`LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6057Trigonometric Identities and EquationsIn 1831 Michael Faraday discovered thatwhen a wire passes by a magnet, a small electric current is produced in the wire. Now we generate massive amounts of electricity by simultaneously rotating thousands of wires near large electromagnets. Because electric current alternates its direction on electrical wires, it is modeled accurately by either the sine or the cosine function. We give many examples of applications of the trigonometric functions to electricity and other phenomena in the examples and exercises in this chapter, including a model of the wattage consumption of a toaster in Section 7.4, Example 5.7.1 Fundamental Identities 7.2 Verifying Trigonometric Identities 7.3 Sum and Difference Identities 7.4 Double-Angle Identities and HalfAngle Identities Summary Exercises on Verifying Trigonometric Identities7.5 Inverse Circular Functions 7.6 Trigonometric Equations 7.7 Equations Involving InverseTrigonometric Functions605LIALMC07_0321227638.QXP2/26/0410:47 AMPage 606606 CHAPTER 7 Trigonometric Identities and Equations7.1 Fundamental IdentitiesNegative-Angle IdentitiesysFundamental IdentitiessUsing the Fundamental Identities(x, y) y ­y r (x, ­y) y sin(­ ) = ­ r = ­ sinFigure 1r x O ­xNegative-Angle Identities As suggested by the circle shown in Figure 1, an angle having the point x, y on its terminal side has a corresponding angle with the point x, y on its terminal side. From the definition of sine,sin so sin y r and sin y , r(Section 5.2)and sin are negatives of each other, or sin sin . in anyFigure 1 shows an angle in quadrant II, but the same result holds for quadrant. Also, by definition, cos so x r and cos cos cos . and cos sin cos tan . to find tan sin cos x , r(Section 5.2)We can use these identities for sin tan :TEACHING TIP Point out that intrigonometric identities, can be an angle in degrees, a real number, or a variable.in terms oftansin cos tanSimilar reasoning gives the following identities. csc csc , sec sec , cot cotThis group of identities is known as the negative-angle or negative-number identities.Fundamental Identities In Chapter 5 we used the definitions of the trigonometric functions to derive the reciprocal, quotient, and Pythagorean identities. Together with the negative-angle identities, these are called the fundamental identities.Fundamental IdentitiesReciprocal Identities cot Quotient Identities tan sin cos cot cos sin(continued)1 tansec1 coscsc1 sinLIALMC07_0321227638.QXP2/26/0410:47 AMPage 6077.1 Fundamental Identities607TEACHING TIP Encourage studentsto memorize the identities presented in this section as well as subsequent sections. Point out that numerical values can be used to help check whether or not an identity was recalled correctly.Pythagorean Identities sin2 cos2 1 tan2 1 sec2 1 cot2 csc2Negative-Angle Identities sin( csc( ) ) sin csc cos( sec( ) ) cos sec tan( cot( ) ) tan cotNOTE The most commonly recognized forms of the fundamental identities are given above. Throughout this chapter you must also recognize alternative cos2 1 forms of these identities. For example, two other forms of sin2 aresin21cos2andcos21sin2 .Using the Fundamental Identities One way we use these identities is to find the values of other trigonometric functions from the value of a given trigonometric function. Although we could find such values using a right triangle, this is a good way to practice using the fundamental identities.EXAMPLE 1 Finding Trigonometric Function Values Given One Value and the QuadrantIf tan (a) secSolution5 3and is in quadrant II, find each function value. (b) sin (c) cot(a) Look for an identity that relates tangent and secant. tan2 5 3TEACHING TIP Warn students thatthe given information in 5 5 Example 1, tan , 3 3 does not mean that sin 5 and cos 3. Ask them why these values cannot be correct.21 1sec2 sec2 sec2 sec2Pythagorean identity tan5 325 91 34 9 34 9 34 3Combine terms.sec secTake the negative square root. (Section 1.4)Simplify the radical. (Section R.7)We chose the negative square root since sec is negative in quadrant II.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 608608 CHAPTER 7 Trigonometric Identities and Equations(b)tan cos tan 1 sec 3 34 34 tan 5 3 sinsin cos sin sin sin 5 1 34 34Quotient identity Multiply by cos . Reciprocal identity1 From part (a), sec 3 34 3 34 34 ;tan5 3(c)cot cot cottan 1 tan 15 3Reciprocal identityNegative-angle identity3 5tan5 3;Simplify. (Section R.5)Now try Exercises 5, 7, and 9.To avoid a common error, when taking the square root, be sure to choose the sign based on the quadrant of and the function being evaluated.CAUTIONAny trigonometric function of a number or angle can be expressed in terms of any other function.EXAMPLE 2 Expressing One Function in Terms of AnotherExpress cos x in terms of tan x.Solution1tan2 x 1 1 1Since sec x is related to both cos x and tan x by identities, start with sec2 x. 1 tan2 x 1 tan2 x 1 tan2 x cos x cos x 1 sec2 x cos2 x cos x 1 tan2 x 1 1Take reciprocals.Reciprocal identityTake square roots. Quotient rule (Section R.7); rewrite.1tan2 x Rationalize the denominator. (Section R.7) tan2 x sign, depending on the quadrant of x.Now try Exercise 43.Choose thesign or theLIALMC07_0321227638.QXP2/26/0410:47 AMPage 6097.1 Fundamental Identities609We can use a graphing calculator to decide whether two functions are identical. See Figure 2, which supports the identity sin2 x cos2 x 1. With an identity, you should see no difference in the two graphs. s All other trigonometric functions can easily be expressed in terms of sin and/or cos . We often make such substitutions in an expression to simplify it.Y1 = Y24EXAMPLE 3 Rewriting an Expression in Terms of Sine and CosineWrite tan expression.­2 2cotin terms of sinand cos , and then simplify theSolution­4Figure 2tancotsin cos sin2 cos sincos sin cos2 cos sinQuotient identities Write each fraction with the LCD. (Section R.5) Add fractions.y2 =y1 = tan x + cot x 1 cos x sin x4sin2 cos2 cos sin tan cot 1 cos sinPythagorean identity­22Now try Exercise 55.­4 The graph supports the result in Example 3. The graphs of y1 and y2 appear to be identical.CAUTION When working with trigonometric expressions and identities, be sure to write the argument of the function. For example, we would not write sin2 cos2 1; an argument such as is necessary in this identity.7.1 Exercises1. 4. 7. 9. 2.6 .75 2 5 105 11 5 2. .65 3. .625 3 10 7 5. 6. 4 10 8. 10. 77 11 5 8Concept Check 1. If tan x 2. If cos x 3. If tan x 4. If cos xFill in the blanks. x x . .6, then tan x . . . .65, then cos2.6, then tan 1.6, then cot x .8 and sin xFind sin s. See Example 1. 5. cos s 7. cos 9. sec s s 3 , s in quadrant I 4 5 , tan s 5 11 , tan s 4 0 0 6. cot s 8. tan s 10. csc s 1 , s in quadrant IV 3 7 , sec s 2 8 5 09 11.Why is it unnecessary to give the quadrant of s in Exercise 10?LIALMC07_0321227638.QXP2/26/0410:47 AMPage 610610 CHAPTER 7 Trigonometric Identities and Equations 12. 