TRIGONOMETRY MICHAEL CORRAL

Transcription

1 TRIGONOMETRY MICHAEL CORRAL

2

3 Trigonometr Michael Corral Schoolcraft College

4 About the author: Michael Corral is an Adjunct Facult member of the Department of Mathematics at Schoolcraft College. He received a B.A. in Mathematics from the Universit of California at Berkele, and received an M.A. in Mathematics and an M.S. in Industrial & Operations Engineering from the Universit of Michigan. This tet was tpeset in LATEX with the KOMA-Script bundle, using the GNU Emacs tet editor on a Fedora Linu sstem. The graphics were created using TikZ and Gnuplot. Copright 009 Michael Corral. Permission is granted to cop, distribute and/or modif this document under the terms of the GNU Free Documentation License, Version. or an later version published b the Free Software Foundation; with no Invariant Sections, no Front-Cover Tets, and no Back-Cover Tets. A cop of the license is included in the section entitled GNU Free Documentation License.

5 Preface This book covers elementar trigonometr. It is suitable for a one-semester course at the college level, though it could also be used in high schools. The prerequisites are high school algebra and geometr. This book basicall consists of m lecture notes from teaching trigonometr at Schoolcraft College over several ears, epanded with some eercises. There are eercises at the end of each section. I have tried to include some more challenging problems, with hints when I felt those were needed. An average student should be able to do most of the eercises. Answers and hints to man of the odd-numbered and some of the even-numbered eercises are provided in Appendi A. This tet probabl has a more geometric feel to it than most current trigonometr tets. Thatwas,infact,oneofthereasonsIwantedtowritethisbook. Ithinkthatapproachingthe subject with too much of an analtic emphasis is a bit confusing to students. It makes much of the material appear unmotivated. This book starts with the old-fashioned right triangle approach to the trigonometric functions, which is more intuitive for students to grasp. In m eperience, presenting the definitions of the trigonometric functions and then immediatel jumping into proving identities is too much of a detour from geometr to analsis for most students. So this book presents material in a ver different order than most books toda. For eample, after starting with the right triangle definitions and some applications, general (oblique) triangles are presented. That seems like a more natural progression of topics, instead of leaving general triangles until the end as is usuall the case. The goal of this book is a bit different, too. Instead of taking the (doomed) approach that students have to be shown that trigonometr is relevant to their everda lives (which inevitabl comes off as artificial), this book has a different mindset: preparing students to use trigonometr as it is used in other courses. Virtuall no students will ever in their everda life figure out the height of a tree with a protractor or determine the angular speed of a Ferris wheel. Students are far more likel to need trigonometr in other courses (e.g. engineering, phsics). I think that math instructors have a dut to prepare students for that. In Chapter 5 students are asked to use the free open-source software Gnuplot to graph some functions. However, an program can be used for those eercises, as long as it produces accurate graphs. Appendi B contains a brief tutorial on Gnuplot. There are a few eercises that require the student to write his or her own computer program to solve some numerical computation problems. There are a few code samples in Chapter 6, written in the Java and Pthon programming languages, hopefull sufficientl clear so that the reader can figure out what is being done even without knowing those languages. iii

6 iv PREFACE Octave and Sage are also mentioned. This book probabl discusses numerical issues more than most tets at this level (e.g. the numerical instabilit of Heron s formula for the area of a triangle, the secant method for solving trigonometric equations). Numerical methods probabl should have been emphasized even more in the tet, since it is rare when even a moderatel complicated trigonometric equation can be solved with elementar methods, and since mathematical software is so readil available. I wanted to keep this book as brief as possible. Someone once joked that trigonometr is two weeks of material spread out over a full semester, and I think that there is some truth to that. However, some decisions had to be made on what material to leave out. I had planned to include sections on vectors, spherical trigonometr- a subject which has basicall vanished from trigonometr tets in the last few decades (wh?) - and a few other topics, but decided against it. The hardest decision was to eclude Paul Rider s clever geometric proof of the Law of Tangents without using an sum-to-product identities, though I do give a reference to it. This book is released under the GNU Free Documentation License (GFDL), which allows others to not onl cop and distribute the book but also to modif it. For more details, see the included cop of the GFDL. So that there is no ambiguit on this matter, anone can make as man copies of this book as desired and distribute it as desired, without needing m permission. The PDF version will alwas be freel available to the public at no cost (go to Feel free to contact me at mcorral@schoolcraft.edu for an questions on this or an other matter involving the book (e.g. comments, suggestions, corrections, etc). I welcome our input. Jul 009 Livonia, Michigan MICHAEL CORRAL

