If we start with a generic comple number in polar form z = z cisθ and solve w n = z in the same manner as above, we arrive at the following theorem. Theorem.7. The n th roots of a Comple Number: Let z 0 be a comple number with polar form z = rcisθ. For each natural number n, z has n distinct n th roots, which we denote b w 0, w,..., w n, and the are given b the formula w k = n θ rcis n + n k The proof of Theorem.7 breaks into to two parts: first, showing that each w k is an n th root, and second, showing that the set {w k k = 0,,..., n } consists of n different comple numbers. To show w k is an n th root of z, we use DeMoivre s Theorem to show w k n = z.

00 Applications of Trigonometr w k n = n rcis θ n + n k n = n r n cis n [ θ n + n k] = rcis θ + k DeMoivre s Theorem Since k is a whole number, cosθ + k = cosθ and sinθ + k = sinθ. Hence, it follows that cisθ + k = cisθ, so w k n = rcisθ = z, as required. To show that the formula in Theorem.7 generates n distinct numbers, we assume n or else there is nothing to prove and note that the modulus of each of the w k is the same, namel n r. Therefore, the onl wa an two of these polar forms correspond to the same number is if their arguments are coterminal that is, if the arguments differ b an integer multiple of. Suppose k and j are whole numbers between 0 and n, inclusive, with k j. Since k and j are different, let s assume for the sake of argument k j n. For this to be an integer multiple of, that k > j. Then θ n + n k θ n + n j = k j must be a multiple of n. But because of the restrictions on k and j, 0 < k j n. Think this through. Hence, k j is a positive number less than n, so it cannot be a multiple of n. As a result, w k and w j are different comple numbers, and we are done. B Theorem., we know there at most n distinct solutions to w n = z, and we have just found all of them. We illustrate Theorem.7 in the net eample. Eample.7.. Use Theorem.7 to find the following:. both square roots of z = + i. the four fourth roots of z = 6. the three cube roots of z = + i. the five fifth roots of z =. Solution.. We start b writing z = + i = cis. To use Theorem.7, we identif r =, θ = and n =. We know that z has two square roots, and in keeping with the notation in Theorem.7, we ll call them w 0 and w. We get w 0 = cis / + 0 = cis and w = cis / + = cis. In rectangular form, the two square roots of z are w 0 = + i and w = i. We can check our answers b squaring them and showing that we get z = + i.. Proceeding as above, we get z = 6 = 6cis. With r = 6, θ = and n =, we get the four fourth roots of z to be w 0 = 6cis + 0 = cis, w = 6cis + = cis, w = 6cis + = cis 5 and w = 6cis + = cis 7. Converting these to rectangular form gives w 0 = +i, w = +i, w = i and w = i.

.7 Polar Form of Comple Numbers 00. For z = +i, we have z = cis. With r =, θ = and n = the usual computations ield w 0 = cis, w = cis 9 = cis and w = cis 7. If we were to convert these to rectangular form, we would need to use either the Sum and Difference Identities in Theorem 0.6 or the Half-Angle Identities in Theorem 0.9 to evaluate w 0 and w. Since we are not eplicitl told to do so, we leave this as a good, but mess, eercise.. To find the five fifth roots of, we write = cis0. We have r =, θ = 0 and n = 5. Since 5 =, the roots are w 0 = cis0 =, w = cis 5, w = cis 5, w = cis 6 5 and w = cis 8 5. The situation here is even graver than in the previous eample, since we have not developed an identities to help us determine the cosine or sine of 5. At this stage, we could approimate our answers using a calculator, and we leave this as an eercise. Now that we have done some computations using Theorem.7, we take a step back to look at things geometricall. Essentiall, Theorem.7 sas that to find the n th roots of a comple number, we first take the n th root of the modulus and divide the argument b n. This gives the first root w 0. Each succeessive root is found b adding n to the argument, which amounts to rotating w 0 b n radians. This results in n roots, spaced equall around the comple plane. As an eample of this, we plot our answers to number in Eample.7. below. Imaginar Ais i w i w 0 0 Real Ais i w w i The four fourth roots of z = 6 equall spaced = around the plane. We have onl glimpsed at the beaut of the comple numbers in this section. The comple plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibl important Science and Engineering applications. For now, the following eercises will have to suffice. For more on this, see the beautifull written epilogue to Section. found on page 9.

00 Applications of Trigonometr.7. Eercises In Eercises - 0, find a polar representation for the comple number z and then identif Rez, Imz, z, argz and Argz.. z = 9 + 9i. z = 5 + 5i. z = 6i. z = +i 5. z = 6 + 6i 6. z = 7. z = i 8. z = i 9. z = 5i 0. z = i. z = 6. z = i 7. z = + i. z = + i 5. z = 7 + i 6. z = + 6i 7. z = 5i 8. z = 5 i 9. z = i 0. z = i In Eercises - 0, find the rectangular form of the given comple number. Use whatever identities are necessar to find the eact values.. z = 6cis0. z = cis. z = 7 cis. z = cis 6 5. z = cis 9. z = 7cis. z = 8cis 5. z = 5cis arctan 6. z = 6cis 0. z = cis 7. z = 9cis 8. z = cis. z = 7 cis. z = cis 7. z = cis 8 6. z = 0cis arctan 7. z = 5cis arctan 8. z = arctan 9. z = 50cis 7 arctan 0. z = 5 cis + arctan For Eercises - 5, use z = + i and w = i to compute the quantit. Epress our answers in polar form using the principal argument.. zw. z w 5. w 6. z 5 w 7. z w 8.. w z. z z w

.7 Polar Form of Comple Numbers 005 9. w z 50. z w 5. w z 5. w z 6 In Eercises 5-6, use DeMoivre s Theorem to find the indicated power of the given comple number. Epress our final answers in rectangular form. 5. + i 57. 5 + 5 i 58. 5. i 55. + i 56. + i 6 i 59. i 60. i 6. + i 6. + i 5 6. i 5 6. i 8 In Eercises 65-76, find the indicated comple roots. Epress our answers in polar form and then convert them into rectangular form. 65. the two square roots of z = i 66. the two square roots of z = 5i 67. the two square roots of z = + i 68. the two square roots of 5 5 i 69. the three cube roots of z = 6 70. the three cube roots of z = 5 7. the three cube roots of z = i 7. the three cube roots of z = 8i 7. the four fourth roots of z = 6 7. the four fourth roots of z = 8 75. the si sith roots of z = 6 76. the si sith roots of z = 79 77. Use the Sum and Difference Identities in Theorem 0.6 or the Half Angle Identities in Theorem 0.9 to epress the three cube roots of z = + i in rectangular form. See Eample.7., number. 78. Use a calculator to approimate the five fifth roots of. See Eample.7., number. 79. According to Theorem.6 in Section., the polnomial p = + can be factored into the product linear and irreducible quadratic factors. In Eercise 8 in Section 8.7, we showed ou how to factor this polnomial into the product of two irreducible quadratic factors using a sstem of non-linear equations. Now that we can compute the comple fourth roots of directl, we can simpl appl the Comple Factorization Theorem, Theorem., to obtain the linear factorization p = + i i + i i. B multipling the first two factors together and then the second two factors together, thus pairing up the comple conjugate pairs of zeros Theorem.5 told us we d get, we have that p = + + +. Use the comple th roots of 096 to factor p = 096 into a product of linear and irreducible quadratic factors.

006 Applications of Trigonometr 80. Complete the proof of Theorem. b showing that if w 0 than = 8. Recall from Section. that given a comple number z = a+bi its comple conjugate, denoted z, is given b z = a bi. a Prove that z = z. b Prove that z = zz c Show that Rez = z + z and Imz = z z i d Show that if θ argz then θ arg z. Interpret this result geometricall. e Is it alwas true that Arg z = Argz? 8. Given an natural number n, the n comple n th roots of the number z = are called the n th Roots of Unit. In the following eercises, assume that n is a fied, but arbitrar, natural number such that n. a Show that w = is an n th root of unit. b Show that if both w j and w k are n th roots of unit then so is their product w j w k. c Show that if w j is an n th root of unit then there eists another n th root of unit w j such that w j w j =. Hint: If w j = cisθ let w j = cis θ. You ll need to verif that w j = cis θ is indeed an n th root of unit. 8. Another wa to epress the polar form of a comple number is to use the eponential function. For real numbers t, Euler s Formula defines e it = cost + i sint. a Use Theorem.6 to show that e i e i = e i+ for all real numbers and. b Use Theorem.6 to show that e i n = e in for an real number and an natural number n. c Use Theorem.6 to show that ei e i = ei for all real numbers and. d If z = rcisθ is the polar form of z, show that z = re it where θ = t radians. e Show that e i + = 0. This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics. f Show that cost = eit + e it w w. and that sint = eit e it for all real numbers t. i

.7 Polar Form of Comple Numbers 007.7. Answers. z = 9 + 9i = 9 cis, Rez = 9, Imz = 9, z = 9 argz = { + k k is an integer} and Argz =.. z = 5 + 5i = 0cis, Rez = 5, Imz = 5, z = 0 argz = { + k k is an integer} and Argz =.. z = 6i = 6cis, Rez = 0, Imz = 6, z = 6 argz = { + k k is an integer} and Argz =.. z = + i = 6cis, Rez =, Imz =, z = 6 argz = { + k k is an integer} and Argz =. 5. z = 6 + 6i = cis 5 6, Rez = 6, Imz = 6, z = argz = { 5 6 + k k is an integer} and Argz = 5 6. 6. z = = cis, Rez =, Imz = 0, z = argz = {k + k is an integer} and Argz =. 7. z = i = cis 7 6, Rez =, Imz =, z = argz = { 7 6 + k k is an integer} and Argz = 5 6., Rez =, Imz =, z = 8. z = i = cis 5 argz = { 5 + k k is an integer} and Argz =., Rez = 0, Imz = 5, z = 5 9. z = 5i = 5cis argz = { + k k is an integer} and Argz =., Rez =, Imz =, z = 0. z = i = cis 7 argz = { 7 + k k is an integer} and Argz =.. z = 6 = 6cis 0, Rez = 6, Imz = 0, z = 6 argz = {k k is an integer} and Argz = 0.. z = i 7 = 7cis, Rez = 0, Imz = 7, z = 7 argz = { + k k is an integer} and Argz =.. z = + i = 5cis arctan, Rez =, Imz =, z = 5 argz = { arctan + k k is an integer } and Argz = arctan.

008 Applications of Trigonometr. z = + i = cis argz = { arctan, Rez =, Imz =, z = } and Argz = arctan arctan + k k is an integer 5. z = 7 + i = 5cis arctan 7, Rez = 7, Imz =, z = 5 argz = { arctan } 7 + k k is an integer and Argz = arctan 7. 6. z = + 6i = 0cis arctan, Rez =, Imz = 6, z = 0 argz = { arctan + k k is an integer} and Argz = arctan. 7. z = 5i = cis + arctan 5, Rez =, Imz = 5, z = argz = { + arctan } 5 + k k is an integer and Argz = arctan 5. 8. z = 5 i = 9cis + arctan 5, Rez = 5, Imz =, z = 9 argz = { + arctan 5 + k k is an integer } and Argz = arctan 5. 9. z = i = 5cis arctan, Rez =, Imz =, z = 5 argz = { arctan + k k is an integer } and Argz = arctan = arctan. 0. z = i = 0cis arctan, Rez =, Imz =, z = 0 argz = {arctan + k k is an integer} and Argz = arctan = arctan.. z = 6cis0 = 6. z = cis 6 = + i. z = 7 cis = 7 + 7i. z = cis = i 5. z = cis = + i 6. z = 6cis = + i 7. z = 9cis = 9 8. z = cis = i 9. z = 7cis = 7 7 i 0. z = cis = i.. z = cis 7 = i. z = cis = 6 6i. z = 8cis = + + i. z = cis 7 8 = + + i 5. z = 5cis arctan = + i 6. z = 0cis arctan = + i 7. z = 5cis arctan = 5 6i 5 8. z = cis arctan = i 9. z = 50cis arctan 7 = 8 + i 0. z = cis + arctan 5 = 6 5i 6

.7 Polar Form of Comple Numbers 009 In Eercises - 5, we have that z = + i = cis 5 6 and w = i = 6cis so we get the following.. zw = 8cis 7. z = 8cis z. w = cis 5. w = 6cis 7. z w z = 97cis0 8. w = cis w. z = cis 6. z 5 w = 878cis 9. w z = cis z 50. = w cis 5. w = z cis 5. w 6 z = 6cis 5. + i = 6 5. i = 8i 55. + i = 56. + i = 8 + 8i 57. 5 + 5 i 6 = 5 + 5 i 58. i = 59. i = 7 7 i 60. i = 8 8 8i 8 6. + i = 6. + i 5 = 8 8i 6. i 5 = 6 6i 6. i 8 = 6 65. Since z = i = cis we have w 0 = cis = + i w = cis 5 = i 66. Since z = 5i = 5cis we have w 0 = 5cis = 5 + 5 i w = 5cis 7 = 5 5 i 67. Since z = + i = cis we have w 0 = cis 6 = 6 + i w = cis 7 6 = 6 i 68. Since z = 5 5 i = 5cis 5 we have w 0 = 5cis 5 6 = 5 + 5 i w = 5cis 6 = 5 5 i 69. Since z = 6 = 6cis 0 we have w 0 = cis 0 = w = cis = + i w = cis = i

00 Applications of Trigonometr 70. Since z = 5 = 5cis we have w 0 = 5cis = 5 + 5 i w = 5cis = 5 w = 5cis 5 = 5 5 i 7. Since z = i = cis we have w 0 = cis 6 = + i w = cis 5 6 = + i w = cis = i 7. Since z = 8i = 8cis we have w 0 = cis = i w = cis 7 6 7. Since z = 6 = 6cis 0 we have = i w = cis 6 = i w 0 = cis 0 = w = cis = w = cis = i w = cis = i 7. Since z = 8 = 8cis we have w 0 = cis = + i w = cis = + w = cis 5 = i w = cis 7 i = i 75. Since z = 6 = 6cis0 we have w 0 = cis0 = w = cis = + i w = cis = + i w = cis = w = cis = i w5 = cis = i 76. Since z = 79 = 79cis we have w 0 = cis 6 = + i w = cis w = cis 7 6 = i w = cis = i w = cis 5 6 = + i = i w5 = cis 6 = i 77. Note: In the answers for w 0 and w the first rectangular form comes from appling the appropriate Sum or Difference Identit = 7 and = +, respectivel and the second comes from using the Half-Angle Identities. w 0 = cis 6+ 6 = + i = + + i w = cis = + i w = cis 7 = 6 + i 6 = + + i

.7 Polar Form of Comple Numbers 0 78. w 0 = cis0 = w = cis 5 0.09 + 0.95i w = cis 5 0.809 + 0.588i w = cis 6 5 0.809 0.588i w = cis 8 5 0.09 0.95i 79. p = 096 = + + + ++ + + +

0 Applications of Trigonometr.8 Vectors As we have seen numerous times in this book, Mathematics can be used to model and solve real-world problems. For man applications, real numbers suffice; that is, real numbers with the appropriate units attached can be used to answer questions like How close is the nearest Sasquatch nest? There are other times though, when these kinds of quantities do not suffice. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. Foreshadowing the use of bearings in the eercises, perhaps? To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects called vectors. A vector is represented geometricall as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrow at one endpoint of the segment. When referring to vectors in this tet, we shall adopt the arrow notation, so the smbol v is read as the vector v. Below is a tpical vector v with endpoints P, and Q, 6. The point P is called the initial point or tail of v and the point Q is called the terminal point or head of v. Since we can reconstruct v completel from P and Q, we write v = P Q, where the order of points P initial point and Q terminal point is important. Think about this before moving on. Q, 6 P, v = P Q While it is true that P and Q completel determine v, it is important to note that since vectors are defined in terms of their two characteristics, magnitude and direction, an directed line segment with the same length and direction as v is considered to be the same vector as v, regardless of its initial point. In the case of our vector v above, an vector which moves three units to the right and four up from its initial point to arrive at its terminal point is considered the same vector as v. The notation we use to capture this idea is the component form of the vector, v =,, where the first number,, is called the -component of v and the second number,, is called the -component of v. If we wanted to reconstruct v =, with initial point P,, then we would find the terminal point of v b adding to the -coordinate and adding to the -coordinate to obtain the terminal point Q, 7, as seen below. The word vector comes from the Latin vehere meaning to conve or to carr. Other tetbook authors use bold vectors such as v. We find that writing in bold font on the chalkboard is inconvenient at best, so we have chosen the arrow notation. If this idea of over and up seems familiar, it should. The slope of the line segment containing v is.