15. 18. 19. 20. sin x 13. even 16. f x f x f x f f x f 3 2 5 3 5 2 5 6 ; sec 12 5 6 12 17 ; 17 4 17 ; cot 17 17 ; csc 4 2 ; 5 21 ; tan 5 21 ; sec 2 3 ; cos 5 2 5 4; 17 5 5 odd 14. cos x tan x 17. odd x x 5 ; 5 2 3 2 2 5; 6; ;Relating ConceptsFor individual or collaborative investigation (Exercises 12­17)21. cos tan sec 22. sin cot csc 23. sin cos sec 24. sin cos cot 25. sin tan; cot ; csc 6 ; tanf x for all x in the domain of f. A funcA function is called an even function if f x f x for all x in the domain of f. Work tion is called an odd function if f x Exercises 12­17 in order, to see the connection between the negative-angle identities and even and odd functions. 12. Complete the statement: sin 13. Is the function defined by f x 14. Complete the statement: cos 15. Is the function defined by f x 16. Complete the statement: tan 17. Is the function defined by f x x x x . sin x even or odd? . cos x even or odd? . tan x even or odd?Concept Check is true. 18.For each graph, determine whether f4xf x or f4xf x19.­22­2221 ; 21 21 21­4 ­44 ; 520.43 5 5 ; sec ; csc 4 4 3 3 4 26. cos ; tan ; 5 3 3 5 cot ; sec ; 4 3 5 csc 4 7 3 27. sin ; cos ; 4 4 tan csc 7 3 4 7 ; cot 7 3 7 7 ;­22­4Find the remaining five trigonometric functions of . See Example 1. 21. sin 23. tan 25. cot 27. sec 2 , 3 1 , 4 4 , sin 3 4 , sin 3 in quadrant II in quadrant IV 0 0 22. cos 24. csc 26. sin 28. cos 1 , 5 5 , 2 in quadrant I in quadrant III 0 04 , cos 5 1 , sin 4LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6117.1 Fundamental Identities61128. sin tan sec 29. B 33. A 37. D 41. sin 42. tan 43. sin x 44. cot x 45. tan x 46. cot x 47. csc x 48. 49. 52. 55. 56. 58. 60. 62.15 ; 4 15; cot 4 15 ; 15Concept Check For each expression in Column I, choose the expression from Column II that completes an identity. I cos x 29. sin x 30. tan x 31. cos 32. tan2 x 33. 1 x 1 II A. sin2 x B. cot x C. sec2 x D. sin x cos x cos2 x15 4; csc 15 30. D 31. E 32. C 34. C 35. A 36. E 38. B 2x 1 x 1 2 1 2p p 4 cos2 xE. cos xConcept Check For each expression in Column I, choose the expression from Column II that completes an identity. You may have to rewrite one or both expressions. I 34. tan x cos x 1 A. B. sin x cos2 x 1 sec2 x x cot 2 x sin2 x21 sin2 x sin x sec2 x csc2 x 1 1 1 1 1IIcos2 x cos2 x35. sec2 x 36. sec x csc xsin2 x sec x 1 sin2 x 50. 1 51. cot cos 53. cos2 54. tan2 csc2 sec cos 57. cot 1 cot tan sin2 59. sin2 cos2 cos 61. cos2 cot tan tan2C. sin D. csc2 x E. tan x cot 237. 12sin2 x38. cos x9 39. 9 40.A student writes &quot;1csc2.&quot; Comment on this student's work.Another student makes the following claim: &quot;Since sin2 cos2 1, I should be able to also say that sin cos 1 if I take the square root of both sides.&quot; Comment on this student's statement. Suppose that cos Find tan if secx x p p 1. 441. Concept Check 42. Concept CheckFind sin ..Write the first trigonometric function in terms of the second trigonometric function. See Example 2. 43. sin x; cos x 46. cot x; csc x 44. cot x; sin x 47. csc x; cos x 45. tan x; sec x 48. sec x; sin xWrite each expression in terms of sine and cosine, and simplify it. See Example 3. 49. cot sin 52. cot 55. 1250. sec cot sin tan 1251. cos csc 1 54. sec sin sin sin2 cot 2 csc tan cot 221 cos53. sin sec2csc21 sec156.coscos2 sin2 57. sin cos 59. sec 61. sin cos csc sin1 58. 1 60. sec 62. 1 1cossinLIALMC07_0321227638.QXP2/26/0410:47 AMPage 612612 CHAPTER 7 Trigonometric Identities and Equations63. sec2 64. sin 25 6 25 6 60 65. ; 12 12 66. 26063. sin2tan2cos2 Let cos x Let csc x1 5.64.tan sec65. Concept Check 66. Concept CheckFind all possible values for sec xsin xtan x . 3. Find all possible values for sin xsec xcos x .2 2 8 2 8 ; 9 9 67. sin 2x 68. It is the negative of sin 2x . 69. cos 4x 70. It is the same function. 71. (a) sin 4x (b) cos 2x (c) 5 sin 3x 72. identity 73. not an identity 74. not an identity 75. identity 76. not an identityRelating ConceptsFor individual or collaborative investigation (Exercises 67­71)In Chapter 6 we graphed functions defined by y with the assumption that b ercises 67 ­ 71 in order. c a f bx d 0, work Ex-0. To see what happens when b sin cos67. Use a negative-angle identity to write y 69. Use a negative-angle identity to write y2x as a function of 2x. sin 2x ? 4x as a function of 4x. cos 4x ?68. How does your answer to Exercise 67 relate to y 70. How does your answer to Exercise 69 relate to y71. Use your results from Exercises 67­70 to rewrite the following with a positive value of b. (a) sin 4x (b) cos 2x (c) 5 sin 3xUse a graphing calculator to decide whether each equation is an identity. (Hint: In Exercise 76, graph the function of x for a few different values of y (in radians).) 72. cos 2x 74. sin x 76. cos x y 1 1 2 sin2 x cos x cos x cos y273. 2 sin s 75. cos 2xsin 2s cos2 x sin2 x7.2 Verifying Trigonometric IdentitiesVerifying Identities by Working with One SidesVerifying Identities by Working with Both SidesRecall that an identity is an equation that is satisfied for all meaningful replacements of the variable. One of the skills required for more advanced work in mathematics, especially in calculus, is the ability to use identities to write expressions in alternative forms. We develop this skill by using the fundamental identities to verify that a trigonometric equation is an identity (for those values of the variable for which it is defined). Here are some hints to help you get started.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6137.2 Verifying Trigonometric Identities613Looking Ahead to CalculusTrigonometric identities are used in calculus to simplify trigonometric expressions, determine derivatives of trigonometric functions, and change the form of some integrals.Hints for Verifying Identities1. Learn the fundamental identities given in the last section. Whenever you see either side of a fundamental identity, the other side should come to mind. Also, be aware of equivalent forms of the fundamental identities. 1 cos2 is an alternative form of the identity For example, sin2 2 2 cos sin 1. 2. Try to rewrite the more complicated side of the equation so that it is identical to the simpler side. 3. It is sometimes helpful to express all trigonometric functions in the equation in terms of sine and cosine and then simplify the result. 4. Usually, any factoring or indicated algebraic operations should be performed. For example, the expression sin2 x 2 sin x 1 can be factored as sin x 1 2. The sum or difference of two trigonometric expressions, 1 1 such as sin cos , can be added or subtracted in the same way as any other rational expression. 1 sin 1 cos cos sin cos cos sin sin cos 5. As you select substitutions, keep in mind the side you are not changing, because it represents your goal. For example, to verify the identity tan2 x 1 1 , cos2 x sin sin cosTEACHING TIP There is no substitute for experience when it comes to verifying identities. Guide students through several examples, giving hints such as &quot;Apply a reciprocal identity,&quot; or &quot;Use a different form of the Pythagorean identity cos2 1.