7 Contents Preface iii Right Triangle Trigonometr. Angles Trigonometric Functions of an Acute Angle Applications and Solving Right Triangles Trigonometric Functions of An Angle Rotations and Reflections of Angles General Triangles 8. The Law of Sines The Law of Cosines The Law of Tangents The Area of a Triangle Circumscribed and Inscribed Circles Identities 65. Basic Trigonometric Identities Sum and Difference Formulas Double-Angle and Half-Angle Formulas Other Identities Radian Measure Radians and Degrees Arc Length Area of a Sector Circular Motion: Linear and Angular Speed Graphing and Inverse Functions 0 5. Graphing the Trigonometric Functions Properties of Graphs of Trigonometric Functions Inverse Trigonometric Functions Additional Topics 9 6. Solving Trigonometric Equations Numerical Methods in Trigonometr v

8 vi CONTENTS 6. Comple Numbers Polar Coordinates Appendi A: Answers and Hints to Selected Eercises 5 Appendi B: Graphing with Gnuplot 55 GNU Free Documentation License 60 Histor 68 Inde 69

9 Right Triangle Trigonometr Trigonometr is the stud of the relations between the sides and angles of triangles. The word trigonometr is derived from the Greek words trigono (τρίγωνo), meaning triangle, and metro (µǫτρώ), meaning measure. Though the ancient Greeks, such as Hipparchus and Ptolem, used trigonometr in their stud of astronom between roughl 50 B.C. - A.D. 00, its histor is much older. For eample, the Egptian scribe Ahmes recorded some rudimentar trigonometric calculations (concerning ratios of sides of pramids) in the famous Rhind Paprus sometime around 650 B.C. Trigonometr is distinguished from elementar geometr in part b its etensive use of certain functions of angles, known as the trigonometric functions. Before discussing those functions, we will review some basic terminolog about angles.. Angles Recall the following definitions from elementar geometr: (a) An angle is acute if it is between 0 and 90. (b) An angle is a right angle if it equals 90. (c) An angle is obtuse if it is between 90 and 80. (d) An angle is a straight angle if it equals 80. (a) acute angle (b) right angle (c) obtuse angle (d) straight angle Figure.. Tpes of angles In elementar geometr, angles are alwas considered to be positive and not larger than 60. For now we will onl consider such angles. The following definitions will be used throughout the tet: Ahmes claimed that he copied the paprus from a work that ma date as far back as 000 B.C. Later in the tet we will discuss negative angles and angles larger than 60.

10 Chapter Right Triangle Trigonometr. (a) Two acute angles are complementar if their sum equals 90. In other words, if 0 A, B 90 then A and B are complementar if A+ B=90. (b) Two angles between 0 and 80 are supplementar if their sum equals 80. In other words, if 0 A, B 80 then A and B are supplementar if A+ B=80. (c) Two angles between 0 and 60 are conjugate (or eplementar) if their sum equals 60. Inotherwords,if0 A, B 60 then A and Bareconjugateif A+ B= 60. B A B A B A (a) complementar (b) supplementar Figure.. Tpes of pairs of angles (c) conjugate Instead of using the angle notation A to denote an angle, we will sometimes use just a capital letter b itself (e.g. A, B, C) or a lowercase variable name (e.g.,, t). It is also common to use letters (either uppercase or lowercase) from the Greek alphabet, shown in the table below, to represent angles: Table. The Greek alphabet Letters Name Letters Name Letters Name A α alpha I ι iota P ρ rho B β beta K κ kappa Σ σ sigma Γ γ gamma Λ λ lambda T τ tau δ delta M µ mu Υ υ upsilon E ǫ epsilon N ν nu Φ φ phi Z ζ zeta Ξ ξ i X χ chi H η eta O o omicron Ψ ψ psi Θ θ theta Π pi Ω ω omega In elementar geometr ou learned that the sum of the angles in a triangle equals 80, and that an isosceles triangle is a triangle with two sides of equal length. Recall that in a right triangle one of the angles is a right angle. Thus, in a right triangle one of the angles is 90 and the other two angles are acute angles whose sum is 90 (i.e. the other two angles are complementar angles).