.8 Vectors 0 Q, 7 up P, over v =, with initial point P,. The component form of a vector is what ties these ver geometric objects back to Algebra and ultimatel Trigonometr. We generalize our eample in our definition below. Definition.5. Suppose v is represented b a directed line segment with initial point P 0, 0 and terminal point Q,. The component form of v is given b v = P Q = 0, 0 Using the language of components, we have that two vectors are equal if and onl if their corresponding components are equal. That is, v, v = v, v if and onl if v = v and v = v. Again, think about this before reading on. We now set about defining operations on vectors. Suppose we are given two vectors v and w. The sum, or resultant vector v + w is obtained as follows. First, plot v. Net, plot w so that its initial point is the terminal point of v. To plot the vector v + w we begin at the initial point of v and end at the terminal point of w. It is helpful to think of the vector v + w as the net result of moving along v then moving along w. v + w w v v, w, and v + w Our net eample makes good use of resultant vectors and reviews bearings and the Law of Cosines. Eample.8.. A plane leaves an airport with an airspeed 5 of 75 miles per hour at a bearing of N0 E. A 5 mile per hour wind is blowing at a bearing of S60 E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. If necessar, review page 905 and Section.. 5 That is, the speed of the plane relative to the air around it. If there were no wind, plane s airspeed would be the same as its speed as observed from the ground. How does wind affect this? Keep reading!

0 Applications of Trigonometr Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed as a magnitude with direction is the concept of velocit which we ve seen a few times before in this tetbook. 6 We let v denote the plane s velocit and w denote the wind s velocit in the diagram below. The true speed and bearing is found b analzing the resultant vector, v + w. From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of v, which is 75, the magnitude of w, which is 5, and the magnitude of v + w, which we ll call c. From the given bearing information, we go through the usual geometr to determine that the angle between the sides of length 5 and 75 measures 00. N N v v + w 00 5 0 0 75 α c 60 w E 60 E From the Law of Cosines, we determine c = 850 50 cos00 8, which means the true speed of the plane is approimatel 8 miles per hour. To determine the true bearing of the plane, we need to determine the angle α. Using the Law of Cosines once more, 7 we find cosα = c +900 50c so that α. Given the geometr of the situation, we add α to the given 0 and find the true bearing of the plane to be approimatel N5 E. Our net step is to define addition of vectors component-wise to match the geometric action. 8 Definition.6. Suppose v = v, v and w = w, w. The vector v + w is defined b v + w = v + w, v + w Eample.8.. Let v =, and suppose w = P Q where P, 7 and Q, 5. Find v + w and interpret this sum geometricall. Solution. Before can add the vectors using Definition.6, we need to write w in component form. Using Definition.5, we get w =, 5 7 =,. Thus 6 See Section 0.., for instance. 7 Or, since our given angle, 00, is obtuse, we could use the Law of Sines without an ambiguit here. 8 Adding vectors component-wise should seem hauntingl familiar. Compare this with how matri addition was defined in section 8.. In fact, in more advanced courses such as Linear Algebra, vectors are defined as n or n matrices, depending on the situation.

.8 Vectors 05 v + w =, +, = +, + =, To visualize this sum, we draw v with its initial point at 0, 0 for convenience so that its terminal point is,. Net, we graph w with its initial point at,. Moving one to the right and two down, we find the terminal point of w to be,. We see that the vector v + w has initial point 0, 0 and terminal point, so its component form is,, as required. v w v + w In order for vector addition to enjo the same kinds of properties as real number addition, it is necessar to etend our definition of vectors to include a zero vector, 0 = 0, 0. Geometricall, 0 represents a point, which we can think of as a directed line segment with the same initial and terminal points. The reader ma well object to the inclusion of 0, since after all, vectors are supposed to have both a magnitude length and a direction. While it seems clear that the magnitude of 0 should be 0, it is not clear what its direction is. As we shall see, the direction of 0 is in fact undefined, but this minor hiccup in the natural flow of things is worth the benefits we reap b including 0 in our discussions. We have the following theorem. Theorem.8. Properties of Vector Addition ˆ ˆ Commutative Propert: For all vectors v and w, v + w = w + v. Associative Propert: For all vectors u, v and w, u + v + w = u + v + w. ˆ Identit Propert: The vector 0 acts as the additive identit for vector addition. That is, for all vectors v, v + 0 = 0 + v = v. ˆ Inverse Propert: Ever vector v has a unique additive inverse, denoted v. That is, for ever vector v, there is a vector v so that v + v = v + v = 0.

06 Applications of Trigonometr The properties in Theorem.8 are easil verified using the definition of vector addition. 9 the commutative propert, we note that if v = v, v and w = w, w then For v + w = v, v + w, w = v + w, v + w = w + v, w + v = w + v Geometricall, we can see the commutative propert b realizing that the sums v + w and w + v are the same directed diagonal determined b the parallelogram below. v w w + v v + w w v Demonstrating the commutative propert of vector addition. The proofs of the associative and identit properties proceed similarl, and the reader is encouraged to verif them and provide accompaning diagrams. The eistence and uniqueness of the additive inverse is et another propert inherited from the real numbers. Given a vector v = v, v, suppose we wish to find a vector w = w, w so that v + w = 0. B the definition of vector addition, we have v + w, v + w = 0, 0, and hence, v + w = 0 and v + w = 0. We get w = v and w = v so that w = v, v. Hence, v has an additive inverse, and moreover, it is unique and can be obtained b the formula v = v, v. Geometricall, the vectors v = v, v and v = v, v have the same length, but opposite directions. As a result, when adding the vectors geometricall, the sum v + v results in starting at the initial point of v and ending back at the initial point of v, or in other words, the net result of moving v then v is not moving at all. v v Using the additive inverse of a vector, we can define the difference of two vectors, v w = v + w. If v = v, v and w = w, w then 9 The interested reader is encouraged to compare Theorem.8 and the ensuing discussion with Theorem 8. in Section 8. and the discussion there.

.8 Vectors 07 v w = v + w = v, v + w, w = v + w, v + w = v w, v w In other words, like vector addition, vector subtraction works component-wise. To interpret the vector v w geometricall, we note w + v w = w + v + w Definition of Vector Subtraction = w + w + v Commutativit of Vector Addition = w + w + v Associativit of Vector Addition = 0 + v Definition of Additive Inverse = v Definition of Additive Identit This means that the net result of moving along w then moving along v w is just v itself. From the diagram below, we see that v w ma be interpreted as the vector whose initial point is the terminal point of w and whose terminal point is the terminal point of v as depicted below. It is also worth mentioning that in the parallelogram determined b the vectors v and w, the vector v w is one of the diagonals the other being v + w. v v w v w w w w v v Net, we discuss scalar multiplication that is, taking a real number times a vector. We define scalar multiplication for vectors in the same wa we defined it for matrices in Section 8.. Definition.7. If k is a real number and v = v, v, we define k v b k v = k v, v = kv, kv Scalar multiplication b k in vectors can be understood geometricall as scaling the vector if k > 0 or scaling the vector and reversing its direction if k < 0 as demonstrated below.

08 Applications of Trigonometr v v v v Note that, b definition.7, v = v, v = v, v = v, v = v. This, and other properties of scalar multiplication are summarized below. Theorem.9. Properties of Scalar Multiplication ˆ Associative Propert: For ever vector v and scalars k and r, kr v = kr v. ˆ ˆ Identit Propert: For all vectors v, v = v. Additive Inverse Propert: For all vectors v, v = v. ˆ Distributive Propert of Scalar Multiplication over Scalar Addition: For ever vector v and scalars k and r, k + r v = k v + r v ˆ Distributive Propert of Scalar Multiplication over Vector Addition: For all vectors v and w and scalars k, k v + w = k v + k w ˆ Zero Product Propert: If v is vector and k is a scalar, then k v = 0 if and onl if k = 0 or v = 0 The proof of Theorem.9, like the proof of Theorem.8, ultimatel boils down to the definition of scalar multiplication and properties of real numbers. For eample, to prove the associative propert, we let v = v, v. If k and r are scalars then kr v = kr v, v = krv, krv Definition of Scalar Multiplication = krv, krv Associative Propert of Real Number Multiplication = k rv, rv Definition of Scalar Multiplication = k r v, v Definition of Scalar Multiplication = kr v

.8 Vectors 09 The remaining properties are proved similarl and are left as eercises. Our net eample demonstrates how Theorem.9 allows us to do the same kind of algebraic manipulations with vectors as we do with variables multiplication and division of vectors notwithstanding. If the pedantr seems familiar, it should. This is the same treatment we gave Eample 8.. in Section 8.. As in that eample, we spell out the solution in ecruciating detail to encourage the reader to think carefull about wh each step is justified. Eample.8.. Solve 5 v v +, = 0 for v. Solution. 5 v v +, = 0 5 v + [ v +, ] = 0 5 v + [ ] v +, = 0 5 v + v +, = 0 5 v + [ v +, ] = 0 5 v + [ v +, ] = 0 [5 v + v] +, = 0 5 + v +, = 0 v +, = 0 v +, +, = 0 +, v + [, +, ] = 0 +, v + 0 = 0 +, v =, [ v =, ] v =, v =, v =, A vector whose initial point is 0, 0 is said to be in standard position. If v = v, v is plotted in standard position, then its terminal point is necessaril v, v. Once more, think about this before reading on. v, v v = v, v in standard position.

00 Applications of Trigonometr Plotting a vector in standard position enables us to more easil quantif the concepts of magnitude and direction of the vector. We can convert the point v, v in rectangular coordinates to a pair r, θ in polar coordinates where r 0. The magnitude of v, which we said earlier was length of the directed line segment, is r = v + v and is denoted b v. From Section., we know v = r cosθ = v cosθ and v = r sinθ = v sinθ. From the definition of scalar multiplication and vector equalit, we get This motivates the following definition. v = v, v = v cosθ, v sinθ = v cosθ, sinθ Definition.8. Suppose v is a vector with component form v = v, v. Let r, θ be a polar representation of the point with rectangular coordinates v, v with r 0. ˆ The magnitude of v, denoted v, is given b v = r = v + v ˆ If v 0, the vector direction of v, denoted ˆv is given b ˆv = cosθ, sinθ Taken together, we get v = v cosθ, v sinθ. A few remarks are in order. First, we note that if v 0 then even though there are infinitel man angles θ which satisf Definition.8, the stipulation r > 0 means that all of the angles are coterminal. Hence, if θ and θ both satisf the conditions of Definition.8, then cosθ = cosθ and sinθ = sinθ, and as such, cosθ, sinθ = cosθ, sinθ making ˆv is well-defined. 0 If v = 0, then v = 0, 0, and we know from Section. that 0, θ is a polar representation for the origin for an angle θ. For this reason, ˆ0 is undefined. The following theorem summarizes the important facts about the magnitude and direction of a vector. Theorem.0. Properties of Magnitude and Direction: Suppose v is a vector. ˆ v 0 and v = 0 if and onl if v = 0 ˆ For all scalars k, k v = k v. ˆ If v 0 then v = v ˆv, so that ˆv = v v. The proof of the first propert in Theorem.0 is a direct consequence of the definition of v. If v = v, v, then v = v + v which is b definition greater than or equal to 0. Moreover, v + v = 0 if and onl of v + v = 0 if and onl if v = v = 0. Hence, v = 0 if and onl if v = 0, 0 = 0, as required. The second propert is a result of the definition of magnitude and scalar multiplication along with a proper of radicals. If v = v, v and k is a scalar then 0 If this all looks familiar, it should. The interested reader is invited to compare Definition.8 to Definition. in Section.7.

.8 Vectors 0 k v = k v, v = kv, kv Definition of scalar multiplication = kv + kv Definition of magnitude = k v + k v = k v + v = k v + v Product Rule for Radicals = k v + v Since k = k = k v The equation v = v ˆv in Theorem.0 is a consequence of the definitions of v and ˆv and was worked out in the discussion just prior to Definition.8 on page 00. In words, the equation v = v ˆv sas that an given vector is the product of its magnitude and its direction an important concept to keep in mind when studing and using vectors. The equation ˆv = v solving v = v ˆv for ˆv b multipling both sides of the equation b of Theorem.9. We are overdue for an eample. Eample.8.. v v is a result of and using the properties. Find the component form of the vector v with v = 5 so that when v is plotted in standard position, it lies in Quadrant II and makes a 60 angle with the negative -ais.. For v =,, find v and θ, 0 θ < so that v = v cosθ, sinθ.. For the vectors v =, and w =,, find the following. Solution. a ˆv b v w c v w d ŵ. We are told that v = 5 and are given information about its direction, so we can use the formula v = v ˆv to get the component form of v. To determine ˆv, we appeal to Definition.8. We are told that v lies in Quadrant II and makes a 60 angle with the negative -ais, so the polar form of the terminal point of v, when plotted in standard position is 5, 0. See the diagram below. Thus ˆv = cos 0, sin 0 = 5, = 5, 5. Of course, to go from v = v ˆv to ˆv = v,, so v = v ˆv = v, we are essentiall dividing both sides of the equation b the scalar v. The authors encourage the reader, however, to work out the details carefull to gain an appreciation of the properties in pla. Due to the utilit of vectors in real-world applications, we will usuall use degree measure for the angle when giving the vector s direction. However, since Carl doesn t want ou to forget about radians, he s made sure there are eamples and eercises which use them.