&quot; sin2try to think of an identity that relates tan x to cos x. In this case, since 1 2 sec x tan2 x 1, the secant function is the best link cos x and sec x between the two sides. 6. If an expression contains 1 sin x, multiplying both numerator and denominator by 1 sin x would give 1 sin2 x, which could be replaced with cos2 x. Similar results for 1 sin x, 1 cos x, and 1 cos x may be useful.CAUTION Verifying identities is not the same as solving equations. Techniques used in solving equations, such as adding the same terms to both sides, or multiplying both sides by the same term, should not be used when working with identities since you are starting with a statement (to be verified) that may not be true.Verifying Identities by Working with One Side To avoid the temptation to use algebraic properties of equations to verify identities, work with only one side and rewrite it to match the other side, as shown in Examples 1 ­4.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 614614 CHAPTER 7 Trigonometric Identities and EquationsEXAMPLE 1 Verifying an Identity (Working with One Side)Verify that the following equation is an identity. cot sSolution1csc s cos ssin sWe use the fundamental identities from Section 7.1 to rewrite one side of the equation so that it is identical to the other side. Since the right side is more complicated, we work with it, using the third hint to change all functions to sine or cosine.StepsFor s = x, cot x + 1 = csc x (cos x + sin x)4ReasonsRight side of given equation        csc s cos s2sin s1 cos s sin s cos s sin s cot s sin s sin s 1sin scsc s1 sin s­2Distributive property (Section R.1)cos s sin scot s; sin s sin s1    ­4 The graphs coincide, supporting the conclusion in Example 1.Left side of given equationThe given equation is an identity since the right side equals the left side.Now try Exercise 33.EXAMPLE 2 Verifying an Identity (Working with One Side)Verify that the following equation is an identity. tan2 x 1Solutioncot 2 x11 sin2 xWe work with the more complicated left side, as suggested in the second hint. Again, we use the fundamental identities from Section 7.1. tan2 x 1 cot 2 x tan2 x tan2 x tan2 x sec x2tan2 x cot 2 x tan2 x 1 1 tan2 xDistributive property cot 2 x tan2 x tan x sec2 x cos2 x2tan2 x(1 +cot2 x)1 = 1 ­ sin2 x 41 tan2 x 1 tan2 x1 sec2 x1­221 cos2 x 1 1 sin2 x1 cos2 x­4 The screen supports the conclusion in Example 2.1sin2 xSince the left side is identical to the right side, the given equation is an identity.Now try Exercise 37.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6157.2 Verifying Trigonometric Identities615EXAMPLE 3 Verifying an Identity (Working with One Side)Verify that the following equation is an identity. tan t cot t sin t cos tSolutionsec2 tcsc2 tWe transform the more complicated left side to match the right side. tan t sin t cos t tan t cot t sin t cos t cot t 1 sin t cos ta c b a c b ctan t cot t sin t cos t(Section R.5)1 sin t cos ta ba1 bsin t 1 cos t sin t cos t 1 cos2 t sec2 t 1 sin2 t csc2 tcos t 1 sin t sin t cos ttan tsin t cos t ;cot tcos t sin t1 cos2 t1 sec2 t; sin2 tcsc2 tThe third hint about writing all trigonometric functions in terms of sine and cosine was used in the third line of the solution.Now try Exercise 41.EXAMPLE 4 Verifying an Identity (Working with One Side)TEACHING TIP Show that the identity in Example 4 can also be verified by multiplying the numerator and denominator of the left side by 1 sin x.Verify that the following equation is an identity. cos x 1 sin xSolution1sin x cos xWe work on the right side, using the last hint in the list given earlier to multiply numerator and denominator on the right by 1 sin x. 1 sin x cos x 1 sin x 1 sin x cos x 1 sin xMultiply by 1. (Section R.1) x y x sin2 x y x2 y 2 (Section R.3)1 sin2 x cos x 1 sin x cos2 x cos x 1 sin x 1 cos x sin x1cos2 xLowest terms (Section R.5)Now try Exercise 47.Verifying Identities by Working with Both Sides If both sides of an identity appear to be equally complex, the identity can be verified by working independently on the left side and on the right side, until each side is changed into some common third result. Each step, on each side, must be reversible.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 616616 CHAPTER 7 Trigonometric Identities and Equationsleftrightcommon third expressionWith all steps reversible, the procedure is as shown in the margin. The left side leads to a common third expression, which leads back to the right side. This procedure is just a shortcut for the procedure used in Examples 1­ 4: one side is changed into the other side, but by going through an intermediate step.EXAMPLE 5 Verifying an Identity (Working with Both Sides)Verify that the following equation is an identity. sec secSolutiontan tan12 sin cos2sin2Both sides appear equally complex, so we verify the identity by changing each side into a common third expression. We work first on the left, multiplying numerator and denominator by cos . sec sec tan tan sec sec sec sec 1 1 1 1 1 1 sin2 cos cos tan tan sin cos sin cos sin sin2tan tancos cos tan tan cos cosMultiply by 1.Distributive propertycos cos cosseccos1tansin coscosOn the right side of the original equation, begin by factoring. 1 2 sin cos2 1 1 1 1 1 1 We have shown thatLeft side of given equation        Common third expression      Right side of given equationsin cos2 sin sin2 1 sin sin sinx22xyy2xy2(Section R.4)2cos21sin2 sin2 .sin 12sinFactor 1Lowest termssec sectan tan1 1sin sinverifying that the given equation is an identity.Now try Exercise 51.        12 sin cos2sin2,LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6177.2 Verifying Trigonometric Identities617CAUTIONUse the method of Example 5 only if the steps are reversible.There are usually several ways to verify a given identity. For instance, another way to begin verifying the identity in Example 5 is to work on the left as follows. 1 cos 1 cos sin cos 1 sin cos 1 1 sin sin 1 sin cos sin cossec sectan tanFundamental identities (Section 7.1)Add and subtract fractions.(Section R.5)Simplify the complex fraction.(Section R.5)Compare this with the result shown in Example 5 for the right side to see that the two sides indeed agree.EXAMPLE 6 Applying a Pythagorean Identity to RadiosTuners in radios select a radio station by adjusting the frequency. A tuner may contain an inductor L and a capacitor C, as illustrated in Figure 3. The energy stored in the inductor at time t is given by Lt k sin2 2 Ft k cos2 2 Ft ,and the energy stored in the capacitor is given by Ctwhere F is the frequency of the radio station and k is a constant. The total energy E in the circuit is given byL CEtAn Inductor and a CapacitorFigure 3LtCt.Show that E is a constant function. (Source: Weidner, R. and R. Sells, Elementary Classical Physics, Vol. 2, Allyn &amp; Bacon, 1973.)SolutionEtLt2 2Ct k cos 2 Ft cos 2 Ft2 2Given equation Substitute. Factor. (Section R.4) sin2 cos2 1 (Here 2 Ft).k sin 2 Ft k sin 2 Ft k1 kSince k is constant, E t is a constant function.Now try Exercise 85.