11 Angles Section. Eample. For each triangle below, determine the unknown angle(s): B 5 E Y α A 5 0 C D F X α α Z Note: We will sometimes refer to the angles of a triangle b their verte points. For eample, in the first triangle above we will simpl refer to the angle BAC as angle A. Solution: For triangle ABC, A=5 and C=0, and we know that A+B+C=80, so 5 + B + 0 = 80 B = B = 5. For the right triangle DEF, E=5 and F =90, and we know that the two acute angles D and E are complementar, so D + E = 90 D = 90 5 D = 7. For triangle XYZ, the angles are in terms of an unknown number α, but we do know that X+Y + Z=80, which we can use to solve for α and then use that to solve for X, Y, and Z: α + α + α = 80 5α = 80 α = 6 X =6, Y = 6 =08, Z=6 Eample. Thales Theorem states that if A, B, and C are (distinct) points on a circle such that the line segment AB is a diameter of the circle, then the angle ACB is a right angle (see Figure..(a)). In other words, the triangle ABC is a right triangle. C C α β A O B A α O β B (a) (b) Figure.. Thales Theorem: ACB=90 To prove this, let O be the center of the circle and draw the line segment OC, as in Figure..(b). Let α = BAC and β = ABC. Since AB is a diameter of the circle, OA and OC have the same length (namel, the circle s radius). This means that OAC is an isosceles triangle, and so OCA= OAC = α. Likewise, OBC is an isosceles triangle and OCB = OBC = β. So we see that ACB= α+β. Andsincetheanglesof ABC mustaddupto80,weseethat80 = α+(α+β)+β= (α+β), so α+β=90. Thus, ACB=90. QED

12 4 Chapter Right Triangle Trigonometr. In a right triangle, the side opposite the right angle is called the hpotenuse, and the other two sides are called its legs. For eample, in Figure..4 the right angle is C, the hpotenuse is the line segment AB, which has length c, and BC and AC are the legs, with lengths a and b, respectivel. The hpotenuse is alwas the longest side of a right triangle (see Eercise ). B knowing the lengths of two sides of a right triangle, the length of the third side can be determined b using the Pthagorean Theorem: A B c a b C Figure..4 Theorem.. Pthagorean Theorem: The square of the length of the hpotenuse of a right triangle is equal to the sum of the squares of the lengths of its legs. Thus, if a right triangle has a hpotenuse of length c and legs of lengths a and b, as in Figure..4, then the Pthagorean Theorem sas: a + b = c (.) Let us prove this. In the right triangle ABC in Figure..5(a) below, if we draw a line segmentfromthevertec tothepointd onthehpotenusesuchthatcd isperpendicular to AB (that is, CD forms a right angle with AB), then this divides ABC into two smaller triangles CBD and ACD, which are both similar to ABC. c c d D d B a a B d b C A b C C D A c d D (a) ABC (b) CBD (c) ACD Figure..5 Similar triangles ABC, CBD, ACD Recall that triangles are similar if their corresponding angles are equal, and that similarit implies that corresponding sides are proportional. Thus, since ABC is similar to CBD, b proportionalit of corresponding sides we see that AB is to CB (hpotenuses) as BC is to BD (vertical legs) c a = a d cd = a. Since ABC is similar to ACD, comparing horizontal legs and hpotenuses gives b c d = c b b = c cd = c a a + b = c. QED Note: The smbols and denote perpendicularit and similarit, respectivel. For eample, in the above proof we had CD AB and ABC CBD ACD.