0 Applications of Trigonometr 5 v 60 θ = 0. For v =,, we get v = + = 6. In light of Definition.8, we can find the θ we re after b converting the point with rectangular coordinates, to polar form r, θ where r = v > 0. From Section., we have tanθ = =. Since, is a point in Quadrant IV, θ is a Quadrant IV angle. Hence, we pick θ = 5. We ma check our answer b verifing v =, = 6 cos 5, sin 5.. a Since we are given the component form of v, we ll use the formula ˆv = v v. For v =,, we have v = + = 5 = 5. Hence, ˆv = 5, = 5, 5. b We know from our work above that v = 5, so to find v w, we need onl find w. Since w =,, we get w = + = 5. Hence, v w = 5 5. c In the epression v w, notice that the arithmetic on the vectors comes first, then the magnitude. Hence, our first step is to find the component form of the vector v w. We get v w =,, =, 8. Hence, v w =, 8 = + 8 = 65. d To find ŵ, we first need ŵ. Using the formula ŵ = w which we found the in the previous problem, we get ŵ = 5, = 5 5 5, 5 5. Hence, ŵ = 5 + 5 5 = 5 5 + 0 5 = =. w along with w = 5, 5, 5 = The process eemplified b number in Eample.8. above b which we take information about the magnitude and direction of a vector and find the component form of a vector is called resolving a vector into its components. As an application of this process, we revisit Eample.8. below. Eample.8.5. A plane leaves an airport with an airspeed of 75 miles per hour with bearing N0 E. A 5 mile per hour wind is blowing at a bearing of S60 E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution: We proceed as we did in Eample.8. and let v denote the plane s velocit and w denote the wind s velocit, and set about determining v + w. If we regard the airport as being

.8 Vectors 0 at the origin, the positive -ais acting as due north and the positive -ais acting as due east, we see that the vectors v and w are in standard position and their directions correspond to the angles 50 and 0, respectivel. Hence, the component form of v = 75 cos50, sin50 = 75 cos50, 75 sin50 and the component form of w = 5 cos 0, 5 sin 0. Since we have no convenient wa to epress the eact values of cosine and sine of 50, we leave both vectors in terms of cosines and sines. Adding corresponding components, we find the resultant vector v + w = 75 cos50 + 5 cos 0, 75 sin50 + 5 sin 0. To find the true speed of the plane, we compute the magnitude of this resultant vector v + w = 75 cos50 + 5 cos 0 + 75 sin50 + 5 sin 0 8 Hence, the true speed of the plane is approimatel 8 miles per hour. To find the true bearing, we need to find the angle θ which corresponds to the polar form r, θ, r > 0, of the point, = 75 cos50 + 5 cos 0, 75 sin50 + 5 sin 0. Since both of these coordinates are positive, we know θ is a Quadrant I angle, as depicted below. Furthermore, tanθ = = 75 sin50 + 5 sin 0 75 cos50 + 5 cos 0, so using the arctangent function, we get θ 9. Since, for the purposes of bearing, we need the angle between v+ w and the positive -ais, we take the complement of θ and find the true bearing of the plane to be approimatel N5 E. N N v v v + w 0 50 θ 60 0 w E w E In part d of Eample.8., we saw that ŵ =. Vectors with length have a special name and are important in our further stud of vectors. Definition.9. Unit Vectors: Let v be a vector. If v =, we sa that v is a unit vector. Keeping things calculator friendl, for once! Yes, a calculator approimation is the quickest wa to see this, but ou can also use good old-fashioned inequalities and the fact that 5 50 60.

0 Applications of Trigonometr If v is a unit vector, then necessaril, v = v ˆv = ˆv = ˆv. Conversel, we leave it as an eercise 5 to show that ˆv = v is a unit vector for an nonzero vector v. In practice, if v is a unit v vector we write it as ˆv as opposed to v because we have reserved the ˆ notation for unit vectors. The process of multipling a nonzero vector b the factor v to produce a unit vector is called normalizing the vector, and the resulting vector ˆv is called the unit vector in the direction of v. The terminal points of unit vectors, when plotted in standard position, lie on the Unit Circle. You should take the time to show this. As a result, we visualize normalizing a nonzero vector v as shrinking 6 its terminal point, when plotted in standard position, back to the Unit Circle. v ˆv Visualizing vector normalization ˆv = v v Of all of the unit vectors, two deserve special mention. Definition.0. The Principal Unit Vectors: ˆ The vector î is defined b î =, 0 ˆ The vector ĵ is defined b î = 0, We can think of the vector î as representing the positive -direction, while ĵ represents the positive -direction. We have the following decomposition theorem. 7 Theorem.. Principal Vector Decomposition Theorem: component form v = v, v. Then v = v î + v ĵ. Let v be a vector with The proof of Theorem. is straightforward. Since î =, 0 and ĵ = 0,, we have from the definition of scalar multiplication and vector addition that v î + v ĵ = v, 0 + v 0, = v, 0 + 0, v = v, v = v 5 One proof uses the properties of scalar multiplication and magnitude. If v 0, consider ˆv = the fact that v 0 is a scalar and consider factoring. 6... if v >... 7 We will see a generalization of Theorem. in Section.9. Sta tuned! v v. Use

.8 Vectors 05 Geometricall, the situation looks like this: v ĵ v = v, v ĵ î v î v = v, v = v î + v ĵ. We conclude this section with a classic eample which demonstrates how vectors are used to model forces. A force is defined as a push or a pull. The intensit of the push or pull is the magnitude of the force, and is measured in Netwons N in the SI sstem or pounds lbs. in the English sstem. 8 The following eample uses all of the concepts in this section, and should be studied in great detail. Eample.8.6. A 50 pound speaker is suspended from the ceiling b two support braces. If one of them makes a 60 angle with the ceiling and the other makes a 0 angle with the ceiling, what are the tensions on each of the supports? Solution. We represent the problem schematicall below and then provide the corresponding vector diagram. 0 60 0 60 T T 0 60 50 lbs. w We have three forces acting on the speaker: the weight of the speaker, which we ll call w, pulling the speaker directl downward, and the forces on the support rods, which we ll call T and T for tensions acting upward at angles 60 and 0, respectivel. We are looking for the tensions on the support, which are the magnitudes T and T. In order for the speaker to remain stationar, 9 we require w + T + T = 0. Viewing the common initial point of these vectors as the 8 See also Section... 9 This is the criteria for static equilbrium.

06 Applications of Trigonometr origin and the dashed line as the -ais, we use Theorem.0 to get component representations for the three vectors involved. We can model the weight of the speaker as a vector pointing directl downwards with a magnitude of 50 pounds. That is, w = 50 and ŵ = ĵ = 0,. Hence, w = 50 0, = 0, 50. For the force in the first support, we get T = T cos 60, sin 60 T =, T For the second support, we note that the angle 0 is measured from the negative -ais, so the angle needed to write T in component form is 50. Hence T = T cos 50, sin 50 = T, T The requirement w + T + T = 0 gives us this vector equation. 0, 50 + T T, T + T w + T + T = 0, T T, T + T 50 = 0, 0 = 0, 0 Equating the corresponding components of the vectors on each side, we get a sstem of linear equations in the variables T and T. E E T T T + T = 0 50 = 0 From E, we get T = T. Substituting that into E gives T + T 50 = 0 which ields T 50 = 0. Hence, T = 5 pounds and T = T = 5 pounds.

.8 Vectors 07.8. Eercises In Eercises - 0, use the given pair of vectors v and w to find the following quantities. State whether the result is a vector or a scalar. v + w w v v + w v + w v w w v w ˆv Finall, verif that the vectors satisf the Parallelogram Law v + w = [ v + w + v w ]. v =, 5, w =,. v = 7,, w = 5,. v =,, w =,. v = 0,, w =, 5 5. v =,, w =, 6. v = 5, 5, w = 5, 5 7. v =,, w =, 8. v =,, w =, 9. v = î + ĵ, w = ĵ 0. v = î + ĵ, w = î ĵ In Eercises - 5, find the component form of the vector v using the information given about its magnitude and direction. Give eact values.. v = 6; when drawn in standard position v lies in Quadrant I and makes a 60 angle with the positive -ais. v = ; when drawn in standard position v lies in Quadrant I and makes a 5 angle with the positive -ais. v = ; when drawn in standard position v lies in Quadrant I and makes a 60 angle with the positive -ais. v = ; when drawn in standard position v lies along the positive -ais 5. v = ; when drawn in standard position v lies in Quadrant II and makes a 0 angle with the negative -ais 6. v = ; when drawn in standard position v lies in Quadrant II and makes a 0 angle with the positive -ais 7. v = 7 ; when drawn in standard position v lies along the negative -ais 8. v = 5 6; when drawn in standard position v lies in Quadrant III and makes a 5 angle with the negative -ais 9. v = 6.5; when drawn in standard position v lies along the negative -ais

08 Applications of Trigonometr 0. v = ; when drawn in standard position v lies in Quadrant IV and makes a 0 angle with the positive -ais. v = 5 ; when drawn in standard position v lies in Quadrant IV and makes a 5 angle with the negative -ais. v = 5; when drawn in standard position v lies in Quadrant I and makes an angle measuring arctan with the positive -ais. v = 0; when drawn in standard position v lies in Quadrant II and makes an angle measuring arctan with the negative -ais. v = 5; when drawn in standard position v lies in Quadrant III and makes an angle measuring arctan with the negative -ais 5. v = 6; when drawn in standard position v lies in Quadrant IV and makes an angle measuring arctan 5 with the positive -ais In Eercises 6 -, approimate the component form of the vector v using the information given about its magnitude and direction. Round our approimations to two decimal places. 6. v = 9; when drawn in standard position v makes a 7 angle with the positive -ais 7. v = 6.9; when drawn in standard position v makes a 78. angle with the positive -ais 8. v = 580; when drawn in standard position v makes a angle with the positive -ais 9. v = 50; when drawn in standard position v makes a 0.75 angle with the positive -ais 0. v = 68.7; when drawn in standard position v makes a 5 angle with the positive -ais. v = 6; when drawn in standard position v makes a 0.5 angle with the positive -ais In Eercises - 5, for the given vector v, find the magnitude v and an angle θ with 0 θ < 60 so that v = v cosθ, sinθ See Definition.8. Round approimations to two decimal places.. v =,. v = 5, 5. v =, 5. v =, 6. v =, 7. v =, 8. v = 6, 0 9. v =.5, 0 0. v = 0, 7. v = 0ĵ. v =,. v =, 5

.8 Vectors 09. v =, 5. v = 7, 6. v =, 7. v =, 6 8. v = î + ĵ 9. v = î ĵ 50. v =., 77.05 5. v = 965.5, 8.6 5. v =.,. 5. A small boat leaves the dock at Camp DuNuthin and heads across the Nessie River at 7 miles per hour that is, with respect to the water at a bearing of S68 W. The river is flowing due east at 8 miles per hour. What is the boat s true speed and heading? Round the speed to the nearest mile per hour and epress the heading as a bearing, rounded to the nearest tenth of a degree. 5. The HMS Sasquatch leaves port with bearing S0 E maintaining a speed of miles per hour that is, with respect to the water. If the ocean current is 5 miles per hour with a bearing of N60 E, find the HMS Sasquatch s true speed and bearing. Round the speed to the nearest mile per hour and epress the heading as a bearing, rounded to the nearest tenth of a degree. 55. If the captain of the HMS Sasquatch in Eercise 5 wishes to reach Chupacabra Cove, an island 00 miles awa at a bearing of S0 E from port, in three hours, what speed and heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and epress the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If v denotes the velocit of the HMS Sasquatch and w denotes the velocit of the current, what does v + w need to be to reach Chupacabra Cove in three hours? 56. In calm air, a plane fling from the Pedimaus International Airport can reach Cliffs of Insanit Point in two hours b following a bearing of N8. E at 96 miles an hour. The distance between the airport and the cliffs is 9 miles. If the wind is blowing from the southeast at 5 miles per hour, what speed and bearing should the pilot take so that she makes the trip in two hours along the original heading? Round the speed to the nearest hundredth of a mile per hour and our angle to the nearest tenth of a degree. 57. The SS Bigfoot leaves Yeti Ba on a course of N7 W at a speed of 50 miles per hour. After traveling half an hour, the captain determines he is 0 miles from the ba and his bearing back to the ba is S0 E. What is the speed and bearing of the ocean current? Round the speed to the nearest mile per hour and epress the heading as a bearing, rounded to the nearest tenth of a degree. 58. A 600 pound Sasquatch statue is suspended b two cables from a gmnasium ceiling. If each cable makes a 60 angle with the ceiling, find the tension on each cable. Round our answer to the nearest pound. 59. Two cables are to support an object hanging from a ceiling. If the cables are each to make a angle with the ceiling, and each cable is rated to withstand a maimum tension of 00 pounds, what is the heaviest object that can be supported? Round our answer down to the nearest pound.

00 Applications of Trigonometr 60. A 00 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 7 angle with the ceiling while the right hand cable makes a 8 angle with the ceiling. What is the tension on each of the cables? Round our answers to three decimal places. 6. Two drunken college students have filled an empt beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 00 pounds at a heading of N77 E and the other pulls at a heading of S68 E. What force should the weaker student appl to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round our answer to the nearest pound. 6. Emboldened b the success of their late night keg pull in Eercise 6 above, our intrepid oung scholars have decided to pa homage to the chariot race scene from the movie Ben-Hur b ting three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N80 W, the second points due west and the third points S80 W. The force applied to the first rope is 00 pounds, the force applied to the second rope is 0 pounds and the force applied b the non-riding friend to the third rope is 60 pounds. The need the resultant force to be at least 00 pounds otherwise the couch won t move. Does it move? If so, is it heading due west? 6. Let v = v, v be an non-zero vector. Show that v has length. v 6. We sa that two non-zero vectors v and w are parallel if the have same or opposite directions. That is, v 0 and w 0 are parallel if either ˆv = ŵ or ˆv = ŵ. Show that this means v = k w for some non-zero scalar k and that k > 0 if the vectors have the same direction and k < 0 if the point in opposite directions. 65. The goal of this eercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line =. Let v 0 = 0, and let s =,. Let t be an real number. Show that the vector defined b v = v 0 + t s, when drawn in standard position, has its terminal point on the line =. Hint: Show that v 0 + t s = t, t for an real number t. Now consider the non-vertical line = m + b. Repeat the previous analsis with v 0 = 0, b and let s =, m. Thus an non-vertical line can be thought of as a collection of terminal points of the vector sum of 0, b the position vector of the -intercept and a scalar multiple of the slope vector s =, m. 66. Prove the associative and identit properties of vector addition in Theorem.8. 67. Prove the properties of scalar multiplication in Theorem.9.