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 618618 CHAPTER 7 Trigonometric Identities and Equations7.2 Exercises1 sin cos 1 2. csc x sec x or sin x cos x 3. 1 sec s 4. 1 cot 2 5. 1 6. or 2 sec2 cos2 1 7. 1 8. or csc sin 9. 2 2 sin t 10. sec2 s 2 cos x 11. or 2 cot x csc x sin2 x 12. 1 2 sin cos 13. sin 1 sin 1 14. sec 1 sec 1 15. 4 sin x 16. 4 tan x cot x or 4 17. 2 sin x 1 sin x 1 18. 4 tan 3 tan 1 19. cos2 x 1 2 20. cot 2 x 2 cot 2 x 1 or csc2 x cot 2 x 2 21. sin x cos x 1 sin x cos x 22. sin cos 1 sin cos 23. sin 24. cos 25. 1 26. 1 27. tan2 28. sec2 29. tan2 x 30. cot 2 t 31. sec2 x 32. csc2 1. csc sec orPerform each indicated operation and simplify the result. 1. cot 4. cos 7. cos x sec x 1 cot sec sin x csc x tan s22. csc 5. 8. 2 tan s 11.sec x csc x 1 csc2 cos sin 1csc x sec x 1 sec2 1 sin cos 1 1 cos x3. tan s cot s 6. 1 sin 1 sin t2csc s 1 sin cos2 t219. 1 12. sin10. 11 cos xcosFactor each trigonometric expression. 13. sin2 15. sin x 17. 2 sin2 x 19. cos x 21. sin3 x41 1214. sec2 sin x 1 121 cot x tan 3 cot 2 x cos321216. tan x 18. 4 tan2 20. cot 4 x 22. sin3tan x 3 2cot x23 sin x 2 cos x cos3 xEach expression simplifies to a constant, a single function, or a power of a function. Use fundamental identities to simplify each expression. 23. tan cos 26. cot t tan t 29. sec2 x 32. 1 tan2 1 cot tan 24. cot 27. sin 25. sec r cos r 28. 31. csc sec cot sin2 x cos2 x sin x csc x sin tan cos 130. csc2 tIn Exercises 33 ­ 68, verify that each trigonometric equation is an identity. See Examples 1­ 5. 33. 35. cot csc 1 sin2 cos tan2 tan s sin csc cos4 cos2 sin2 1 cos cos 1 1 34. 36. tan sec tan2 sec 1 sin 1 sec cot 2 tan2 sec 1 1 cos2 cos sec237. cos2 39. cot s 41. cos sec38. sin2 40. sin2 tan2 1 2 sin2 cos2 1 42. 44. sin4 sin cos2sec s csc s sec2 2 sin2 cos2 tan243. sin4 45. 1 46. tan2cos sin cotLIALMC07_0321227638.QXP2/26/0410:47 AMPage 6197.2 Verifying Trigonometric Identities61947. 49. 50. 51. 52. 53. 54.cos tan2 1 sec 1 sin 11cos sec 1 1 sin sec1 2 sec2 tan cot s248.sec sec csctan 2 1 tan csc2 tantantan s 1 cos s 1 1 cot cot 1 tan csc tan 1 1 sin sin cos sec sec2 cot sin cos x cos x 1 1sin s 1 cos s cot x 1 1 tan tan 1 tan sin2 csc2sec s csc scsc xsec sec22 tan55. sin2 56.cot csc tan4 x tan2 x tan257. sec4 x 58.sec2 x sec22 sec tan59. sinsin cos 1 sin sin cos 1 coscos sin 1 cos csc tan2 s 1 cos260. 61. 62. 63.1sin cossec4 s sec2 stan4 s tan2 s 1sec2 s 2 sin2 t cot t cot t2cot 2 t 1 1 cot 2 t tan2 t 1 sec2 t sin x cos x cos x tan csc cos cos24tan t tan t cos x 1 1264. 1 65. 1 121sin x 1cos xcos x cos x 1 1 cos 14 cot x csc x sin sin sin cot tan66. sec 67. sec 68. sin sin249 69. 9 70.A student claims that the equation cos 90 or 2 radians we get 0 1 dent's reasoning.sin 1 is an identity, since by letting 1, a true statement. Comment on this stu-An equation that is an identity has an infinite number of solutions. If an equation has an infinite number of solutions, is it necessarily an identity? Explain.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 620620 CHAPTER 7 Trigonometric Identities and Equations71. sec tan 1 sin cos 72. csc cot sec 1 tan cos 1 73. cot sin tan 74. tan sin cos sec 75. identity 76. identity 77. not an identity 78. not an identity 83. It is true when sin x 0. 84. (a) I k 1 sin2 (b) For 2 n for all integers n, cos2 1, its maximum value, and I attains a maximum value of k. 85. (a) P 16k cos2 2 t (b) P 16k 1 sin2 2 tConcept Check Graph each expression and conjecture an identity. Then verify your conjecture algebraically. 71. sec 73. cos sin tan 1 tan 1 sin 72. csc 74. tan sin cot sec cos 1Graph the expressions on each side of the equals sign to determine whether the equation might be an identity. (Note: Use a domain whose length is at least 2 .) If the equation looks like an identity, prove it algebraically. See Example 1. 75. 77. 2 5 cos s sin s cot s cot s 2 csc s 2 sin2 s 5 cot s 76. 1 78. cot 2 s 1 sin s sec2 s sec2 s 1 1 1 sin s sec2 stan s tan s1By substituting a number for s or t, show that the equation is not an identity. 79. sin csc s 81. csc t 1 1 cot t 1280. cos2 x a true statement?cos2 scos s 1 sin2 t82. cos t83. When is sin x(Modeling) Work each problem. 84. Intensity of a Lamp According to Lambert's law, the intensity of light from a single source on a flat surface at point P is given by I k cos2 ,where k is a constant. (Source: Winter, C., Solar Power Plants, Springer-Verlag, 1991.)9(b) Explain(a) Write I in terms of the sine function. why the maximum value of I occurs when 0.yP85. Oscillating Spring The distance or displacement y of a weight attached to an oscillating spring from its natural position is modeled by y 4 cos 2 t ,where t is time in seconds. Potential energy is the energy of position and is given by P ky 2,xwhere k is a constant. The weight has the greatest potential energy when the spring is stretched the most. (Source: Weidner, R. and R. Sells, Elementary Classical Physics, Vol. 1, Allyn &amp; Bacon, 1973.) (a) Write an expression for P that involves the cosine function. (b) Use a fundamental identity to write P in terms of sin 2 t . 86. Radio Tuners Refer to Example 6. Let the energy stored in the inductor be given by Lt 3 cos2 6,000,000tand the energy in the capacitor be given by Ct 3 sin2 6,000,000t ,LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6217.3 Sum and Difference Identities62186. (a) The sum of L and C equals 3.4where t is time in seconds. The total energy E in the circuit is given by Et Lt Ct. (a) Graph L, C, and E in the window 0, 10 6 by 1, 4 , with Xscl 10 7 and Yscl 1. Interpret the graph. (b) Make a table of values for L, C, and E starting at t 0, incrementing by 10 7. Interpret your results. (c) Use a fundamental identity to derive a simplified expression for E t .(b) Let Y1 L t , Y2 (c) E t 3 C t , and Y3 E t . Y3 3 for all inputs.0 ­110 ­67.3 Sum and Difference IdentitiesCosine Sum and Difference Identities Difference IdentitiessCofunction IdentitiessSine and Tangent Sum andCosine Sum and Difference Identities Several examples presented earlier should have convinced you by now that cos A B does not equal cos A cos B. For example, if A 2 and B 0, thencos A while cos A B cos B cos cos 2 2 0 cos 0 cos 0 2 1 0, 1.We can now derive a formula for cos A B . We start by locating angles A and B in standard position on a unit circle, with B A. Let S and Q be the points where the terminal sides of angles A and B, respectively, intersect the circle. Locate point R on the unit circle so that angle POR equals the difference A B. See Figure 4.y(cos(A ­ B), sin(A ­ B)) R Q (cos B, sin B (cos A, sin A) S A A ­ B P (1, 0) x BOFigure 4Point Q is on the unit circle, so by the work with circular functions in Chapter 6, the x-coordinate of Q is the cosine of angle B, while the y-coordinate of Q is the sine of angle B. Q has coordinates cos B, sin B . In the same way, S has coordinates cos A, sin A , and R has coordinates cos A B , sin A B .LIALMC07_0321227638.QXP2/26/0410:47 AMPage 622622 CHAPTER 7 Trigonometric Identities and EquationsAngle SOQ also equals A B. Since the central angles SOQ and POR are equal, chords PR and SQ are equal. By the distance formula, since PR SQ, cos A B 12sin A cos AB0 cos B2 2sin Asin B 2.