13 Angles Section. 5 Eample. For each right triangle below, determine the length of the unknown side: 5 B a E z Y A 4 C D e F X Z Solution: For triangle ABC, the Pthagorean Theorem sas that a + 4 = 5 a = 5 6 = 9 a =. For triangle DEF, the Pthagorean Theorem sas that e + = e = 4 = e =. For triangle XYZ, the Pthagorean Theorem sas that Eample.4 + = z z = z =. A 7 ft ladder leaning against a wall has its foot 8 ft from the base of the wall. At what height is the top of the ladder touching the wall? Solution: Let h be the height at which the ladder touches the wall. We can assume that the ground makes a right angle with the wall, as in the picture on the right. Then we see that the ladder, ground, and wall form a right triangle with a hpotenuse of length 7 ft (the length of the ladder) and legs with lengths 8 ft and h ft. So b the Pthagorean Theorem, we have h + 8 = 7 h = = 5 h = 5 ft h Eercises For Eercises -4, find the numeric value of the indicated angle(s) for the triangle ABC.. Find B if A=5 and C=50.. Find C if A=0 and B=.. Find A and B if C=4, A= α, and B=α. 4. Find A, B, and C if A= β and B=C=4β. For Eercises 5-8, find the numeric value of the indicated angle(s) for the right triangle ABC, with C being the right angle. 5. Find B if A= Find A and B if A= α and B=α. 7. Find A and B if A= φ and B= φ. 8. Find A and B if A= θ and B=/θ. 9. A car goes 4 miles due north then 7 miles due east. What is the straight distance between the car s starting point and end point?

14 6 Chapter Right Triangle Trigonometr. 0. One end of a rope is attached to the top of a pole 00 ft high. If the rope is 50 ft long, what is the maimum distance along the ground from the base of the pole to where the other end can be attached? You ma assume that the pole is perpendicular to the ground.. Prove that the hpotenuse is the longest side in ever right triangle. (Hint: Is a +b >a?). Can a right triangle have sides with lengths, 5, and 6? Eplain our answer.. If the lengths a, b, and c of the sides of a right triangle are positive integers, with a +b = c, then the form what is called a Pthagorean triple. The triple is normall written as (a,b,c). For eample, (,4,5) and (5,,) are well-known Pthagorean triples. (a) Show that (6,8,0) is a Pthagorean triple. (b) Show that if (a,b,c) is a Pthagorean triple then so is (ka,kb,kc) for an integer k>0. How would ou interpret this geometricall? (c) Show that (mn,m n,m +n ) is a Pthagorean triple for all integers m>n>0. (d) The triple in part(c) is known as Euclid s formula for generating Pthagorean triples. Write down the first ten Pthagorean triples generated b this formula, i.e. use: m= and n=; m= and n=, ; m=4 and n=,, ; m=5 and n=,,, This eercise will describe how to draw a line through an point outside a circle such that the lineintersectsthecircleatonlonepoint. Thisiscalledatangentlinetothecircle(seethepicture on the left in Figure..6), a notion which we will use throughout the tet. tangent line A not tangent O C P Figure..6 On a sheet of paper draw a circle of radius inch, and call the center of that circle O. Pick a point P which is.5 inches awa from O. Draw the circle which has OP as a diameter, as in the picture on the right in Figure..6. Let A be one of the points where this circle intersects the first circle. Draw the line through P and A. In general the tangent line through a point on a circle is perpendicular to the line joining that point to the center of the circle (wh?). Use this fact to eplain wh the line ou drew is the tangent line through A and to calculate the length of PA. Does it match the phsical measurement of PA? 5. Suppose that ABC is a triangle with side AB a diameter of a circle with center O, as in the picture on the right, and suppose that the verte C lies on the circle. Now imagine that ou rotate the circle 80 around its center, so that ABC is in a new position, as indicated b the dashed lines in the picture. Eplain how this picture proves Thales Theorem. A O C B