.8 Vectors 0.8. Answers. ˆ v + w = 5,, vector ˆ w v =,, vector ˆ v + w = 6, scalar ˆ v + w = 8, scalar ˆ v w w v =, 77, vector ˆ w ˆv = 60, 5, vector. ˆ v + w =,, vector ˆ w v = 9, 60, vector ˆ v + w =, scalar ˆ v + w = 8, scalar ˆ v w w v =, 6, vector ˆ w ˆv = 9 5, 5, vector. ˆ v + w = 0,, vector ˆ w v = 6, 6, vector ˆ v + w =, scalar ˆ v + w = 5, scalar ˆ v w w v = 6 5, 6 5, vector ˆ w ˆv =,, vector. ˆ v + w = 8, 9, vector ˆ w v =,, vector ˆ v + w = 5, scalar ˆ v + w = 9, scalar ˆ v w w v = 9, 6 9, vector ˆ w ˆv = 5,, vector 5. ˆ v + w =,, vector ˆ w v =, 0, vector ˆ v + w =, scalar ˆ v + w = 6, scalar ˆ v w w v = 8, 0, vector ˆ w ˆv =,, vector 6. ˆ v + w = 5, 7 5, vector ˆ w v =,, vector ˆ v + w =, scalar ˆ v + w =, scalar ˆ v w w v = 7 5, 5, vector ˆ w ˆv = 5, 5, vector 7. ˆ v + w = 0, 0, vector ˆ w v =,, vector ˆ v + w = 0, scalar ˆ v + w =, scalar ˆ v w w v =,, vector ˆ w ˆv =,, vector

0 Applications of Trigonometr 8. ˆ v + w =,, vector ˆ w v =,, vector ˆ v + w =, scalar ˆ v + w =, scalar ˆ v w w v =,, vector ˆ w ˆv =,, vector 9. ˆ v + w =,, vector ˆ w v = 6, 0, vector ˆ v + w =, scalar ˆ v + w = 7, scalar ˆ v w w v = 6, 8, vector ˆ w ˆv = 6 5, 8 5, vector 0. ˆ v + w =, 0, vector ˆ w v =,, vector ˆ v + w =, scalar ˆ v + w =, scalar ˆ v w w v = 0,, vector ˆ w ˆv =,, vector. v =,. v =,. v =. v = 0, 5. v =, 6. v =, 7. v = 7, 0 8. v = 5, 5 9. v = 0, 6.5 0. v = 6,. v = 5, 5. v =,. v =,. v =, 5. v =, 0 6. v 77.96, 9.7 7. v.96, 6.59 8. v 56.6, 097.77 9. v 86.7, 0.08 0. v 5., 60.. v.7,.. v =, θ = 60. v = 5, θ = 5. v =, θ = 50 5. v =, θ = 5 6. v =, θ = 5 7. v =, θ = 0 8. v = 6, θ = 0 9. v =.5, θ = 80 0. v = 7, θ = 90. v = 0, θ = 70. v = 5, θ 5.. v =, θ.6. v = 5, θ. 5. v = 5, θ 06.6 6. v = 5, θ 06.57,

.8 Vectors 0 7. v = 0, θ 5.57 8. v =, θ 5 9. v = 7, θ 8.0 50. v 5.8, θ 8.0 5. v 7.00, θ 0.75 5. v.69, θ 59.66 5. The boat s true speed is about 0 miles per hour at a heading of S50.6 W. 5. The HMS Sasquatch s true speed is about miles per hour at a heading of S6.8 E. 55. She should maintain a speed of about 5 miles per hour at a heading of S.8 E. 56. She should fl at 8.6 miles per hour with a heading of N. E 57. The current is moving at about 0 miles per hour bearing N5.6 W. 58. The tension on each of the cables is about 6 pounds. 59. The maimum weight that can be held b the cables in that configuration is about pounds. 60. The tension on the left hand cable is 85.7 lbs. and on the right hand cable is 9.705 lbs. 6. The weaker student should pull about 60 pounds. The net force on the keg is about 5 pounds. 6. The resultant force is onl about 96 pounds so the couch doesn t budge. Even if it did move, the stronger force on the third rope would have made the couch drift slightl to the south as it traveled down the street.

0 Applications of Trigonometr.9 The Dot Product and Projection In Section.8, we learned how add and subtract vectors and how to multipl vectors b scalars. In this section, we define a product of vectors. We begin with the following definition. Definition.. Suppose v and w are vectors whose component forms are v = v, v and w = w, w. The dot product of v and w is given b v w = v, v w, w = v w + v w For eample, let v =, and w =,. Then v w =,, = + = 5. Note that the dot product takes two vectors and produces a scalar. For that reason, the quantit v w is often called the scalar product of v and w. The dot product enjos the following properties. Theorem.. Properties of the Dot Product ˆ ˆ ˆ Commutative Propert: For all vectors v and w, v w = w v. Distributive Propert: For all vectors u, v and w, u v + w = u v + u w. Scalar Propert: For all vectors v and w and scalars k, k v w = k v w = v k w. ˆ Relation to Magnitude: For all vectors v, v v = v. Like most of the theorems involving vectors, the proof of Theorem. amounts to using the definition of the dot product and properties of real number arithmetic. To show the commutative propert for instance, let v = v, v and w = w, w. Then v w = v, v w, w = v w + v w Definition of Dot Product = w v + w v Commutativit of Real Number Multiplication = w, w v, v Definition of Dot Product = w v The distributive propert is proved similarl and is left as an eercise. For the scalar propert, assume that v = v, v and w = w, w and k is a scalar. Then k v w = k v, v w, w = kv, kv w, w Definition of Scalar Multiplication = kv w + kv w Definition of Dot Product = kv w + kv w Associativit of Real Number Multiplication = kv w + v w Distributive Law of Real Numbers = k v, v w, w Definition of Dot Product = k v w We leave the proof of k v w = v k w as an eercise.

.9 The Dot Product and Projection 05 For the last propert, we note that if v = v, v, then v v = v, v v, v = v + v = v, where the last equalit comes courtes of Definition.8. The following eample puts Theorem. to good use. As in Eample.8., we work out the problem in great detail and encourage the reader to suppl the justification for each step. Eample.9.. Prove the identit: v w = v v w + w. Solution. We begin b rewriting v w in terms of the dot product using Theorem.. v w = v w v w = v + [ w] v + [ w] = v + [ w] v + v + [ w] [ w] = v v + [ w] + [ w] v + [ w] = v v + v [ w] + [ w] v + [ w] [ w] = v v + v [ w] + [ w] v + [ w] [ w] = v v + v w + w v + [ ] w w = v v + v w + v w + w w = v v v w + w w = v v w + w Hence, v w = v v w + w as required. If we take a step back from the pedantr in Eample.9., we see that the bulk of the work is needed to show that v w v w = v v v w+ w w. If this looks familiar, it should. Since the dot product enjos man of the same properties enjoed b real numbers, the machinations required to epand v w v w for vectors v and w match those required to epand v wv w for real numbers v and w, and hence we get similar looking results. The identit verified in Eample.9. plas a large role in the development of the geometric properties of the dot product, which we now eplore. Suppose v and w are two nonzero vectors. If we draw v and w with the same initial point, we define the angle between v and w to be the angle θ determined b the ras containing the vectors v and w, as illustrated below. We require 0 θ. Think about wh this is needed in the definition. w v w θ v v w θ = 0 0 < θ < θ = The following theorem gives us some insight into the geometric role the dot product plas. Theorem.. Geometric Interpretation of Dot Product: If v and w are nonzero vectors then v w = v w cosθ, where θ is the angle between v and w.

06 Applications of Trigonometr We prove Theorem. in cases. If θ = 0, then v and w have the same direction. It follows that there is a real number k > 0 so that w = k v. Hence, v w = v k v = k v v = k v = k v v. Since k > 0, k = k, so k v = k v = k v b Theorem.0. Hence, k v v = v k v = v k v = v w. Since cos0 =, we get v w = k v v = v w = v w cos0, proving that the formula holds for θ = 0. If θ =, we repeat the argument with the difference being w = k v where k < 0. In this case, k = k, so k v = k v = k v = w. Since cos =, we get v w = v w = v w cos, as required. Net, if 0 < θ <, the vectors v, w and v w determine a triangle with side lengths v, w and v w, respectivel, as seen below. w θ v w v w v w θ v The Law of Cosines ields v w = v + w v w cosθ. From Eample.9., we know v w = v v w + w. Equating these two epressions for v w gives v + w v w cosθ = v v w+ w which reduces to v w cosθ = v w, or v w = v w cosθ, as required. An immediate consequence of Theorem. is the following. Theorem.. Let v and w be nonzero vectors and let θ the angle between v and w. Then v w θ = arccos = arccosˆv ŵ v w We obtain the formula in Theorem. b solving the equation given in Theorem. for θ. Since v and w are nonzero, so are v and w. Hence, we ma divide both sides of v w = v w cosθ v w b v w to get cosθ = v w. Since 0 θ b definition, the values of θ eactl match the range of the arccosine function. Hence, θ = arccos v w v w. Using Theorem., we can rewrite v w v w = v v w w = ˆv ŵ, giving us the alternative formula θ = arccosˆv ŵ. We are overdue for an eample. Eample.9.. Find the angle between the following pairs of vectors.. v =,, and w =,. v =,, and w = 5, 5. v =,, and w =, Solution. We use the formula θ = arccos v w v w from Theorem. in each case below. Since v = v ˆv and w = w ŵ, if ˆv = ŵ then w = w ˆv = w w w v ˆv = v. In this case, k = > 0. v v v

.9 The Dot Product and Projection 07. We have v w =,, = = 6. Since v = + = 6 = 6 and w = + = =, θ = arccos = arccos = 5 6. 6. For v =, and w = 5, 5, we find v w =, 5, 5 = 0 0 = 0. Hence, it doesn t matter what v and w are, θ = arccos v w v w = arccos0 =.. We find v w =,, = 6 =. Also v = + = 5 = 5 and w = + = 5, so θ = arccos 5 = arccos 5 5 5. Since 5 5 isn t the cosine of one of the common angles, we leave our answer as θ = arccos. The vectors v =,, and w = 5, 5 in Eample.9. are called orthogonal and we write v w, because the angle between them is radians = 90. Geometricall, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. w 5 5 v v and w are orthogonal, v w We state the relationship between orthogonal vectors and their dot product in the following theorem. Theorem.5. The Dot Product Detects Orthogonalit: Let v and w be nonzero vectors. Then v w if and onl if v w = 0. To prove Theorem.5, we first assume v and w are nonzero vectors with v w. B definition, the angle between v and w is. B Theorem., v w = v w cos = 0. Conversel, if v and w are nonzero vectors and v w = 0, then Theorem. gives θ = arccos v w v w = arccos 0 v w = arccos0 =, so v w. We can use Theorem.5 in the following eample to provide a different proof about the relationship between the slopes of perpendicular lines. Eample.9.. Let L be the line = m + b and let L be the line = m + b. Prove that L is perpendicular to L if and onl if m m =. Solution. Our strateg is to find two vectors: v, which has the same direction as L, and v, which has the same direction as L and show v v if and onl if m m =. To that end, we substitute = 0 and = into = m + b to find two points which lie on L, namel P 0, b Note that there is no zero product propert for the dot product since neither v nor w is 0, et v w = 0. See Eercise.. in Section..

08 Applications of Trigonometr and Q, m + b. We let v = P Q = 0, m + b b =, m, and note that since v is determined b two points on L, it ma be viewed as ling on L. Hence it has the same direction as L. Similarl, we get the vector v =, m which has the same direction as the line L. Hence, L and L are perpendicular if and onl if v v. According to Theorem.5, v v if and onl if v v = 0. Notice that v v =, m, m = + m m. Hence, v v = 0 if and onl if + m m = 0, which is true if and onl if m m =, as required. While Theorem.5 certainl gives us some insight into what the dot product means geometricall, there is more to the stor of the dot product. Consider the two nonzero vectors v and w drawn with a common initial point O below. For the moment, assume that the angle between v and w, which we ll denote θ, is acute. We wish to develop a formula for the vector p, indicated below, which is called the orthogonal projection of v onto w. The vector p is obtained geometricall as follows: drop a perpendicular from the terminal point T of v to the vector w and call the point of intersection R. The vector p is then defined as p = OR. Like an vector, p is determined b its magnitude p and its direction ˆp according to the formula p = p ˆp. Since we want ˆp to have the same direction as w, we have ˆp = ŵ. To determine p, we make use of Theorem 0. as applied to the right triangle ORT. We find cosθ = p v, or p = v cosθ. To get things in terms of v w cosθ v w just v and w, we use Theorem. to get p = v cosθ = w = w. Using Theorem v w., we rewrite w = v w w = v ŵ. Hence, p = v ŵ, and since ˆp = ŵ, we now have a formula for p completel in terms of v and w, namel p = p ˆp = v ŵŵ. v v T T w w v O θ O θ p = OR R O θ p R Now suppose that the angle θ between v and w is obtuse, and consider the diagram below. In this case, we see that ˆp = ŵ and using the triangle ORT, we find p = v cosθ. Since θ+θ =, it follows that cosθ = cosθ, which means p = v cosθ = v cosθ. Rewriting this last equation in terms of v and w as before, we get p = v ŵ. Putting this together with ˆp = ŵ, we get p = p ˆp = v ŵ ŵ = v ŵŵ in this case as well.

.9 The Dot Product and Projection 09 T v θ w R θ O p = OR If the angle between v and w is then it is eas to show that p = 0. Since v w in this case, v w = 0. It follows that v ŵ = 0 and p = 0 = 0ŵ = v ŵŵ in this case, too. This gives us Definition.. Let v and w be nonzero vectors. The orthogonal projection of v onto w, denoted proj w v is given b proj w v = v ŵŵ. Definition. gives us a good idea what the dot product does. The scalar v ŵ is a measure of how much of the vector v is in the direction of the vector w and is thus called the scalar projection of v onto w. While the formula given in Definition. is theoreticall appealing, because of the presence of the normalized unit vector ŵ, computing the projection using the formula proj w v = v ŵŵ can be mess. We present two other formulas that are often used in practice. Theorem.6. Alternate Formulas for Vector Projections: If v and w are nonzero vectors then v w v w proj w v = v ŵŵ = w w = w w w The proof of Theorem.6, which we leave to the reader as an eercise, amounts to using the formula ŵ = w w and properties of the dot product. It is time for an eample. Eample.9.. Let v =, 8 and w =,. Find p = proj w v, and plot v, w and p in standard position. Solution. We find v w =, 8, = + 6 = 5 and w w =,, = + = 5. v w Hence, p = w w w = 5 5, =, 6. We plot v, w and p below. In this case, the point R coincides with the point O, so p = OR = OO = 0.

00 Applications of Trigonometr p w 8 7 6 5 v Suppose we wanted to verif that our answer p in Eample.9. is indeed the orthogonal projection of v onto w. We first note that since p is a scalar multiple of w, it has the correct direction, so what remains to check is the orthogonalit condition. Consider the vector q whose initial point is the terminal point of p and whose terminal point is the terminal point of v. p w q 8 7 6 5 v From the definition of vector arithmetic, p+ q = v, so that q = v p. In the case of Eample.9., v =, 8 and p =, 6, so q =, 8, 6 =,. Then q w =,, = + = 0, which shows q w, as required. This result is generalized in the following theorem. Theorem.7. Generalized Decomposition Theorem: Let v and w be nonzero vectors. There are unique vectors p and q such that v = p + q where p = k w for some scalar k, and q w = 0. Note that if the vectors p and q in Theorem.7 are nonzero, then we can sa p is parallel 5 to w and q is orthogonal to w. In this case, the vector p is sometimes called the vector component of v parallel to w and q is called the vector component of v orthogonal to w. To prove Theorem.7, we take p = proj w v and q = v p. Then p is, b definition, a scalar multiple of w. Net, we compute q w. 5 See Eercise 6 in Section.8.