(Section 2.1)Squaring both sides and clearing parentheses gives cos2 A B22 cos A cos2 x B BB1sin2 A cos B2B sin2 A 2 sin A sin B sin2 B.cos A Since sin2 x 22 cos A cos B1 for any value of x, we can rewrite the equation as 2 2 cos A cos B 2 sin A sin B cos A cos B sin A sin B. Subtract 2; divide by 2.2 cos A cos AAlthough Figure 4 shows angles A and B in the second and first quadrants, respectively, this result is the same for any values of these angles. B To find a similar expression for cos A B , rewrite A B as A and use the identity for cos A B . cos A B cos A cos A cos cos A cos B cos A B cos A cos B B B sin A sin sin A sin B BCosine difference identity Negative-angle identities(Section 7.1)sin A sin BCosine of a Sum or Differencecos(A cos(A B) B) cos A cos B cos A cos B sin A sin B sin A sin BThese identities are important in calculus and useful in certain applications. Although a calculator can be used to find an approximation for cos 15°, for example, the method shown below can be applied to get an exact value, as well as to practice using the sum and difference identities.EXAMPLE 1 Finding Exact Cosine Function ValuesFind the exact value of each expression. (a) cos 15°Solution(b) cos5 12(c) cos 87 cos 93sin 87 sin 93(a) To find cos 15°, we write 15° as the sum or difference of two angles with known function values. Since we know the exact trigonometric function values of 45° and 30°, we write 15° as 45 30 . (We could also use 60 45 .) Then we use the identity for the cosine of the difference of two angles.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6237.3 Sum and Difference Identities623TEACHING TIP In Example 1(b),students may benefit from con5 verting radians to 75° in 12 order to realize that 6 4 30° 45°cos 15cos 45 2 2 6 4 2 330 sin 45 sin 30 2 2 2 1 2Cosine difference identity(Section 5.3)cos 45 cos 305 can be used in place of . 12(b) cos5 12cos cos 3 2 66 6 cos 2 2 2 44 4 1 2 sin 6 2 2 sin 462 12 ; 43 12Cosine sum identity(Section 6.2)The screen supports the solution in Example 1(b) by showing that cos 5 = 6 ­ 2 . 4 12(c) cos 87 cos 93sin 87 sin 93cos 87 cos 180 193Cosine sum identity(Section 5.2)Now try Exercises 7, 9, and 11.Cofunction Identities We can use the identity for the cosine of the difference of two angles and the fundamental identities to derive cofunction identities.TEACHING TIP Mention that theseidentities state that the trigonometric function of an acute angle is the same as the cofunction of its complement. Verify the cofunction identities for acute angles using complementary angles in a right triangle along with the righttriangle-based definitions of the trigonometric functions. Emphasize that these identities apply to any angle , not just acute angles.Cofunction Identitiescos(90 sin(90 tan(90 ) ) ) sin cos cot cot(90 sec(90 csc(90 ) ) ) tan csc secSimilar identities can be obtained for a real number domain by replacing 90° with 2 .Substituting 90° for A and for B in the identity for cos A cos 90 cos 90 cos 0 cos sin . This result is true for any value of any values of A and B. sin 90 sin 1 sinB givessince the identity for cos AB is true forLIALMC07_0321227638.QXP2/26/0410:47 AMPage 624624 CHAPTER 7 Trigonometric Identities and EquationsEXAMPLE 2 Using Cofunction Identities to FindFind an angleTEACHING TIP Verify results fromExample 2 using a calculator.that satisfies each of the following. (b) sin cos 30 (c) csc 3 4 sec(a) cotSolutiontan 25(a) Since tangent and cotangent are cofunctions, tan 90 cot tan 90 90 (b) cos 90 90 3 4 sin cos cos 30 120 (c) sec csc sec sec secCofunction identitycot .tan 25 tan 25 25 65Cofunction identity Set angle measures equal.30 30Cofunction identity2 sec3 4 4 4Combine terms.Now try Exercises 25 and 27.Because trigonometric (circular) functions are periodic, the solutions in Example 2 are not unique. We give only one of infinitely many possibilities.NOTEIf one of the angles A or B in the identities for cos A B and cos A B is a quadrantal angle, then the identity allows us to write the expression in terms of a single function of A or B.EXAMPLE 3 Reducing cos A B to a Function of a Single VariableWrite cos 180Solutionas a trigonometric function of .Use the difference identity. Replace A with 180° and B with . cos 180 cos 1 cos cosNow try Exercise 39.cos 180sin 180 sin 0 sin(Section 5.2)LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6257.3 Sum and Difference Identities625Sine and Tangent Sum and Difference Identities We can use the cosine sum and difference identities to derive similar identities for sine and cos 90 , we replace with A B to get tangent. Since sinsin A B cos 90 cos 90 cos 90 sin A B A A B B sin 90 A sin BCosine difference identity Cofunction identityA cos Bsin A cos Bcos A sin B. Cofunction identities B and use the identity forNow we write sin A sin A B . sin A sin A B B sin A sin A cosB as sin A B B cos A sin cos A sin BBSine sum identity Negative-angle identitiessin A cos BSine of a Sum or Differencesin(A sin(A B) B) sin A cos B sin A cos B cos A sin B cos A sin BTo derive the identity for tan A tan A B sin A cos A B BB , we start withFundamental identity(Section 7.1)sin A cos B cos A cos Bcos A sin B . sin A sin BSum identitiesWe express this result in terms of the tangent function by multiplying both nu1 merator and denominator by cos A cos B . 1 1 cos A cos B cos A cos B sin A sin B 1 1 cos A cos B sin A cos B cos A sin B sin A cos B cos A cos B cos A cos B cos A cos B cos A sin B cos A cos B sin A sin B cos A cos BSimplify the complex fraction.(Section R.5)tan ABMultiply numerators; multiply denominators.sin B sin A cos A cos B sin A sin B 1 cos A cos B tan A B tan A tan B 1 tan A tan Btansin cosReplacing B with B and using the fact that tan B identity for the tangent of the difference of two angles.tan B gives theLIALMC07_0321227638.QXP2/26/0410:47 AMPage 626626 CHAPTER 7 Trigonometric Identities and EquationsTangent of a Sum or Differencetan(A B) tan A tan B 1 tan A tan B tan(A B) tan A tan B 1 tan A tan BEXAMPLE 4 Finding Exact Sine and Tangent Function ValuesFind the exact value of each expression. (a) sin 75°Solution(b) tan7 12(c) sin 40 cos 160cos 40 sin 160(a) sin 75sin 45 2 2 6 4 2 330 cos 45 sin 30 2 2 2 1 2Sine sum identity(Section 5.3)sin 45 cos 30(b) tan7 12tan34Tangent sum identity(Section 6.2)tan 4 tan 3 1 tan 3 tan 4 3 1 3 1 3 4 1 3 1 1 1 3 1 3 1 1 3 3 3 3Rationalize the denominator. (Section R.7) Multiply. (Section R.7) Combine terms. Factor out 2. (Section R.5) Lowest terms2 3 2 3 22 2 2 (c) sin 40 cos 160 3cos 40 sin 160sin 40 sin 120160Sine difference identitysin 120 3 2Negative-angle identity(Section 5.3)Now try Exercises 29, 31, and 35.