15 Trigonometric Functions of an Acute Angle Section. 7. Trigonometric Functions of an Acute Angle Consider a right triangle ABC, with the right angle at C and with lengths a, b, and c, as in the figure on the right. For the acute angle A, call the leg BC its opposite side, and call the leg AC its adjacent side. Recall that the hpotenuse of the triangle is the side AB. The ratios of sides of a right triangle occur often enough in practical applications to warrant their own names, so we define the si trigonometric functions of A as follows: A c hpotenuse b adjacent B a C opposite Table. The si trigonometric functions of A Name of function Abbreviation Definition sine A sin A = cosine A cos A = tangent A tan A = cosecant A csc A = secant A sec A = cotangent A cot A = opposite side hpotenuse adjacent side hpotenuse opposite side adjacent side hpotenuse opposite side hpotenuse adjacent side adjacent side opposite side = a c = b c = a b = c a = c b = b a We will usuall use the abbreviated names of the functions. Notice from Table. that the pairs sin A and csc A, cos A and sec A, and tan A and cot A are reciprocals: csc A = sin A sec A = cos A cot A = tan A sin A = csc A cos A = sec A tan A = cot A

16 8 Chapter Right Triangle Trigonometr. Eample.5 Fortherighttriangle ABC shownontheright,findthevaluesofallsitrigonometric functions of the acute angles A and B. Solution: The hpotenuse of ABC has length 5. For angle A, the opposite side BC has length and the adjacent side AC has length 4. Thus: A 5 4 B C sin A = opposite hpotenuse = 5 cos A = adjacent hpotenuse = 4 5 tan A = opposite adjacent = 4 csc A = hpotenuse opposite = 5 sec A = hpotenuse adjacent = 5 4 cot A = adjacent opposite = 4 For angle B, the opposite side AC has length 4 and the adjacent side BC has length. Thus: sin B = opposite hpotenuse = 4 5 cos B = adjacent hpotenuse = 5 tan B = opposite adjacent = 4 csc B = hpotenuse opposite = 5 4 sec B = hpotenuse adjacent = 5 cot B = adjacent opposite = 4 Notice in Eample.5 that we did not specif the units for the lengths. This raises the possibilit that our answers depended on a triangle of a specific phsical size. For eample, suppose that two different students are reading this tetbook: one in the United States and one in German. The American student thinks that the lengths, 4, and 5 in Eample.5 are measured in inches, while the German student thinks that the are measured in centimeters. Since in.54 cm, the students are using triangles of different phsical sizes (see Figure.. below, not drawn to scale). B B 5 B B A 4 C A 5 4 C A A C C (a) Inches (b) Centimeters (c) Similar triangles Figure.. ABC A B C If the American triangle is ABC and the German triangle is A B C, then we see from Figure.. that ABC is similar to A B C, and hence the corresponding angles

17 Trigonometric Functions of an Acute Angle Section. 9 are equal and the ratios of the corresponding sides are equal. In fact, we know that common ratio: the sides of ABC are approimatel.54 times longer than the corresponding sides of A B C. So when the American student calculates sin A and the German student calculates sin A, the get the same answer: ABC A B C BC B C = AB A B BC AB = B C A B sin A = sin A Likewise, the other values of the trigonometric functions of A and A are the same. In fact, our argument was general enough to work with an similar right triangles. This leads us to the following conclusion: When calculating the trigonometric functions of an acute angle A, ou ma use an right triangle which has A as one of the angles. Since we defined the trigonometric functions in terms of ratios of sides, ou can think of the units of measurement for those sides as canceling out in those ratios. This means that the values of the trigonometric functions are unitless numbers. So when the American student calculated /5 as the value of sin A in Eample.5, that is the same as the /5 that the German student calculated, despite the different units for the lengths of the sides. Eample.6 Find the values of all si trigonometric functions of 45. Solution: Since we ma use an right triangle which has 45 as one of the angles, use the simplest one: take a square whose sides are all unit long and divide it in half diagonall, as in the figure on the right. Since the two legs of the triangle ABC have the same length, ABC is an isosceles triangle, which means that the angles A and B are equal. So since A+B = 90, this means that we must have A = B = 45. B the Pthagorean Theorem, the length c of the hpotenuse is given b A 45 B C c = + = c =. Thus, using the angle A we get: sin 45 = opposite hpotenuse = cos 45 = adjacent hpotenuse = tan 45 = opposite adjacent = = csc 45 = hpotenuse opposite = sec 45 = hpotenuse adjacent = cot 45 = adjacent opposite = = Note that we would have obtained the same answers if we had used an right triangle similar to ABC. For eample, if we multipl each side of ABC b, then we would have a similar triangle with legs of length and hpotenuse of length. This would give us sin 45 =, which equals = as before. The same goes for the other functions. We will use the notation AB to denote the length of a line segment AB.