.9 The Dot Product and Projection 0 q w = v p w Definition of q. = v w p w Properties of Dot Product v w = v w w w w w Since p = proj w v. v w = v w w w Properties of Dot Product. w w = v w v w = 0 Hence, q w = 0, as required. At this point, we have shown that the vectors p and q guaranteed b Theorem.7 eist. Now we need to show that the are unique. Suppose v = p + q = p + q where the vectors p and q satisf the same properties described in Theorem.7 as p and q. Then p p = q q, so w p p = w q q = w q w q = 0 0 = 0. Hence, w p p = 0. Now there are scalars k and k so that p = k w and p = k w. This means w p p = w k w k w = w [k k ] w = k k w w = k k w. Since w 0, w 0, which means the onl wa w p p = k k w = 0 is for k k = 0, or k = k. This means p = k w = k w = p. With q q = p p = p p = 0, it must be that q = q as well. Hence, we have shown there is onl one wa to write v as a sum of vectors as described in Theorem.7. We close this section with an application of the dot product. In Phsics, if a constant force F is eerted over a distance d, the work W done b the force is given b W = F d. Here, we assume the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to find the work done. Consider the scenario below where the constant force F is applied to move an object from the point P to the point Q. F F P θ Q θ To find the work W done in this scenario, we need to find how much of the force F is in the direction of the motion P Q. This is precisel what the dot product F P Q represents. Since the distance the object travels is P Q, we get W = F P Q P Q. Since P Q = P Q P Q, W = F P Q P Q = F P Q P Q = F P Q = F P Q cosθ, where θ is the angle between the applied force F and the trajector of the motion P Q. We have proved the following.

0 Applications of Trigonometr Theorem.8. Work as a Dot Product: Suppose a constant force F is applied along the vector P Q. The work W done b F is given b W = F P Q = F P Q cosθ, where θ is the angle between F and P Q. Eample.9.5. Talor eerts a force of 0 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a 0 angle with the horizontal, how much work did Talor do pulling the wagon? Assume Talor eerts the force of 0 pounds at a 0 angle for the duration of the 50 feet. 0 Solution. There are two was to attack this problem. One wa is to find the vectors F and P Q mentioned in Theorem.8 and compute W = F P Q. To do this, we assume the origin is at the point where the handle of the wagon meets the wagon and the positive -ais lies along the dashed line in the figure above. Since the force applied is a constant 0 pounds, we have F = 0. Since it is being applied at a constant angle of θ = 0 with respect to the positive -ais, Definition.8 gives us F = 0 cos0, sin0 = 5, 5. Since the wagon is being pulled along 50 feet in the positive direction, the displacement vector is P Q = 50î = 50, 0 = 50, 0. We get W = F P Q = 5, 5 50, 0 = 50. Since force is measured in pounds and distance is measured in feet, we get W = 50 foot-pounds. Alternativel, we can use the formulation W = F P Q cosθ to get W = 0 pounds50 feet cos 0 = 50 foot-pounds of work.

.9 The Dot Product and Projection 0.9. Eercises In Eercises - 0, use the pair of vectors v and w to find the following quantities. ˆ ˆ v w The angle θ in degrees between v and w ˆ proj w v ˆ q = v proj w v Show that q w = 0.. v =, 7 and w = 5, 9. v = 6, 5 and w = 0,. v =, and w =,. v =, and w = 6, 8 5. v =, and w =, 6 6. v =, and w =, 7. v =, 7 and w =, 0 8. v =, and w = 5, 9. v =, and w =, 5 0. v = 5, 6 and w =, 7. v = 8, and w =, 6. v =, 9 and w = 0,. v = î ĵ and w = ĵ. v = î + 7ĵ and w = î 5. v = î + ĵ and w = î ĵ 6. v = 5î + ĵ and w = î + ĵ 7. v =, and w =, 8. v =, and w =, 9. v =, and w =, 0. v =, and w =,. A force of 500 pounds is required to tow a trailer. Find the work done towing the trailer along a flat stretch of road 00 feet. Assume the force is applied in the direction of the motion.. Find the work done lifting a 0 pound book feet straight up into the air. Assume the force of gravit is acting straight downwards.. Suppose Talor fills her wagon with rocks and must eert a force of pounds to pull her wagon across the ard. If she maintains a 5 angle between the handle of the wagon and the horizontal, compute how much work Talor does pulling her wagon 5 feet. Round our answer to two decimal places.. In Eercise 6 in Section.8, two drunken college students have filled an empt beer keg with rocks which the drag down the street b pulling on two attached ropes. The stronger of the two students pulls with a force of 00 pounds on a rope which makes a angle with the direction of motion. In this case, the keg was being pulled due east and the student s heading was N77 E. Find the work done b this student if the keg is dragged feet.

0 Applications of Trigonometr 5. Find the work done pushing a 00 pound barrel 0 feet up a.5 incline. Ignore all forces acting on the barrel ecept gravit, which acts downwards. Round our answer to two decimal places. HINT: Since ou are working to overcome gravit onl, the force being applied acts directl upwards. This means that the angle between the applied force in this case and the motion of the object is not the.5 of the incline! 6. Prove the distributive propert of the dot product in Theorem.. 7. Finish the proof of the scalar propert of the dot product in Theorem.. 8. Use the identit in Eample.9. to prove the Parallelogram Law v + w = [ v + w + v w ] 9. We know that + + for all real numbers and b the Triangle Inequalit established in Eercise 6 in Section.. We can now establish a Triangle Inequalit for vectors. In this eercise, we prove that u + v u + v for all pairs of vectors u and v. a Step Show that u + v = u + u v + v. b Step Show that u v u v. This is the celebrated Cauch-Schwarz Inequalit. 6 Hint: To show this inequalit, start with the fact that u v = u v cosθ and use the fact that cosθ for all θ. c Step Show that u + v = u + u v + v u + u v + v u + u v + v = u + v. d Step Use Step to show that u + v u + v for all pairs of vectors u and v. e As an added bonus, we can now show that the Triangle Inequalit z + w z + w holds for all comple numbers z and w as well. Identif the comple number z = a + bi with the vector u = a, b and identif the comple number w = c + di with the vector v = c, d and just follow our nose! 6 It is also known b other names. Check out this site for details.

.9 The Dot Product and Projection 05.9. Answers. v =, 7 and w = 5, 9 v w = 5 θ = 5 proj w v = 5, 9 q = 9, 5. v =, and w =, v w = θ = 0 proj w v =, q =, 5. v =, and w =, 6 v w = 0 θ = 90 proj w v = 0, 0 q =,. v = 6, 5 and w = 0, v w = 0 θ = 90 proj w v = 0, 0 q = 6, 5. v =, and w = 6, 8 v w = 50 θ = 80 proj w v =, q = 0, 0 6. v =, and w =, v w = 6 θ = 60 proj w v =, q =, 9 7. v =, 7 and w =, 0 v w = θ 9.7 proj w v =, 0 q = 0, 7 8. v =, and w = 5, v w = 6 θ.5 proj w v = 5 q = 9 69, 80 69 69, 756 69 9. v =, and w =, 5 v w = 6 θ 7.7 proj w v =, 5 q = 55, 0. v = 5, 6 and w =, 7 v w = 6 θ 69.9 proj w v = 8 q = 77 65, 65 65, 65

06 Applications of Trigonometr. v = 8, and w =, 6 v w = θ 87.88 proj w v = q = 8 0, 7 0 0, 0. v = î ĵ and w = ĵ v w = θ 08. proj w v = 0, q =, 0 5. v = î + ĵ and w = î ĵ v w = 0 θ = 90 proj w v = 0, 0 q =, 7. v =, and w = v w = θ = 75 6 proj w v = q = +, + 9. v =,, and w = v w = 6+ θ = 65 proj w v = q = +,,,, +. v =, 9 and w = 0, v w = 9 θ 59.5 proj w v = 0, 9 q =, 0. v = î + 7ĵ and w = î v w = 8 θ 6.7 proj w v =, 0 q = 0, 7 6. v = 5î + ĵ and w = î + ĵ v w = θ 59.9 proj w v = 99 q = 5, 68 5 8. v = v w =, 6 θ = 05 proj w v = q = + 6 8, 5, 5 and w = 6 8, 6 8 + 6 8 0. v =, and w = v w = θ = 5 6+ + proj w v =, q =,, +,. 500 pounds00 feet cos 0 = 50, 000 foot-pounds. 0 pounds feet cos 0 = 0 foot-pounds

.9 The Dot Product and Projection 07. pounds5 feet cos 5.9 foot-pounds. 00 pounds feet cos 09.5 foot-pounds 5. 00 pounds0 feet cos 77.5.88 foot-pounds

08 Applications of Trigonometr.0 Parametric Equations As we have seen in Eercises 5-56 in Section., Chapter 7 and most recentl in Section.5, there are scores of interesting curves which, when plotted in the -plane, neither represent as a function of nor as a function of. In this section, we present a new concept which allows us to use functions to stud these kinds of curves. To motivate the idea, we imagine a bug crawling across a table top starting at the point O and tracing out a curve C in the plane, as shown below. 5 P, = ft, gt Q O 5 The curve C does not represent as a function of because it fails the Vertical Line Test and it does not represent as a function of because it fails the Horizontal Line Test. However, since the bug can be in onl one place P, at an given time t, we can define the -coordinate of P as a function of t and the -coordinate of P as a usuall, but not necessaril different function of t. Traditionall, ft is used for and gt is used for. The independent variable t in this case is called a parameter and the sstem of equations { = ft = gt is called a sstem of parametric equations or a parametrization of the curve C. The parametrization of C endows it with an orientation and the arrows on C indicate motion in the direction of increasing values of t. In this case, our bug starts at the point O, travels upwards to the left, then loops back around to cross its path at the point Q and finall heads off into the first quadrant. It is important to note that the curve itself is a set of points and as such is devoid of an orientation. The parametrization determines the orientation and as we shall see, different parametrizations can determine different orientations. If all of this seems hauntingl familiar, it should. B definition, the sstem of equations { = cost, = sint parametrizes the Unit Circle, giving it a counter-clockwise orientation. More generall, the equations of circular motion { = r cosωt, = r sinωt developed on page 7 in Section 0.. are parametric equations which trace out a circle of radius r centered at the origin. If ω > 0, the orientation is counterclockwise; if ω < 0, the orientation is clockwise. The angular frequenc ω determines how fast the Note the use of the indefinite article a. As we shall see, there are infinitel man different parametric representations for an given curve. Here, the bug reaches the point Q at two different times. While this does not contradict our claim that ft and gt are functions of t, it shows that neither f nor g can be one-to-one. Think about this before reading on.

.0 Parametric Equations 09 object moves around the circle. In particular, the equations { = 960 cos t, = 960 sin t that model the motion of Lakeland Communit College as the earth rotates see Eample 0..7 in Section 0. parameterize a circle of radius 960 with a counter-clockwise rotation which completes one revolution as t runs through the interval [0,. It is time for another eample. { = t Eample.0.. Sketch the curve described b for t. = t Solution. We follow the same procedure here as we have time and time again when asked to graph anthing new choose friendl values of t, plot the corresponding points and connect the results in a pleasing fashion. Since we are told t, we start there and as we plot successive points, we draw an arrow to indicate the direction of the path for increasing values of t. t t t t, t 5, 5, 0,,, 6 5 6, 5 5 5 6 5 The curve sketched out in Eample.0. certainl looks like a parabola, and the presence of the { t term in the equation = t reinforces this hunch. Since the parametric equations = t, = t given to describe this curve are a sstem of equations, we can use the technique of substitution as described in Section 8.7 to eliminate the parameter t and get an equation involving just and. To do so, we choose to solve the equation = t for t to get t = +. Substituting this into the equation = t ields = + or, after some rearrangement, + = +. Thinking back to Section 7., we see that the graph of this equation is a parabola with verte, which opens to the right, as required. Technicall speaking, the equation + = + describes the entire parabola, while the parametric equations { = t, = t for t describe onl a portion of the parabola. In this case, we can remed this situation b restricting the bounds on. Since the portion of the parabola we want is eactl the part where 5, the equation + = + coupled with the restriction 5 describes the same curve as the given parametric equations. The one piece of information we can never recover after eliminating the parameter is the orientation of the curve. Eliminating the parameter and obtaining an equation in terms of and, whenever possible, can be a great help in graphing curves determined b parametric equations. If the sstem of parametric equations contains algebraic functions, as was the case in Eample.0., then the usual techniques of substitution and elimination as learned in Section 8.7 can be applied to the We will have an eample shortl where no matter how we restrict and, we can never accuratel describe the curve once we ve eliminated the parameter.

050 Applications of Trigonometr sstem { = ft, = gt to eliminate the parameter. If, on the other hand, the parametrization involves the trigonometric functions, the strateg changes slightl. In this case, it is often best to solve for the trigonometric functions and relate them using an identit. We demonstrate these techniques in the following eample. Eample.0.. Sketch the curves described b the following parametric equations.. { = t = t for t. { = sint = csct for 0 < t <. { = e t = e t for t 0. { = + cost = sint for 0 t Solution.. To get a feel for the curve described b the sstem { = t, = t we first sketch the graphs of = t and = t over the interval [, ]. We note that as t takes on values in the interval [, ], = t ranges between and, and = t ranges between 0 and. This means that all of the action is happening on a portion of the plane, namel {,, 0 }. Net, we plot a few points to get a sense of the position and orientation of the curve. Certainl, t = and t = are good values to pick since these are the etreme values of t. We also choose t = 0, since that corresponds to a relative minimum on the graph of = t. Plugging in t = gives the point,, t = 0 gives 0, 0 and t = gives,. More generall, we see that = t is increasing over the entire interval [, ] whereas = t is decreasing over the interval [, 0] and then increasing over [0, ]. Geometricall, this means that in order to trace out the path described b the parametric equations, we start at, where t =, then move to the right since is increasing and down since is decreasing to 0, 0 where t = 0. We continue to move to the right since is still increasing but now move upwards since is now increasing until we reach, where t =. Finall, to get a good sense of the shape of the curve, we eliminate the parameter. Solving = t for t, we get t =. Substituting this into = t gives = = /. Our eperience in Section 5. ields the graph of our final answer below. t t = t, t = t, t { = t, = t, t You should review Section.6. if ou ve forgotten what increasing, decreasing and relative minimum mean.