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6277.3 Sum and Difference Identities627EXAMPLE 5 Writing Functions as Expressions Involving Functions ofWrite each function as an expression involving functions of . (a) sin 30Solution(b) tan 45(c) sin 180(a) Using the identity for sin A sin 30B, sin 30 cos 1 cos 2 3 2 1 1 1 sin tan tan cos 30 sin sin .(b) tan 45 (c) sin 180tan 45 tan 1 tan 45 tan sin 180 cos 0 cos sincos 180 sinNow try Exercises 43 and 47.EXAMPLE 6 Finding Function Values and the Quadrant of AB4 5,Suppose that A and B are angles in standard position, with sin A 5 A , and cos B B 32 . Find each of the following. 2 13 , (a) sin ASolutionB(b) tan AB(c) the quadrant of AB(a) The identity for sin A B requires sin A, cos A, sin B, and cos B. We are given values of sin A and cos B. We must find values of cos A and sin B. sin2 A 16 25 cos2 A cos2 A cos2 A cos A In the same way, sin B sin A B 1 1 9 25 3 512 13 .Fundamental identity sin A4 5Subtract 16 . 25 Since A is in quadrant II, cos A 0.Now use the formula for sin A 5 13 3 5 16 65 12 13B.4 5 20 6536 65(b) To find tan A get tan A4 3B , first use the values of sine and cosine from part (a) to 12 and tan B 5 . B4 3tan A112 5 4 12 3 516 15148 1516 15 63 1516 63LIALMC07_0321227638.QXP2/26/0410:47 AMPage 628628 CHAPTER 7 Trigonometric Identities and Equations16 16B (c) From parts (a) and (b), sin A B 65 and tan A 63 , both positive. Therefore, A B must be in quadrant I, since it is the only quadrant in which both sine and tangent are positive.Now try Exercise 51.EXAMPLE 7 Applying the Cosine Difference Identity to VoltageCommon household electric current is called alternating current because the current alternates direction within the wires. The voltage V in a typical 115-volt 163 sin t, where is the anguoutlet can be expressed by the function V t lar speed (in radians per second) of the rotating generator at the electrical plant and t is time measured in seconds. (Source: Bell, D., Fundamentals of Electric Circuits, Fourth Edition, Prentice-Hall, 1988.) (a) It is essential for electric generators to rotate at precisely 60 cycles per sec so household appliances and computers will function properly. Determine for these electric generators. (b) Graph V in the window 0, .05 by 200, 200 . 163 cos t is the(c) Determine a value of so that the graph of V t 163 sin t. same as the graph of V tSolution(a) Each cycle is 2 radians at 60 cycles per sec, so the angular speed is 60 2 120 radians per sec. 163 sin t 163 sin 120 t. Because the amplitude of the function (b) V t 200, 200 is an appropriate interval for the is 163 (from Section 6.3), range, as shown in Figure 5.For x = t, V(t) = 163 sin 120 t2000.05­200Figure 5(c) Using the negative-angle identity for cosine and a cofunction identity, cos x Therefore, if Vt 22,cos then 163 cos2xcos2xsin x.t2163 sin t.Now try Exercise 81.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6297.3 Sum and Difference Identities6297.3 Exercises1. F 2. A 3. C 6 2 5. 6. 4 7. 9. 2 4 2 6 6 8. 10. 4. D 6 4 2 4 6 2 2 6Concept Check Match each expression in Column I with the correct expression in Column II to form an identity. I 1. cos x 2. cos x 3. sin x 4. sin x y y y y II A. cos x cos y B. sin x sin y C. sin x cos y D. sin x cos y E. cos x sin y F. cos x cos y sin x sin y cos x cos y cos x sin y cos x sin y sin x cos y sin x sin y4 11. 0 12. 14. cos 75 16. cos 18. tan4 1 13. cot 3 5 15. sin 12 17. cos 810Use identities to find each exact value. (Do not use a calculator.) See Example 1. 5. cos 75 7. cos 105 (Hint: 105 9. cos 7 12 sin 40 sin 50 60 45 ) 6. cos 15 60 45 ) 8. cos 105 (Hint: 105 10. cos 12. cos 12 7 2 cos 9 9 sin 7 2 sin 9 92 19. csc 56 42 5 20. cot 84 3 21. tan 22. cos 23. cos 24. tan 100° 25. 15 26. 27. 20 3 80° 2 6 28. 29. 3 4 30. 2 3 3 31. 2 32. 34. 2 37. 1 6 4 3 38. 2 33. 35. 1 2 2 6 4 36. 1 211. cos 40 cos 50Write each function value in terms of the cofunction of a complementary angle. See Example 2. 13. tan 87 5 17. sin 8 14. sin 15 9 18. cot 10 15. cos 12 16. sin 2 519. sec 146 4220. tan 174 3Use the cofunction identities to fill in each blank with the appropriate trigonometric function name. See Example 2. 21. cot 23. 3 33 6 sin 57 22. sin 24. 2 3 72 cot 18 6Find an angle that makes each statement true. See Example 2. 25. tan 27. sin 3 cot 45 15 2 cos 25 26. sin 28. cot cos 2 10 10 tan 2 20Use identities to find the exact value of each of the following. See Example 4. 29. sin 32. sin 5 12 12 30. tan 33. sin cos 76 sin 31 5 12 7 12 31. tan 34. tan 36. sin 40 cos 50 38. 12 7 12 cos 40 sin 5035. sin 76 cos 31 37.tan 80 tan 55 1 tan 80 tan 55tan 80 tan 55 1 tan 80 tan 55LIALMC07_0321227638.QXP2/26/0410:47 AMPage 630630 CHAPTER 7 Trigonometric Identities and Equations 39. sin 40. cos 41. sin x 42. sin x 2 cos sin 43. 2 44. sin 1 45. cos x 3 sin x 2 2 cos x sin x 46. 2 3 tan 47. 1 3 tan 3 tan 1 48. 3 tan 3 tan x 1 1 tan x 49. 50. 1 tan x 3 tan x 16 33 63 51. (a) (b) (c) (d) I 65 65 16 4 6 6 8 6 3 52. (a) (b) 25 25 8 6 3 (c) (d) IV 4 6 6 2 10 2 53. (a) 9 4 2 5 (b) 9 8 5 5 2 (c) (d) II 20 2 10 16 33 63 54. (a) (b) (c) (d) IV 65 65 16 36 13 77 55. (a) (b) (c) 85 85 36 (d) II 77 84 36 56. (a) (b) (c) (d) III 85 85 77 2 6 58. 4 2 6 59. 4 6 2 60. (a) 4 2 6 (b) 4 6 2 61. 62. 2 3 4 6 2 63. 64. 2 3 4 6 2 65. 2 3 66. 4Use identities to write each expression as a function of x or . See Examples 3 and 5. 39. cos 90 42. cos 45. sin 48. tan x x 30 40. cos 180 43. sin 45 46. sin x 41. cos x 44. sin 180 47. tan 60 50. tan x t , and (d) the 3 22 5 6449. tan x6 t , (b) sin s4Use the given information to find (a) cos s quadrant of s t. See Example 6. 51. cos s 52. cos s 53. sin s 54. sin s 55. cos s 56. cos s 3 and sin t 5 1 and sin t 5 2 and sin t 3 3 and sin t 5 8 and cos t 17 15 and sin t 17t , (c) tan s5 , s and t in quadrant I 13 3 , s and t in quadrant II 5 1 , s in quadrant II and t in quadrant IV 3 12 , s in quadrant I and t in quadrant III 13 3 , s and t in quadrant III 5 4 , s in quadrant II and t in quadrant I 5Relating ConceptsFor individual or collaborative investigation (Exercises 57­60)The identities for cos A B and cos A B can be used to find exact values of expressions like cos 195 and cos 255 , where the angle is not in the first quadrant. Work Exercises 57­60 in order, to see how this is done. 57. By writing 195 as 180 as cos 15 . 58. Use the identity for cos A 15 , use the identity for cos A B to find cos 15 . . B to express cos 19559. By the results of Exercises 57 and 58, cos 195 11 1260. Find each exact value using the method shown in Exercises 57­59. (a) cos 255 (b) cosFind each exact value. Use the technique developed in Relating Concepts Exercises 57­ 60. 61. sin 165 64. tan 285 62. tan 165 65. tan 11 12 63. sin 255 66. sin 13 12LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6317.3 Sum and Difference Identities63171. sin 72. 