18 0 Chapter Right Triangle Trigonometr. Eample.7 Find the values of all si trigonometric functions of 60. Solution: Since we ma use an right triangle which has 60 as one of the angles, we will use a simple one: take a triangle whose sides are all units long and divide it in half b drawing the bisector from one verte to the opposite side, as in the figure on the right. Since the original triangle was an equilateral triangle (i.e. all three sides had the same length), its three angles were all the same, namel 60. Recall from elementar geometr that the bisector from the verte angle of an equilateral triangle to its opposite side bisects both the verte angle and the opposite side. So as in the figure on the right, the triangle ABC has angle A = 60 and angle B=0, which forces the angle C to be 90. Thus, ABC is a right A B C triangle. We see that the hpotenuse has length c= AB= and the leg AC has length b= AC =. B the Pthagorean Theorem, the length a of the leg BC is given b Thus, using the angle A we get: a + b = c a = = a =. sin 60 = opposite hpotenuse = cos 60 = adjacent hpotenuse = tan 60 = opposite adjacent = = csc 60 = hpotenuse opposite = sec 60 = hpotenuse adjacent = cot 60 = adjacent opposite = Notice that, as a bonus, we get the values of all si trigonometric functions of 0, b using angle B=0 in the same triangle ABC above: sin 0 = opposite hpotenuse = cos 0 = adjacent hpotenuse = tan 0 = opposite adjacent = csc 0 = hpotenuse opposite = sec 0 = hpotenuse adjacent = cot 0 = adjacent opposite = = Eample.8 A isanacuteanglesuchthatsin A=. Findthevaluesoftheothertrigonometric B functions of A. Solution: In general it helps to draw a right triangle to solve problems of this tpe. The reason is that the trigonometric functions were defined in terms of ratios of sides of a right triangle, and ou are given one such function (the sine, A b C in this case) alread in terms of a ratio: sin A =. Since sin A is defined as opposite hpotenuse, use as the length of the side opposite A and use as the length of the hpotenuse in a right triangle ABC (see the figure above), so that sin A=. The adjacent side to A has unknown length b, but we can use the Pthagorean Theorem to find it: + b = b = 9 4 = 5 b = 5

19 Trigonometric Functions of an Acute Angle Section. We now know the lengths of all sides of the triangle ABC, so we have: cos A = adjacent hpotenuse = 5 tan A = opposite adjacent = 5 csc A = hpotenuse opposite = sec A = hpotenuse adjacent = 5 cot A = adjacent opposite = 5 You ma have noticed the connections between the sine and cosine, secant and cosecant, and tangent and cotangent of the complementar angles in Eamples.5 and.7. Generalizing those eamples gives us the following theorem: Theorem.. Cofunction Theorem: If A and B are the complementar acute angles in a right triangle ABC, then the following relations hold: sin A = cos B sec A = csc B tan A = cot B sin B = cos A sec B = csc A tan B = cot A We sa that the pairs of functions {sin,cos}, {sec,csc}, and {tan,cot} are cofunctions. So sine and cosine are cofunctions, secant and cosecant are cofunctions, and tangent and cotangent are cofunctions. That is how the functions cosine, cosecant, and cotangent got the co in their names. The Cofunction Theorem sas that an trigonometric function of an acute angle is equal to its cofunction of the complementar angle. Eample.9 Write each of the following numbers as trigonometric functions of an angle less than 45 : (a) sin 65 ; (b) cos 78 ; (c) tan 59. Solution: (a) The complement of 65 is = 5 and the cofunction of sin is cos, so b the Cofunction Theorem we know that sin 65 =cos 5. (b) The complement of 78 is = and the cofunction of cos is sin, so cos 78 =sin. (c) The complement of 59 is = and the cofunction of tan is cot, so tan 59 =cot. a 45 a 45 a (a) a 60 a 0 a (b) Figure.. Two general right triangles (an a > 0) The angles 0, 45, and 60 arise often in applications. We can use the Pthagorean Theorem to generalize the right triangles in Eamples.6 and.7 and see what an and right triangles look like, as in Figure.. above.