.0 Parametric Equations 05. For the sstem { = e t, = e t for t 0, we proceed as in the previous eample and graph = e t and = e t over the interval [0,. We find that the range of in this case is 0, ] and the range of is 0, ]. Net, we plug in some friendl values of t to get a sense of the orientation of the curve. Since t lies in the eponent here, friendl values of t involve natural logarithms. Starting with t = ln = 0 we get 5,, for t = ln we get, and for t = ln we get, 9. Since t is ranging over the unbounded interval [0,, we take the time to analze the end behavior of both and. As t, = e t 0 + and = e t 0 + as well. This means the graph of { = e t, = e t approaches the point 0, 0. Since both = e t and = e t are alwas decreasing for t 0, we know that our final graph will start at, where t = 0, and move consistentl to the left since is decreasing and down since is decreasing to approach the origin. To eliminate the parameter, one wa to proceed is to solve = e t for t to get t = ln. Substituting = this for t in = e t gives = e ln/ = e ln/ = e ln/ =. Or, we could recognize that = e t = e t, and since = e t means e t =, we get = = this wa as well. Either wa, the graph of { = e t, = e t for t 0 is a portion of the parabola = which starts at the point, and heads towards, but never reaches,6 0, 0. t t = e t, t 0 = e t, t 0 { = e t, = e t, t 0. For the sstem { = sint, = csct for 0 < t <, we start b graphing = sint and = csct over the interval 0,. We find that the range of is 0, ] while the range of is [,. Plotting a few friendl points, we see that t = 6 gives the point,, t = gives, and t = 5 6 returns us to,. Since t = 0 and t = aren t included in the domain for t, because = csct is undefined at these t-values, we analze the behavior of the sstem as t approaches 0 and. We find that as t 0 + as well as when t, we get = sint 0 + and = csct. Piecing all of this information together, we get that for t near 0, we have points with ver small positive -values, but ver large positive -values. As t ranges through the interval 0, ], = sint is increasing and = csct is decreasing. This means that we are moving to the right and downwards, through, when t = 6 to, when t =. Once t =, the orientation reverses, and we start to head to the left, since = sint is now decreasing, and up, since = csct is now increasing. We pass back through, when t = 5 6 back to the points with small positive -coordinates and large 5 The reader is encouraged to review Sections 6. and 6. as needed. 6 Note the open circle at the origin. See the solution to part in Eample.. on page and Theorem. in Section. for a review of this concept.

05 Applications of Trigonometr positive -coordinates. To better eplain this behavior, we eliminate the parameter. Using a reciprocal identit, we write = csct = sint. Since = sint, the curve traced out b this parametrization is a portion of the graph of =. We now can eplain the unusual behavior as t 0 + and t for these values of t, we are hugging the vertical asmptote = 0 of the graph of =. We see that the parametrization given above traces out the portion of = for 0 < twice as t runs through the interval 0,. t t = sint, 0 < t < = csct, 0 < t < { = sint, = csct, 0 < t <. Proceeding as above, we set about graphing { = + cost, = sint for 0 t b first graphing = + cost and = sint on the interval [ 0, ]. We see that ranges from to and ranges from to. Plugging in t = 0,, and gives the points, 0,,,, 0 and,, respectivel. As t ranges from 0 to, = + cost is decreasing, while = sint is increasing. This means that we start tracing out our answer at, 0 and continue moving to the left and upwards towards,. For t, is decreasing, as is, so [ the ] motion is still right to left, but now is downwards from, to, 0. On the interval,, begins to increase, while continues to decrease. Hence, the motion becomes left to right but continues downwards, connecting, 0 to,. To eliminate the parameter here, we note that the trigonometric functions involved, namel cost and sint, are related b the Pthagorean Identit cos t + sin t =. Hence, we solve = + cost for cost to get cost =, and we solve = sint for sint to get sint =. Substituting these epressions into cos t+sin t = gives + =, or 9 + =. From Section 7., we know that the graph of this equation is an ellipse centered at, 0 with vertices at, 0 and, 0 with a minor ais of length. Our parametric equations here are tracing out three-quarters of this ellipse, in a counter-clockwise direction. t t = + cost, 0 t = sint, 0 t { = + cost, = sint, 0 t

.0 Parametric Equations 05 Now that we have had some good practice sketching the graphs of parametric equations, we turn to the problem of finding parametric representations of curves. We start with the following. ˆ ˆ Parametrizations of Common Curves To parametrize = f as runs through some interval I, let = t and = ft and let t run through I. To parametrize = g as runs through some interval I, let = gt and = t and let t run through I. ˆ To parametrize a directed line segment with initial point 0, 0 and terminal point,, let = 0 + 0 t and = 0 + 0 t for 0 t. ˆ To parametrize h + k = where a, b > 0, let = h + a cost and = k + b sint a b for 0 t <. This will impart a counter-clockwise orientation. The reader is encouraged to verif the above formulas b eliminating the parameter and, when indicated, checking the orientation. We put these formulas to good use in the following eample. Eample.0.. Find a parametrization for each of the following curves and check our answers.. = from = to =. = f where f = 5 + +. The line segment which starts at, and ends at, 5. The circle + + = 5. The left half of the ellipse + 9 = Solution.. Since = is written in the form = f, we let = t and = ft = t. Since = t, the bounds on t match precisel the bounds on so we get { = t, = t for t. The check is almost trivial; with = t we have = t = as t = runs from to.. We are told to parametrize = f for f = 5 + + so it is safe to assume that f is one-to-one. Otherwise, f would not eist. To find a formula = f, we follow the procedure outlined on page 8 we start with the equation = f, interchange and and solve for. Doing so gives us the equation = 5 + +. While we could attempt to solve this equation for, we don t need to. We can parametrize = f = 5 + + b setting = t so that = t 5 + t +. We know from our work in Section. that since f = 5 + + is an odd-degree polnomial, the range of = f = 5 + + is,. Hence, in order to trace out the entire graph of = f = 5 + +, we need to let run through all real numbers. Our final answer to this problem is { = t 5 + t +, = t for < t <. As in the previous problem, our solution is trivial to check. 7 7 Provided ou followed the inverse function theor, of course.

05 Applications of Trigonometr. To parametrize line segment which starts at, and ends at, 5, we make use of the formulas = 0 + 0 t and = 0 + 0 t for 0 t. While these equations at first glance are quite a handful, 8 the can be summarized as starting point + displacementt. To find the equation for, we have that the line segment starts at = and ends at =. This means the displacement in the -direction is =. Hence, the equation for is = + t = t. For, we note that the line segment starts at = and ends at = 5. Hence, the displacement in the -direction is 5 = 8, so we get = + 8t. Our final answer is { = t, = + 8t for 0 t. To check, we can solve = t for t to get t =. Substituting this into = + 8t gives = + 8t = + 8, or = 8 +. We know this is the graph of a line, so all we need to check is that it starts and stops at the correct points. When t = 0, = t =, and when t =, = t =. Plugging in = gives = 8 + =, for an initial point of,. Plugging in = gives = 8 + = 5 for an ending point of, 5, as required.. In order to use the formulas above to parametrize the circle ++ =, we first need to put it into the correct form. After completing the squares, we get + + = 9, or + 9 + 9 =. Once again, the formulas = h + a cost and = k + b sint can be a challenge to memorize, but the come from the Pthagorean Identit cos t + sin t =. In the equation + 9 + 9 =, we identif cost = + and sint =. Rearranging these last two equations, we get = + cost and = + sint. In order to complete one revolution around the circle, we let t range through the interval [0,. We get as our final answer { = + cost, = + sint for 0 t <. To check our answer, we could eliminate the parameter b solving = + cost for cost and = + sint for sint, invoking a Pthagorean Identit, and then manipulating the resulting equation in and into the original equation + + =. Instead, we opt for a more direct approach. We substitute = + cost and = + sint into the equation + + = and show that the latter is satisfied for all t such that 0 t <. + + = + cost + + cost + + sint + sint 6 cost + 9 cos t + 6 cost + + sint + 9 sin t 8 sint 9 cos t + 9 sin t 5 9 cos t + sin t 5 9 5? =? =? =? =? = = Now that we know the parametric equations give us points on the circle, we can go through the usual analsis as demonstrated in Eample.0. to show that the entire circle is covered as t ranges through the interval [0,. 8 Compare and contrast this with Eercise 65 in Section.8.

.0 Parametric Equations 055 5. In the equation + 9 =, we can either use the formulas above or think back to the Pthagorean Identit to get = cost and = sint. The normal range on the parameter in this case is 0 t <, but since we are interested in onl the left half of the ellipse, we restrict t to the values which correspond to Quadrant II and Quadrant III angles, namel t. Our final answer is { = cost, = sint for t. Substituting = cost and = sint into + 9 = gives cos t + 9 sin t 9 =, which reduces to the Pthagorean Identit cos t + sin t =. This proves that the points generated b the parametric equations { = cost, = sint lie on the ellipse + 9 =. Emploing the techniques demonstrated in Eample.0., we find that the restriction t generates the left half of the ellipse, as required. We note that the formulas given on page 05 offer onl one of literall infinitel man was to parametrize the common curves listed there. At times, the formulas offered there need to be altered to suit the situation. Two eas was to alter parametrizations are given below. ˆ ˆ Adjusting Parametric Equations Reversing Orientation: Replacing ever occurrence of t with t in a parametric description for a curve including an inequalities which describe the bounds on t reverses the orientation of the curve. Shift of Parameter: Replacing ever occurrence of t with t c in a parametric description for a curve including an inequalities which describe the bounds on t shifts the start of the parameter t ahead b c units. We demonstrate these techniques in the following eample. Eample.0.. Find a parametrization for the following curves.. The curve which starts at, and follows the parabola = to end at,. Shift the parameter so that the path starts at t = 0.. The two part path which starts at 0, 0, travels along a line to,, then travels along a line to 5, 0.. The Unit Circle, oriented clockwise, with t = 0 corresponding to 0,. Solution.. We can parametrize = from = to = using the formula given on Page 05 as { = t, = t for t. This parametrization, however, starts at, and ends at,. Hence, we need to reverse the orientation. To do so, we replace ever occurrence of t with t to get { = t, = t for t. After simplifing, we get { = t, = t for t. We would like t to begin at t = 0 instead of t =. The problem here is that the parametrization we have starts units too soon, so we need to introduce a time dela of. Replacing ever occurrence of t with t gives { = t, = t for t. Simplifing ields { = t, = t t + for 0 t.

056 Applications of Trigonometr. When parameterizing line segments, we think: starting point + displacementt. For the first part of the path, we get { = t, = t for 0 t, and for the second part we get { = + t, = t for 0 t. Since the first parametrization leaves off at t =, we shift the parameter in the second part so it starts at t =. Our current description of the second part starts at t = 0, so we introduce a time dela of unit to the second set of parametric equations. Replacing t with t in the second set of parametric equations gives { = + t, = t for 0 t. Simplifing ields { = + t, = 8 t for t. Hence, we ma parametrize the path as { = ft, = gt for 0 t where { ft = t, for 0 t + t, for t { and gt = t, for 0 t 8 t, for t. We know that { = cost, = sint for 0 t < gives a counter-clockwise parametrization of the Unit Circle with t = 0 corresponding to, 0, so the first order of business is to reverse the orientation. Replacing t with t gives { = cos t, = sin t for 0 t <, which simplifies 9 to { = cost, = sint for < t 0. This parametrization gives a clockwise orientation, but t = 0 still corresponds to the point, 0; the point 0, is reached when t =. Our strateg is to first get the parametrization to start at the point 0, and then shift the parameter accordingl so the start coincides with t = 0. We know that an interval of length will parametrize the entire circle, so we keep the equations { = cost, = sint, but start the parameter t at, and find the upper bound b adding so t <. The reader can verif that { = cost, = sint for t < traces out the Unit Circle clockwise starting at the point 0,. We now shift the parameter b introducing a time dela of units b replacing ever occurrence of t with t {. We get = cos t, = sin t for t <. This simplifies 0 to { = sint, = cost for 0 t <, as required. We put our answer to Eample.0. number to good use to derive the equation of a ccloid. Suppose a circle of radius r rolls along the positive -ais at a constant velocit v as pictured below. Let θ be the angle in radians which measures the amount of clockwise rotation eperienced b the radius highlighted in the figure. P, r θ 9 courtes of the Even/Odd Identities 0 courtes of the Sum/Difference Formulas

.0 Parametric Equations 057 Our goal is to find parametric equations for the coordinates of the point P, in terms of θ. From our work in Eample.0. number, we know that clockwise motion along the Unit Circle starting at the point 0, can be modeled b the equations { = sinθ, = cosθ for 0 θ <. We have renamed the parameter θ to match the contet of this problem. To model this motion on a circle of radius r, all we need to do is multipl both and b the factor r which ields { = r sinθ, = r cosθ. We now need to adjust for the fact that the circle isn t stationar with center 0, 0, but rather, is rolling along the positive -ais. Since the velocit v is constant, we know that at time t, the center of the circle has traveled a distance vt down the positive -ais. Furthermore, since the radius of the circle is r and the circle isn t moving verticall, we know that the center of the circle is alwas r units above the -ais. Putting these two facts together, we have that at time t, the center of the circle is at the point vt, r. From Section 0.., we know v = rθ t, or vt = rθ. Hence, the center of the circle, in terms of the parameter θ, is rθ, r. As a result, we need to modif the equations { = r sinθ, = r cosθ b shifting the -coordinate to the right rθ units b adding rθ to the epression for and the -coordinate up r units b adding r to the epression for. We get { = r sinθ + rθ, = r cosθ + r, which can be written as { = rθ sinθ, = r cosθ. Since the motion starts at θ = 0 and proceeds indefinitel, we set θ 0. We end the section with a demonstration of the graphing calculator. Eample.0.5. Find the parametric equations of a ccloid which results from a circle of radius rolling down the positive -ais as described above. Graph our answer using a calculator. Solution. We have r = which gives the equations { = t sint, = cost for t 0. Here we have returned to the convention of using t as the parameter. Sketching the ccloid b hand is a wonderful eercise in Calculus, but for the purposes of this book, we use a graphing utilit. Using a calculator to graph parametric equations is ver similar to graphing polar equations on a calculator. Ensuring that the calculator is in Parametric Mode and radian mode we enter the equations and advance to the Window screen. As alwas, the challenge is to determine appropriate bounds on the parameter, t, as well as for and. We know that one full revolution of the circle occurs over the interval 0 t <, so If we replace with and with in the equation for the Unit Circle r r + =, we obtain + = r r which reduces to + = r. In the language of Section.7, we are stretching the graph b a factor of r in both the - and -directions. Hence, we multipl both the - and -coordinates of points on the graph b r. Does this seem familiar? See Eample.. in Section.. See page 959 in Section.5.

058 Applications of Trigonometr it seems reasonable to keep these as our bounds on t. The Tstep seems reasonabl small too large a value here can lead to incorrect graphs. We know from our derivation of the equations of the ccloid that the center of the generating circle has coordinates rθ, r, or in this case, t,. Since t ranges between 0 and, we set to range between 0 and 6. The values of go from the bottom of the circle to the top, so ranges between 0 and 6. Below we graph the ccloid with these settings, and then etend t to range from 0 to 6 which forces to range from 0 to 8 ielding three arches of the ccloid. It is instructive to note that keeping the settings between 0 and 6 messes up the geometr of the ccloid. The reader is invited to use the Zoom Square feature on the graphing calculator to see what window gives a true geometric perspective of the three arches. Again, see page 959 in Section.5.