1 12 tan x tan xx tancos x x67. Use the identity cos 90 identity cos A sin 90sin , and replace A.with 90A, to derive the B can be found9 68. 9 69.4Explain how the identities for sec A B , csc A by using the sum identities given in this section.B , and cot A79. 3 80. 163 and 81. (a) 425 lb (c) 0163; noWhy is it not possible to use a method similar to that of Example 5(c) to find a formula for tan 270 ? Show that if A, B, and C are angles of a triangle, then 0.70. Concept Check sin A B CGraph each expression and use the graph to conjecture an identity. Then verify your conjecture algebraically. 71. sin 2 x 72. 1 1 tan x tan xVerify that each equation is an identity. 73. sin x 74. tan x 75. 76. 77. 78. cos cos y y sin x tan y tan tan s tan x tan x y x cot tan t tan y tan y sin s sin t cos t 2 sin x cos y 2 tan x tan y 1 tan x tan ysinsin s t cos s cos t sin x sin x y ysin s t sin tcos s t cos tExercises 79 and 80 refer to Example 7. 79. How many times does the current oscillate in .05 sec? 80. What are the maximum and minimum voltages in this outlet? Is the voltage always equal to 115 volts? (Modeling) Solve each problem. 81. Back Stress If a person bends at the waist with a straight back making an angle of degrees with the horizontal, then the force F exerted on the back muscles can be modeled by the equation F .6W sin 90 , sin 12where W is the weight of the person. (Source: Metcalf, H., Topics in Classical Biophysics, Prentice-Hall, 1980.) (a) Calculate F when W 170 lb and 30 . (b) Use an identity to show that F is approximately equal to 2.9W cos . (c) For what value of is F maximum?LIALMC07_0321227638.QXP2/26/0410:47 AMPage 632632 CHAPTER 7 Trigonometric Identities and Equations 82. (a) 408 lb (b) 46.1 83. (a) The pressure P is oscillating. For x = t, P(t) =.05.4 cos 1082. Back StressRefer to Exercise 81.[20 ­ 1026t] 4.9.0545 . Estimate the force (a) Suppose a 200-lb person bends at the waist so that exerted on the person's back muscles. (b) Approximate graphically the value of that results in the back muscles exerting a force of 400 lb. 83. Sound Waves Sound is a result of waves applying pressure to a person's eardrum. For a pure sound wave radiating outward in a spherical shape, the trigonometric function defined by P a 2 r cos r ct0­.05(b) The pressure oscillates and amplitude decreases as r increases. For x = r, P(r) = 3 cos 2 r ­ 10,260 r 4.9 2[]can be used to model the sound pressure at a radius of r feet from the source, where t is time in seconds, is length of the sound wave in feet, c is speed of sound in feet per second, and a is maximum sound pressure at the source measured in pounds per square foot. (Source: Beranek, L., Noise and Vibration Control, Institute of Noise Control Engineering, Washington, D.C., 1988.) Let 4.9 ft and c 1026 ft per sec. (a) Let a .4 lb per ft 2. Graph the sound pressure at distance r 10 ft from its source in the window 0, .05 by .05, .05 . Describe P at this distance. (b) Now let a 3 and t 10. Graph the sound pressure in the window 0, 20] by 2, 2 . What happens to pressure P as radius r increases? (c) Suppose a person stands at a radius r so that r n , where n is a positive integer. Use the difference identity for cosine to simplify P in this situation. 84. Voltage of a Circuit V1 When the two voltages and V2 40 cos 120 t 30 sin 120 t020­2(c) Pa cos ct n84. (a) For x = t, V = V1 + V2 = 30 sin 120 t + 40 cos 120 t60are applied to the same circuit, the resulting voltage V will be equal to their sum. (Source: Bell, D., Fundamentals of Electric Circuits, Second Edition, Reston Publishing Company, 1981.) (a) Graph the sum in the window 0, .05 by 60, 60 . (b) Use the graph to estimate values for a and so that V a sin 120 t (c) Use identities to verify that your expression for V is valid. .0.05­60(b) a50;5.3537.4 Double-Angle Identities and Half-Angle IdentitiesDouble-Angle IdentitiessProduct-to-Sum and Sum-to-Product IdentitiessHalf-Angle IdentitiesDouble-Angle Identities When A B in the identities for the sum of two angles, these identities are called the double-angle identities. For example, to derive an expression for cos 2A, we let B A in the identity cos A B cos A cos B sin A sin B.TEACHING TIP A common error isto write cos 2 A as 2 cos A.cos 2A cos 2Acos A cos A2A sin A sin A Cosine sum identity (Section 7.3)2cos A cos Asin ALIALMC07_0321227638.QXP2/26/0410:47 AMPage 6337.4 Double-Angle Identities and Half-Angle Identities633Two other useful forms of this identity can be obtained by substituting either cos2 A 1 sin2 A or sin2 A 1 cos2 A. Replace cos2 A with the expression 1 sin2 A to get cos 2A cos 2A cos2 A 1 1 sin2 A sin2 A Fundamental identity (Section 7.1) sin2 A 2 sin2 A, cos2 A to get cos2 A cos2 A cos A cos 2A 2 cos A2 2or replace sin2 A with 1 cos 2Asin2 A 1 1 1. B sin A cos B cos A sin B, cos2 A cos A2Fundamental identityWe find sin 2A with the identity sin A letting B A. sin 2A sin 2ATEACHING TIP Students might findit helpful to see each formula illustrated with a concrete example that they can check. For instance, you might show that cos 60 cos 2 30° cos2 30 sin2 30 .sin AA cos A sin A Sine sum identity B , we find tan 2 A. Asin A cos A 2 sin A cos A Using the identity for tan A tan 2A tan Atan A tan A Tangent sum identity 1 tan A tan A tan 2A 1 2 tan A tan2 ALooking Ahead to CalculusThe identities cos 2A and cos 2A 1 2 sin2 A2Double-Angle Identitiescos 2A cos 2A cos2 A 2 cos2 A sin2 A 1 cos 2A sin 2A 2 tan A 1 tan2 A 1 2 sin2 A 2 sin A cos A2 cos A1tan 2Acan be rewritten as sin2 A21 1 2 1 1 2cos 2 AEXAMPLE 1 Finding Function Values of 2 Given Information aboutcos 2 A .andcos AGiven cosSolution3 5and sin20, find sin 2 , cos 2 , and tan 2 .These identities are used to integrate the functions f A sin2 A and 2 cos A. gATo find sin 2 , we must first find the value of sin . 3 5 sin2 sin 1 16 25 4 5sin2 cos2 1; cos3 5sin2Simplify.Choose the negative square root since sin0.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 634634 CHAPTER 7 Trigonometric Identities and EquationsUsing the double-angle identity for sine, sin 2 2 sin cos 2 4 5 3 5 24 . sin 254 5;cos3 5Now we find cos 2 , using the first of the double-angle identities for cosine. (Any of the three forms may be used.) cos 2 cos2 sin2 9 254 516 257 25The value of tan 2 can be found in either of two ways. We can use the doubleangle identity and the fact that tan 2 tan 1 tan2sin cos 3 5 4 3.tan 28 24 3 7 7 1 9 Simplify. (Section R.5)24 3 16 9Alternatively, we can find tan 2 by finding the quotient of sin 2 and cos 2 . tan 2 sin 2 cos 224 25 7 2524 7Now try Exercise 9.EXAMPLE 2 Verifying a Double-Angle IdentityVerify that the following equation is an identity. cot x sin 2xSolution1cos 2xWe start by working on the left side, using the hint from Section 7.1 about writing all functions in terms of sine and cosine. cot x sin 2x cos x sin 2x sin x cos x 2 sin x cos x sin x 2 cos2 x 1 cos 2xcos 2x 2 cos2 x 1, so 2 cos2 x 1 cos 2x Quotient identityDouble-angle identityThe final step illustrates the importance of being able to recognize alternative forms of identities.Now try Exercise 27.