College Trigonometry Version π Corrected Edition. July 4, 2013

Transcription

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

2 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.

3 Table of Contents vii 0 Foundations of Trigonometr Angles and their Measure Applications of Radian Measure: Circular Motion Eercises Answers The Unit Circle: Cosine and Sine Beond the Unit Circle Eercises Answers The Si Circular Functions and Fundamental Identities Beond the Unit Circle Eercises Answers Trigonometric Identities Eercises Answers Graphs of the Trigonometric Functions Graphs of the Cosine and Sine Functions Graphs of the Secant and Cosecant Functions Graphs of the Tangent and Cotangent Functions Eercises Answers The Inverse Trigonometric Functions Inverses of Secant and Cosecant: Trigonometr Friendl Approach Inverses of Secant and Cosecant: Calculus Friendl Approach Calculators and the Inverse Circular Functions Solving Equations Using the Inverse Trigonometric Functions Eercises Answers Trigonometric Equations and Inequalities Eercises Answers Applications of Trigonometr 88. Applications of Sinusoids Harmonic Motion Eercises Answers The Law of Sines Eercises Answers The Law of Cosines

4 viii Table of Contents.. Eercises Answers Polar Coordinates Eercises Answers Graphs of Polar Equations Eercises Answers Hooked on Conics Again Rotation of Aes The Polar Form of Conics Eercises Answers Polar Form of Comple Numbers Eercises Answers Vectors Eercises Answers The Dot Product and Projection Eercises Answers Parametric Equations Eercises Answers Inde 069

5 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

6 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 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

7 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

8 ii Preface

9 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

10 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 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.

11 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 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, = 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 = Converting the partial amount of degrees to minutes, we find = 7.5 = Converting the partial amount of minutes to seconds gives = 0. Putting it all together ields.5 = = = = = 7 0 On the other hand, to convert to decimal degrees, we first compute 5 60 = and = 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 Does this kind of sstem seem familiar?

12 696 Foundations of Trigonometr = = = = 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 = 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 = Net we convert =.6. Writing.6 = + 0.6, we convert = 5.6. Hence,.7 = = +.6 = = = 5.6 Rounding to seconds, we obtain α 6.

13 0. Angles and their Measure 697. To convert β to decimal degrees, we convert 8 60 = 7 it all together, we have = = = and = Putting 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 = To find a complementar angle for β, we seek an angle γ so that β + γ = 90. We get γ = 90 β = 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 = = = In essence, we are borrowing = 60 from the degree place, and then borrowing = 60 from the minutes place. 8 This ields, γ = = = 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 Back then, ou were working in a base ten sstem; here, it is base sit.

14 698 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, ±, ±,....

15 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 = 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 = = 0. Substituting k = gives θ = = 00. Finall, if we let k =, we get θ = = 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 = 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.

16 700 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 = k, where k is an integer. Working through the arithmetic, we find three such angles: 80, 80 and 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 = 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

17 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

18 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. Substituting k = gives θ = 6 6 = 6 and when we let k =, we get θ = = 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.

19 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 = 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

20 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 = Of particular interest is the fact that an angle which measures in radian measure is equal to 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.

21 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 which tells us our arc etends just a bit beond the quarter mark into Quadrant III.

22 706 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 = 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.

23 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 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.

24 708 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.

25 0. Angles and their Measure Eercises In Eercises -, convert the angles into the DMS sstem. Round each of our answers to the nearest second In Eercises 5-8, convert the angles into decimal degrees. Round each of our answers to three decimal places 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 In Eercises 9-6, convert the angle from degree measure into radian measure, giving the eact value in terms of In Eercises 7 -, convert the angle from radian measure into degree measure

26 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 = 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.

27 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 = θ = 0, r = 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.

28 7 Foundations of Trigonometr 0.. Answers is a Quadrant IV angle coterminal with 690 and is a Quadrant III angle coterminal with 5 and is a Quadrant II angle coterminal with 80 and is a Quadrant I angle coterminal with 5 and lies on the positive -ais coterminal with 90 and is a Quadrant II angle 6 coterminal with 7 6 and 7 6

29 0. Angles and their Measure 7 5. is a Quadrant I angle coterminal with and 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 lies on the negative -ais 0. is a Quadrant I angle coterminal with and coterminal with 9 and 7

30 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 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

31 0. Angles and their Measure is a Quadrant IV angle coterminal with 7 and 8. is a Quadrant IV angle 6 coterminal with 6 and t = t = 7. t = 6 8. t =

32 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 miles per hour 55. About. miles per hour 57. square units square units square units square units square units square units

33 0. The Unit Circle: Cosine and Sine 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..

34 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 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?