.0 Parametric Equations 059.0. Eercises In Eercises - 0, plot the set of parametric equations b hand. Be sure to indicate the orientation imparted on the curve b the parametrization.. { = t = 6t for 0 t. { = t = t for 0 t. 5. { = t = t for t. { = t + t + = t + for t 6. { = t = + t t for 0 t { = 9 8 t = t for t 7. { = t = t for < t < 8. { = t = t for < t < 9. { = cost = sint for t 0. { = cost = sint for 0 t. { = + cost = sint for 0 t. { = cost = sint + for t. { = cost = sect for 0 t <. { = tant = cott for 0 < t < 5. { = sect = tant for < t < 6. { = sect = tant for < t < 7. { = tant = sect for < t < 8. { = tant = sect for < t < 9. { = cost = t for 0 t 0. { = sint = t for t In Eercises -, plot the set of parametric equations with the help of a graphing utilit. Be sure to indicate the orientation imparted on the curve b the parametrization.. { = t t = t for t. { = cos t = sin t for 0 t. { = e t + e t = e t e t for t. { = cost = sint for 0 t

060 Applications of Trigonometr In Eercises 5-9, find a parametric description for the given oriented curve. 5. the directed line segment from, 5 to, 6. the directed line segment from, to, 7. the curve = from, 0 to, 0. 8. the curve = from, 0 to, 0 Shift the parameter so t = 0 corresponds to, 0. 9. the curve = 9 from 5, to 0,. 0. the curve = 9 from 0, to 5,. Shift the parameter so t = 0 corresponds to 0,.. the circle + = 5, oriented counter-clockwise. the circle + =, oriented counter-clockwise. the circle + 6 = 0, oriented counter-clockwise. the circle + 6 = 0, oriented clockwise Shift the parameter so t begins at 0. 5. the circle + + = 7, oriented counter-clockwise 6. the ellipse + 9 = 9, oriented counter-clockwise 7. the ellipse 9 + + = 0, oriented counter-clockwise 8. the ellipse 9 + + = 0, oriented clockwise Shift the parameter so t = 0 corresponds to 0, 0. 9. the triangle with vertices 0, 0,, 0, 0,, oriented counter-clockwise Shift the parameter so t = 0 corresponds to 0, 0. 0. Use parametric equations and a graphing utilit to graph the inverse of f = +.. Ever polar curve r = fθ can be translated to a sstem of parametric equations with parameter θ b { = r cosθ = fθ cosθ, = r sinθ = fθ sinθ. Convert r = 6 cosθ to a sstem of parametric equations. Check our answer b graphing r = 6 cosθ b hand using the techniques presented in Section.5 and then graphing the parametric equations ou found using a graphing utilit.. Use our results from Eercises and in Section. to find the parametric equations which model a passenger s position as the ride the London Ee.

.0 Parametric Equations 06 Suppose an object, called a projectile, is launched into the air. Ignoring everthing ecept the force gravit, the path of the projectile is given b 5 = v 0 cosθ t = gt + v 0 sinθ t + s 0 for 0 t T where v 0 is the initial speed of the object, θ is the angle from the horizontal at which the projectile is launched, 6 g is the acceleration due to gravit, s 0 is the initial height of the projectile above the ground and T is the time when the object returns to the ground. See the figure below. s 0 θ T, 0. Carl s friend Jason competes in Highland Games Competitions across the countr. In one event, the hammer throw, he throws a 56 pound weight for distance. If the weight is released 6 feet above the ground at an angle of with respect to the horizontal with an initial speed of feet per second, find the parametric equations for the flight of the hammer. Here, use g = ft.. When will the hammer hit the ground? How far awa will it hit the ground? s Check our answer using a graphing utilit.. Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve = g sec θ v + tanθ + s 0 0 Use the verte formula Equation. to show the maimum height of the projectile is = v 0 sin θ g + s 0 when = v 0 sinθ g 5 A nice mi of vectors and Calculus are needed to derive this. 6 We ve seen this before. It s the angle of elevation which was defined on page 75.

06 Applications of Trigonometr 5. In another event, the sheaf toss, Jason throws a 0 pound weight for height. If the weight is released 5 feet above the ground at an angle of 85 with respect to the horizontal and the sheaf reaches a maimum height of.5 feet, use our results from part to determine how fast the sheaf was launched into the air. Once again, use g = ft. s. 6. Suppose θ =. The projectile was launched verticall. Simplif the general parametric formula given for t above using g = 9.8 m and compare that to the formula for st given s in Eercise 5 in Section.. What is t in this case? In Eercises 7-5, we eplore the hperbolic cosine function, denoted cosht, and the hperbolic sine function, denoted sinht, defined below: cosht = et + e t and sinht = et e t 7. Using a graphing utilit as needed, verif that the domain of cosht is, and the range of cosht is [,. 8. Using a graphing utilit as needed, verif that the domain and range of sinht are both,. 9. Show that {t = cosht, t = sinht parametrize the right half of the unit hperbola =. Hence the use of the adjective hperbolic. 50. Compare the definitions of cosht and sinht to the formulas for cost and sint given in Eercise 8f in Section.7. 5. Four other hperbolic functions are waiting to be defined: the hperbolic secant secht, the hperbolic cosecant cscht, the hperbolic tangent tanht and the hperbolic cotangent cotht. Define these functions in terms of cosht and sinht, then convert them to formulas involving e t and e t. Consult a suitable reference a Calculus book, or this entr on the hperbolic functions and spend some time reliving the thrills of trigonometr with these hperbolic functions. 5. If these functions look familiar, the should. Enjo some nostalgia and revisit Eercise 5 in Section 6.5, Eercise 7 in Section 6. and the answer to Eercise 8 in Section 6..

.0 Parametric Equations 06.0. Answers. { = t = 6t for 0 t. { = t = t for 0 t. { = t = t for t. { = t = + t t for 0 t 5. { = t + t + = t + for t 6. { = 8 t 9 = t for t 5

06 Applications of Trigonometr 7. { = t = t for < t < 8. { = t = t for < t < 9. { = cost = sint for t 0. { = cost = sint for 0 t. { = + cost = sint for 0 t. { = cost = sint + for t

.0 Parametric Equations 065. { = cost = sect for 0 t <. { = tant = cott for 0 < t < 5. { = sect = tant for < t < 6. { = sect = tant for < t <

066 Applications of Trigonometr 7. { = tant = sect for < t < 8. { = tant = sect for < t < 9. { = cost for 0 < t < = t 0. { = sint = t for < t <. { = t t = t for t. { = cos t = sin t for 0 t

.0 Parametric Equations 067. { = e t + e t = e t e t for t. { = cost = sint for 0 t 7 5 5 6 7 5 7 5. { = 5t = 5 + 7t for 0 t 6. { = 5t = t for 0 t 7. { = t = t for t 8. { = t = t t for 0 t 9. { = t 9 = t for t 0. { = t 6t = t for 0 t 5. { = 5 cost = 5 sint for 0 t <. { = + cost = sint for 0 t <. { = cost = + sint for 0 t <. { = cost = sint for 0 t < 5. { = + 7 cost = + 7 sint { = cost for 0 t < 6. { = + cost = sint for 0 t < 7. for 0 t < = sint = cos t = sint 8. = sin t = + cost for 0 t < 9. {t, t where: t, 0 t t = 6 t, t 0, t 0, 0 t t = t, t t, t

068 Applications of Trigonometr { = t 0. The parametric equations for the inverse are + t = t for < t <. r = 6 cosθ translates to { = 6 cosθ cosθ = 6 cosθ sinθ for 0 θ <.. { The parametric equations which describe the locations of passengers on the London Ee are = 67.5 cos 5 t = 67.5 sin 5 t = 67.5 sin 5 t + 67.5 = 67.5 67.5 cos 5 t for < t < { = cos. The parametric equations for the hammer throw are t = 6t + sin for t + 6 t 0. To find when the hammer hits the ground, we solve t = 0 and get t 0. or.6. Since t 0, the hammer hits the ground after approimatel t =.6 seconds after it was launched into the air. To find how far awa the hammer hits the ground, we find.6 9.8 feet from where it was thrown into the air. 5. We solve = v 0 sin θ + s 0 = v 0 sin 85 + 5 =.5 to get v 0 = ±.. The initial speed g of the sheaf was approimatel. feet per second.

Inde n th root of a comple number, 000, 00 principal, 97 n th Roots of Unit, 006 u-substitution, 7 -ais, 6 -coordinate, 6 -intercept, 5 -ais, 6 -coordinate, 6 -intercept, 5 abscissa, 6 absolute value definition of, 7 inequalit, properties of, 7 acidit of a solution ph, acute angle, 69 adjoint of a matri, 6 alkalinit of a solution ph, amplitude, 79, 88 angle acute, 69 between two vectors, 05, 06 central angle, 70 complementar, 696 coterminal, 698 decimal degrees, 695 definition, 69 degree, 69 DMS, 695 initial side, 698 measurement, 69 negative, 698 obtuse, 69 of declination, 76 of depression, 76 of elevation, 75 of inclination, 75 oriented, 697 positive, 698 quadrantal, 698 radian measure, 70 reference, 7 right, 69 standard position, 698 straight, 69 supplementar, 696 terminal side, 698 verte, 69 angle side opposite pairs, 896 angular frequenc, 708 annuit annuit-due, 667 ordinar definition of, 666 future value, 667 applied domain of a function, 60 arccosecant calculus friendl definition of, 8 graph of, 80 properties of, 8 069

070 Inde trigonometr friendl definition of, 88 graph of, 87 properties of, 88 arccosine definition of, 80 graph of, 89 properties of, 80 arccotangent definition of, 8 graph of, 8 properties of, 8 arcsecant calculus friendl definition of, 8 graph of, 80 properties of, 8 trigonometr friendl definition of, 88 graph of, 87 properties of, 88 arcsine definition of, 80 graph of, 80 properties of, 80 arctangent definition of, 8 graph of, 8 properties of, 8 argument of a comple number definition of, 99 properties of, 995 of a function, 55 of a logarithm, 5 of a trigonometric function, 79 arithmetic sequence, 65 associative propert for function composition, 66 matri addition, 579 matri multiplication, 585 scalar multiplication, 58 vector addition, 05 scalar multiplication, 08 asmptote horizontal formal definition of, 0 intuitive definition of, 0 location of, 08 of a hperbola, 5 slant determination of, formal definition of, slant oblique, vertical formal definition of, 0 intuitive definition of, 0 location of, 06 augmented matri, 568 average angular velocit, 707 average cost, 6 average cost function, 8 average rate of change, 60 average velocit, 706 ais of smmetr, 9 back substitution, 560 bearings, 905 binomial coefficient, 68 Binomial Theorem, 68 Bisection Method, 77 BMI, bod mass inde, 55 Bole s Law, 50 buffer solution, 78 cardioid, 95 Cartesian coordinate plane, 6 Cartesian coordinates, 6 Cauch s Bound, 69 center of a circle, 98 of a hperbola, 5 of an ellipse, 56

Inde 07 central angle, 70 change of base formulas, characteristic polnomial, 66 Charles s Law, 55 circle center of, 98 definition of, 98 from slicing a cone, 95 radius of, 98 standard equation, 98 standard equation, alternate, 59 circular function, 7 cisθ, 995 coefficient of determination, 6 cofactor, 66 Cofunction Identities, 77 common base, 0 common logarithm, commutative propert function composition does not have, 66 matri addition, 579 vector addition, 05 dot product, 0 complementar angles, 696 Comple Factorization Theorem, 90 comple number n th root, 000, 00 n th Roots of Unit, 006 argument definition of, 99 properties of, 995 conjugate definition of, 88 properties of, 89 definition of,, 87, 99 imaginar part, 99 imaginar unit, i, 87 modulus definition of, 99 properties of, 99 polar form cis-notation, 995 principal argument, 99 real part, 99 rectangular form, 99 set of, comple plane, 99 component form of a vector, 0 composite function definition of, 60 properties of, 67 compound interest, 70 conic sections definition, 95 conjugate ais of a hperbola, 5 conjugate of a comple number definition of, 88 properties of, 89 Conjugate Pairs Theorem, 9 consistent sstem, 55 constant function as a horizontal line, 56 formal definition of, 0 intuitive definition of, 00 constant of proportionalit, 50 constant term of a polnomial, 6 continuous, continuousl compounded interest, 7 contradiction, 59 coordinates Cartesian, 6 polar, 99 rectangular, 99 correlation coefficient, 6 cosecant graph of, 80 of an angle, 7, 75 properties of, 80 cosine graph of, 79 of an angle, 77, 70, 7 properties of, 79

07 Inde cost average, 8, 6 fied, start-up, 8 variable, 59 cost function, 8 cotangent graph of, 805 of an angle, 7, 75 properties of, 806 coterminal angle, 698 Coulomb s Law, 55 Cramer s Rule, 69 curve orientated, 08 ccloid, 056 decibel, decimal degrees, 695 decreasing function formal definition of, 0 intuitive definition of, 00 degree measure, 69 degree of a polnomial, 6 DeMoivre s Theorem, 997 dependent sstem, 55 dependent variable, 55 depreciation, 0 Descartes Rule of Signs, 7 determinant of a matri definition of, 6 properties of, 66 Difference Identit for cosine, 77, 775 for sine, 77, 775 for tangent, 775 difference quotient, 79 dimension of a matri, 567 direct variation, 50 directri of a conic section in polar form, 98 of a parabola, 505 discriminant of a conic, 979 of a quadratic equation, 95 trichotom, 95 distance definition, 0 distance formula, distributive propert matri matri multiplication, 585 scalar multiplication, 58 vector dot product, 0 scalar multiplication, 08 DMS, 695 domain applied, 60 definition of, 5 implied, 58 dot product commutative propert of, 0 definition of, 0 distributive propert of, 0 geometric interpretation, 05 properties of, 0 relation to orthogonalit, 07 relation to vector magnitude, 0 work, 0 Double Angle Identities, 776 earthquake Richter Scale, eccentricit, 5, 98 eigenvalue, 66 eigenvector, 66 ellipse center, 56 definition of, 56 eccentricit, 5 foci, 56 from slicing a cone, 96 guide rectangle, 59 major ais, 56 minor ais, 56