20 Chapter Right Triangle Trigonometr. Eample.0 Find the sine, cosine, and tangent of 75. Solution: Since75 =45 +0,placea righttriangle ADB with legs of length and on top of the hpotenuse of a right triangle ABC whose hpotenuse has length, asinthefigureontheright. FromFigure..(a)we know that the length of each leg of ABC is the length of the hpotenuse divided b. So AC =BC = =. Draw DE perpendicular to AC, so that ADE is a right triangle. Since BAC =45 and DAB=0, we see that DAE =75 since it is the sum of those two angles. Thus, we need to find the sine, cosine, and tangent of DAE. Noticethat ADE=5,sinceitisthecomplementof DAE. And ADB =60, since it is the complement of DAB. Draw BF perpendicular to DE, so that DFB is a right triangle. Then BDF =45, since it is the difference of ADB=60 and ADE =5. Also, DBF =45 since it is the complement of BDF. The hpotenuse BD of DFB has length and DFB is a right triangle, so we know that DF =FB=. Now, we know that DE AC and BC AC, so FE and BC are parallel. Likewise, FB and EC are both perpendicular to DE and hence FB is parallel to EC. Thus, FBCE is a rectangle, since BCE is a right angle. So EC=FB= and FE=BC=. Hence, DE = DF + FE = + = + and AE = AC EC = = sin 75 = DE AD = + = 6+ 4, cos 75 = AE AD = = 6 4, and tan 75 = DE AE = Note: Taking reciprocals, we get csc 75 = 4 6+, sec 75 = 4 6, and cot 75 = A 0 45 F D E B C. Thus, + = Eercises For Eercises -0, find the values of all si trigonometric functions of angles A and B in the right triangle ABC in Figure.... a=5, b=, c=. a=8, b=5, c=7. a=7, b=4, c=5 4. a=0, b=, c=9 5. a=9, b=40, c=4 6. a=, b=, c= 5 7. a=, b= B c a A b C Figure.. 8. a=, b=5 9. a=5, c=6 0. b=7, c=8 For Eercises -8, find the values of the other five trigonometric functions of the acute angle A given the indicated value of one of the functions.

21 Trigonometric Functions of an Acute Angle Section.. sin A= 4. cos A=. cos A= 0 4. sin A= 4 5. tan A= tan A= 7. sec A= 7 8. csc A= For Eercises 9-, write the given number as a trigonometric function of an acute angle less than sin sin 5. cos 46. tan 66. sec 77 For Eercises 4-8, write the given number as a trigonometric function of an acute angle greater than sin 5. cos 6. tan 6 7. cot 0 8. csc 4 9. In Eample.7 we found the values of all si trigonometric functions of 60 and 0. (a) Does sin 0 + sin 0 = sin 60? (b) Does cos 0 + cos 0 = cos 60? (c) Does tan 0 + tan 0 = tan 60? (d) Does sin 0 cos 0 = sin 60? 0. For an acute angle A, can sin A be larger than? Eplain our answer.. For an acute angle A, can cos A be larger than? Eplain our answer.. For an acute angle A, can sin A be larger than tan A? Eplain our answer.. If A and B are acute angles and A<B, eplain wh sin A<sin B. 4. If A and B are acute angles and A<B, eplain wh cos A>cos B. 5. Prove the Cofunction Theorem (Theorem.). (Hint: Draw a right triangle and label the angles and sides.) 6. Use Eample.0 to find all si trigonometric functions of In Figure..4, CB is a diameter of a circle with a radius of cm and center O, ABC is a right triangle, and CD has length cm. (a) Find sin A. (Hint: Use Thales Theorem.) (b) Find the length of AC. (c) Find the length of AD. (d) Figure..4 is drawn to scale. Use a protractor to measure the angle A, then use our calculator to find the sine of that angle. Is the calculator result close to our answer from part(a)? Note: Make sure that our calculator is in degree mode. A C D O Figure..4 B 8. In Eercise 7, verif that the area of ABC equals AB CD. Wh does this make sense? 9. In Eercise 7, verif that the area of ABC equals AB AC sin A. 40. In Eercise 7, verif that the area of ABC equals (BC) cot A.