Inde 07 reflective propert, 5 standard equation, 59 vertices, 56 ellipsis...,, 65 empt set, end behavior of f = a n, n even, 0 of f = a n, n odd, 0 of a function graph, 9 polnomial, entr in a matri, 567 equation contradiction, 59 graph of, identit, 59 linear of n variables, 55 linear of two variables, 59 even function, 95 Even/Odd Identities, 770 eponential function algebraic properties of, 7 change of base formula, common base, 0 definition of, 8 graphical properties of, 9 inverse properties of, 7 natural base, 0 one-to-one properties of, 7 solving equations with, 8 etended interval notation, 756 Factor Theorem, 58 factorial, 65, 68 fied cost, 8 focal diameter of a parabola, 507 focal length of a parabola, 506 focus of a conic section in polar form, 98 focus foci of a hperbola, 5 of a parabola, 505 of an ellipse, 56 free variable, 55 frequenc angular, 708, 88 of a sinusoid, 795 ordinar, 708, 88 function absolute maimum, 0 absolute, global minimum, 0 absolute value, 7 algebraic, 99 argument, 55 arithmetic, 76 as a process, 55, 78 average cost, 8 circular, 7 composite definition of, 60 properties of, 67 constant, 00, 56 continuous, cost, 8 decreasing, 00 definition as a relation, dependent variable of, 55 difference, 76 difference quotient, 79 domain, 5 even, 95 eponential, 8 Fundamental Graphing Principle, 9 identit, 68 increasing, 00 independent variable of, 55 inverse definition of, 79 properties of, 79 solving for, 8 uniqueness of, 80 linear, 56 local relative maimum, 0 local relative minimum, 0 logarithmic,

07 Inde notation, 55 odd, 95 one-to-one, 8 periodic, 790 piecewise-defined, 6 polnomial, 5 price-demand, 8 product, 76 profit, 8 quadratic, 88 quotient, 76 range, 5 rational, 0 revenue, 8 smooth, sum, 76 transformation of graphs, 0, 5 zero, 95 fundamental ccle of = cos, 79 Fundamental Graphing Principle for equations, for functions, 9 for polar equations, 98 Fundamental Theorem of Algebra, 90 Gauss-Jordan Elimination, 57 Gaussian Elimination, 557 geometric sequence, 65 geometric series, 669 graph hole in, 05 horizontal scaling, horizontal shift, of a function, 9 of a relation, 0 of an equation, rational function, reflection about an ais, 6 transformations, 5 vertical scaling, 0 vertical shift, greatest integer function, 67 growth model limited, 75 logistic, 75 uninhibited, 7 guide rectangle for a hperbola, 5 for an ellipse, 59 Half-Angle Formulas, 779 harmonic motion, 885 Henderson-Hasselbalch Equation, 6 Heron s Formula, 9 hole in a graph, 05 location of, 06 Hooke s Law, 50 horizontal asmptote formal definition of, 0 intuitive definition of, 0 location of, 08 horizontal line, Horizontal Line Test HLT, 8 hperbola asmptotes, 5 branch, 5 center, 5 conjugate ais, 5 definition of, 5 foci, 5 from slicing a cone, 96 guide rectangle, 5 standard equation horizontal, 5 vertical, 5 transverse ais, 5 vertices, 5 hperbolic cosine, 06 hperbolic sine, 06 hperboloid, 5 identit function, 67 matri, additive, 579

Inde 075 matri, multiplicative, 585 statement which is alwas true, 59 imaginar ais, 99 imaginar part of a comple number, 99 imaginar unit, i, 87 implied domain of a function, 58 inconsistent sstem, 55 increasing function formal definition of, 0 intuitive definition of, 00 independent sstem, 55 independent variable, 55 inde of a root, 97 induction base step, 67 induction hpothesis, 67 inductive step, 67 inequalit absolute value, graphical interpretation, 09 non-linear, 6 quadratic, 5 sign diagram, inflection point, 77 information entrop, 77 initial side of an angle, 698 instantaneous rate of change, 6, 7, 707 integer definition of, greatest integer function, 67 set of, intercept definition of, 5 location of, 5 interest compound, 70 compounded continuousl, 7 simple, 69 Intermediate Value Theorem polnomial zero version, interrobang, intersection of two sets, interval definition of, notation for, notation, etended, 756 inverse matri, additive, 579, 58 matri, multiplicative, 60 of a function definition of, 79 properties of, 79 solving for, 8 uniqueness of, 80 inverse variation, 50 invertibilit function, 8 invertible function, 79 matri, 60 irrational number definition of, set of, irreducible quadratic, 9 joint variation, 50 Kepler s Third Law of Planetar Motion, 55 Kirchhoff s Voltage Law, 605 latus rectum of a parabola, 507 Law of Cosines, 90 Law of Sines, 897 leading coefficient of a polnomial, 6 leading term of a polnomial, 6 Learning Curve Equation, 5 least squares regression line, 5 lemniscate, 950 limaçon, 950 line horizontal, least squares regression, 5 linear function, 56 of best fit, 5 parallel, 66

076 Inde perpendicular, 67 point-slope form, 55 slope of, 5 slope-intercept form, 55 vertical, linear equation n variables, 55 two variables, 59 linear function, 56 local maimum formal definition of, 0 intuitive definition of, 0 local minimum formal definition of, 0 intuitive definition of, 0 logarithm algebraic properties of, 8 change of base formula, common, general, base b, graphical properties of, inverse properties of, 7 natural, one-to-one properties of, 7 solving equations with, 59 logarithmic scales, logistic growth, 75 LORAN, 58 lower triangular matri, 59 main diagonal, 585 major ais of an ellipse, 56 Markov Chain, 59 mathematical model, 60 matri addition associative propert, 579 commutative propert, 579 definition of, 578 properties of, 579 additive identit, 579 additive inverse, 579 adjoint, 6 augmented, 568 characteristic polnomial, 66 cofactor, 66 definition, 567 determinant definition of, 6 properties of, 66 dimension, 567 entr, 567 equalit, 578 invertible, 60 leading entr, 569 lower triangular, 59 main diagonal, 585 matri multiplication associative propert of, 585 definition of, 58 distributive propert, 585 identit for, 585 properties of, 585 minor, 66 multiplicative inverse, 60 product of row and column, 58 reduced row echelon form, 570 rotation, 986 row echelon form, 569 row operations, 568 scalar multiplication associative propert of, 58 definition of, 580 distributive properties, 58 identit for, 58 properties of, 58 zero product propert, 58 size, 567 square matri, 586 sum, 578 upper triangular, 59 maimum formal definition of, 0 intuitive definition of, 0 measure of an angle, 69

Inde 077 midpoint definition of, midpoint formula, minimum formal definition of, 0 intuitive definition of, 0 minor, 66 minor ais of an ellipse, 56 model mathematical, 60 modulus of a comple number definition of, 99 properties of, 99 multiplicit effect on the graph of a polnomial, 5, 9 of a zero, natural base, 0 natural logarithm, natural number definition of, set of, negative angle, 698 Newton s Law of Cooling,, 7 Newton s Law of Universal Gravitation, 5 oblique asmptote, obtuse angle, 69 odd function, 95 Ohm s Law, 50, 605 one-to-one function, 8 ordered pair, 6 ordinar frequenc, 708 ordinate, 6 orientation, 08 oriented angle, 697 oriented arc, 70 origin, 7 orthogonal projection, 08 orthogonal vectors, 07 overdetermined sstem, 55 parabola ais of smmetr, 9 definition of, 505 directri, 505 focal diameter, 507 focal length, 506 focus, 505 from slicing a cone, 96 graph of a quadratic function, 88 latus rectum, 507 reflective propert, 50 standard equation horizontal, 508 vertical, 506 verte, 88, 505 verte formulas, 9 paraboloid, 50 parallel vectors, 00 parameter, 08 parametric equations, 08 parametric solution, 55 parametrization, 08 partial fractions, 68 Pascal s Triangle, 688 password strength, 77 period circular motion, 708 of a function, 790 of a sinusoid, 88 periodic function, 790 ph, phase, 795, 88 phase shift, 795, 88 pi,, 700 piecewise-defined function, 6 point of diminishing returns, 77 point-slope form of a line, 55 polar coordinates conversion into rectangular, 9 definition of, 99 equivalent representations of, 9 polar ais, 99 pole, 99

078 Inde polar form of a comple number, 995 polar rose, 950 polnomial division dividend, 58 divisor, 58 factor, 58 quotient, 58 remainder, 58 snthetic division, 60 polnomial function completel factored over the comple numbers, 9 over the real numbers, 9 constant term, 6 definition of, 5 degree, 6 end behavior, 9 leading coefficient, 6 leading term, 6 variations in sign, 7 zero lower bound, 7 multiplicit, upper bound, 7 positive angle, 698 Power Reduction Formulas, 778 power rule for absolute value, 7 for comple numbers, 997 for eponential functions, 7 for logarithms, 8 for radicals, 98 for the modulus of a comple number, 99 price-demand function, 8 principal, 69 principal n th root, 97 principal argument of a comple number, 99 principal unit vectors, î, ĵ, 0 Principle of Mathematical Induction, 67 product rule for absolute value, 7 for comple numbers, 997 for eponential functions, 7 for logarithms, 8 for radicals, 98 for the modulus of a comple number, 99 Product to Sum Formulas, 780 profit function, 8 projection ais, 5 ais, 6 orthogonal, 08 Pthagorean Conjugates, 75 Pthagorean Identities, 79 quadrantal angle, 698 quadrants, 8 quadratic formula, 9 quadratic function definition of, 88 general form, 90 inequalit, 5 irreducible quadratic, 9 standard form, 90 quadratic regression, 8 Quotient Identities, 75 quotient rule for absolute value, 7 for comple numbers, 997 for eponential functions, 7 for logarithms, 8 for radicals, 98 for the modulus of a comple number, 99 radian measure, 70 radical properties of, 98 radicand, 97 radioactive deca, 7 radius of a circle, 98 range definition of, 5 rate of change average, 60

Inde 079 instantaneous, 6, 7 slope of a line, 5 rational eponent, 98 rational functions, 0 rational number definition of, set of, Rational Zeros Theorem, 69 ra definition of, 69 initial point, 69 real ais, 99 Real Factorization Theorem, 9 real number definition of, set of, real part of a comple number, 99 Reciprocal Identities, 75 rectangular coordinates also known as Cartesian coordinates, 99 conversion into polar, 9 rectangular form of a comple number, 99 recursion equation, 65 reduced row echelon form, 570 reference angle, 7 Reference Angle Theorem for cosine and sine, 7 for the circular functions, 77 reflection of a function graph, 6 of a point, 0 regression coefficient of determination, 6 correlation coefficient, 6 least squares line, 5 quadratic, 8 total squared error, 5 relation algebraic description, definition, 0 Fundamental Graphing Principle, Remainder Theorem, 58 revenue function, 8 Richter Scale, right angle, 69 root inde, 97 radicand, 97 Roots of Unit, 006 rotation matri, 986 rotation of aes, 97 row echelon form, 569 row operations for a matri, 568 scalar multiplication matri associative propert of, 58 definition of, 580 distributive properties of, 58 properties of, 58 vector associative propert of, 08 definition of, 07 distributive properties of, 08 properties of, 08 scalar projection, 09 secant graph of, 800 of an angle, 7, 75 properties of, 80 secant line, 60 sequence n th term, 65 alternating, 65 arithmetic common difference, 65 definition of, 65 formula for n th term, 656 sum of first n terms, 666 definition of, 65 geometric common ratio, 65 definition of, 65 formula for n th term, 656 sum of first n terms, 666

080 Inde recursive, 65 series, 668 set definition of, empt, intersection, roster method, set-builder notation, sets of numbers, union, verbal description, set-builder notation, Side-Angle-Side triangle, 90 Side-Side-Side triangle, 90 sign diagram algebraic function, 99 for quadratic inequalit, polnomial function, rational function, simple interest, 69 sine graph of, 79 of an angle, 77, 70, 7 properties of, 79 sinusoid amplitude, 79, 88 baseline, 88 frequenc angular, 88 ordinar, 88 graph of, 795, 88 period, 88 phase, 88 phase shift, 795, 88 properties of, 88 vertical shift, 88 slant asmptote, slant asmptote determination of, formal definition of, slope definition, 5 of a line, 5 rate of change, 5 slope-intercept form of a line, 55 smooth, sound intensit level decibel, square matri, 586 standard position of a vector, 09 standard position of an angle, 698 start-up cost, 8 stead state, 59 stochastic process, 59 straight angle, 69 Sum Identit for cosine, 77, 775 for sine, 77, 775 for tangent, 775 Sum to Product Formulas, 78 summation notation definition of, 66 inde of summation, 66 lower limit of summation, 66 properties of, 66 upper limit of summation, 66 supplementar angles, 696 smmetr about the -ais, 9 about the -ais, 9 about the origin, 9 testing a function graph for, 95 testing an equation for, 6 snthetic division tableau, 60 sstem of equations back-substitution, 560 coefficient matri, 590 consistent, 55 constant matri, 590 definition, 59 dependent, 55 free variable, 55 Gauss-Jordan Elimination, 57 Gaussian Elimination, 557

Inde 08 inconsistent, 55 independent, 55 leading variable, 556 linear n variables, 55 two variables, 550 linear in form, 66 non-linear, 67 overdetermined, 55 parametric solution, 55 triangular form, 556 underdetermined, 55 unknowns matri, 590 tangent graph of, 80 of an angle, 7, 75 properties of, 806 terminal side of an angle, 698 Thurstone, Louis Leon, 5 total squared error, 5 transformation non-rigid, 9 rigid, 9 transformations of function graphs, 0, 5 transverse ais of a hperbola, 5 Triangle Inequalit, 8 triangular form, 556 underdetermined sstem, 55 uninhibited growth, 7 union of two sets, Unit Circle definition of, 50 important points, 7 unit vector, 0 Upper and Lower Bounds Theorem, 7 upper triangular matri, 59 variable dependent, 55 independent, 55 variable cost, 59 variation constant of proportionalit, 50 direct, 50 inverse, 50 joint, 50 variations in sign, 7 vector -component, 0 -component, 0 addition associative propert, 05 commutative propert, 05 definition of, 0 properties of, 05 additive identit, 05 additive inverse, 05, 08 angle between two, 05, 06 component form, 0 Decomposition Theorem Generalized, 00 Principal, 0 definition of, 0 direction definition of, 00 properties of, 00 dot product commutative propert of, 0 definition of, 0 distributive propert of, 0 geometric interpretation, 05 properties of, 0 relation to magnitude, 0 relation to orthogonalit, 07 work, 0 head, 0 initial point, 0 magnitude definition of, 00 properties of, 00 relation to dot product, 0 normalization, 0 orthogonal projection, 08

08 Inde orthogonal vectors, 07 parallel, 00 principal unit vectors, î, ĵ, 0 resultant, 0 scalar multiplication associative propert of, 08 definition of, 07 distributive properties, 08 identit for, 08 properties of, 08 zero product propert, 08 scalar product definition of, 0 properties of, 0 scalar projection, 09 standard position, 09 tail, 0 terminal point, 0 triangle inequalit, 0 unit vector, 0 velocit average angular, 707 instantaneous, 707 instantaneous angular, 707 verte of a hperbola, 5 of a parabola, 88, 505 of an angle, 69 of an ellipse, 56 vertical asmptote formal definition of, 0 intuitive definition of, 0 location of, 06 vertical line, Vertical Line Test VLT, multiplicit of, of a function, 95 upper and lower bounds, 7 whole number definition of, set of, work, 0 wrapping function, 70 zero