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6357.4 Double-Angle Identities and Half-Angle Identities635EXAMPLE 3 Simplifying Expressions Using Double-Angle IdentitiesSimplify each expression. (a) cos2 7xSolutionsin2 7x(b) sin 15 cos 15(a) This expression suggests one of the double-angle identities for cosine: cos 2A cos2 A sin2 A. Substituting 7x for A gives cos2 7x sin2 7x cos 2 7x cos 14x.(b) If this expression were 2 sin 15° cos 15°, we could apply the identity for sin 2A directly since sin 2A 2 sin A cos A. We can still apply the identity 1 with A 15 by writing the multiplicative identity element 1 as 2 2 . sin 15 cos 15 1 2 sin 15 cos 15 2 1 2 sin 15 cos 15 2 1 sin 2 15 2 1 sin 30 2 1 2 1 2 1 4sin 301 2Multiply by 1 in the form 1 2 . 2 Associative property (Section R.1) 2 sin A cos A sin 2A, with A 15(Section 5.3)Now try Exercises 13 and 15.Identities involving larger multiples of the variable can be derived by repeated use of the double-angle identities and other identities.EXAMPLE 4 Deriving a Multiple-Angle IdentityWrite sin 3x in terms of sin x.Solutionsin 3xsin 2xx cos 2x sin x cos2 x2sin 2x cos xSine sum identity(Section 7.3)2 sin x cos x cos x 2 sin x cos x 2 sin x 1 2 sin x 3 sin x2 2sin2 x sin x sin x2 3 3Double-angle identities Multiply.cos x sin x 1 sin xsin x 2 sin3 x 4 sin x3sin x sin x sin3 xsin x cos2 x1sin2 xsin3 xDistributive property(Section R.1)Combine terms. Now try Exercise 21.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 636636 CHAPTER 7 Trigonometric Identities and EquationsThe next example applies a multiple-angle identity to answer a question about electric current.EXAMPLE 5 Determining Wattage ConsumptionIf a toaster is plugged into a common household outlet, the wattage consumed is not constant. Instead, it varies at a high frequency according to the model W V2 , Rwhere V is the voltage and R is a constant that measures the resistance of the toaster in ohms. (Source: Bell, D., Fundamentals of Electric Circuits, Fourth Edition, Prentice-Hall, 1998.) Graph the wattage W consumed by a typical toaster with R 15 and V 163 sin 120 t in the window 0, .05 by 500, 2000 . How many oscillations are there?Solution For x = t, (163 sin 120 t)2 W(t) = 15 2000Substituting the given values into the wattage equation gives W V2 R 163 sin 120 t 2 . 15To determine the range of W, we note that sin 120 t has maximum value 1, so 1632 1771. The minimum value is 0. the expression for W has maximum value 15 The graph in Figure 6 shows that there are six oscillations.0 ­500Figure 6.05Now try Exercise 81.Product-to-Sum and Sum-to-Product Identities Because they make it possible to rewrite a product as a sum, the identities for cos A B and cos A B are used to derive a group of identities useful in calculus. Adding the identities for cos A B and cos A B givescos A cos A cos A B B B cos A cos B sin A sin B cos A cos B sin A sin B 2 cos A cos B B cos A B gives cos A B . B .cos A orBcos A cos B1 cos A 2 1 cos A 2Similarly, subtracting cos A sin A sin BLooking Ahead to CalculusThe product-to-sum identities are used in calculus to find integrals of functions that are products of trigonometric functions. One classic calculus text includes the following example: Evaluate cos 5x cos 3x dx.B from cos A BUsing the identities for sin A B and sin A B in the same way, we get two more identities. Those and the previous ones are now summarized.Product-to-Sum Identitiescos A cos B sin A sin B 1 [cos(A 2 1 [cos(A 2 B) B) cos(A cos(A B)] B)](continued)The first solution line reads: &quot;We may write cos 5x cos 3x 1 cos 8x 2 cos 2x .&quot;LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6377.4 Double-Angle Identities and Half-Angle Identities637sin A cos B cos A sin B1 [sin(A 2 1 [sin(A 2B) B)sin(A sin(AB)] B)]EXAMPLE 6 Using a Product-to-Sum IdentityWrite cos 2 sin as the sum or difference of two functions.SolutionUse the identity for cos A sin B, with 2 cos 2 sin 1 sin 2 2 1 sin 3 2 1 sin 2A and sin 2B.Now try Exercise 41.From these new identities we can derive another group of identities that are used to write sums of trigonometric functions as products.Sum-to-Product Identitiessin A sin A cos A cos A sin B sin B cos B cos B 2 sin 2 cos 2 cos A 2 A 2 A 2 A 2 B B B B cos sin cos sin A 2 A 2 A 2 A 2 B B B B2 sinEXAMPLE 7 Using a Sum-to-Product IdentityWrite sin 2Solutionsin 4 as a product of two functions. sin B, with 2 4 2 6 sin 2 2 2sin sin(Section 7.1)Use the identity for sin A sin 4 2 cos 2 cos 2A and 4 4B.sin 2sin2 22 cos 3 sin 2 cos 3 sinNow try Exercise 45.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 638638 CHAPTER 7 Trigonometric Identities and EquationsHalf-Angle Identities From the alternative forms of the identity for cos 2A, A A A we derive three additional identities for sin 2 , cos 2 , and tan 2 . These are known as half-angle identities. A To derive the identity for sin 2 , start with the following double-angle identity for cosine and solve for sin x.cos 2x 2 sin2 x sin x A 2 1 1 2 sin2 x cos 2x 1 cos 2x 2 cos A 2Add 2 sin2 x; subtract cos 2x. Divide by 2; take square roots. (Section 1.4)sin1Let 2xA, so xA 2;substitute.The sign in this identity indicates that the appropriate sign is chosen deA A pending on the quadrant of 2 . For example, if 2 is a quadrant III angle, we choose the negative sign since the sine function is negative in quadrant III. A We derive the identity for cos 2 using the double-angle identity 2 cos 2x 2 cos x 1. 1 cos 2x cos2 x cos x A 2A 22 cos2 x 1 cos 2x 2 1Add 1. Rewrite; divide by 2.cos 2x Take square roots. 2 cos A 2Replace x with A . 2A Acos An identity for tan1comes from the identities for sin 2 and cos 2 . 1 cos A 2 cos A 2A 2 A 2tanA 2sin A 2 cos A 211 1cos A cos AWe derive an alternative identity for tan tan A 2 sin cosA 2 A 2using double-angle identities.Multiply by 2 cos 2 in numerator and denominator.A2 sinA 2cosA 22 cos2A 2 A 2sin 2 1 tan A 2cos 2Double-angle identitiessin A 1 cos AA From this identity for tan 2 , we can also derivetanA 21cos A . sin ALIALMC07_0321227638.QXP2/26/0410:47 AMPage 6397.4 Double-Angle Identities and Half-Angle Identities639Half-Angle IdentitiesTEACHING TIP Point out that theA follows 2 directly from the cosine and sine half-angle identities; however, the A other two identities for tan are 2 more useful. first identity for tancos A 2A 2 1 1 cos A cos A1cos A 2 tan A 2sinA 21cos A 2 A 2 1 cos A sin Atansin A 1 cos AtanThe last two identities for tan 2 do not require a sign choice. When using the other half-angle identities, select the plus or minus sign according to A the quadrant in which 2 terminates. For example, if an angle A 324°, then A A A 162 , which lies in quadrant II. In quadrant II, cos 2 and tan 2 are negative, 2 A while sin 2 is positive.NOTEAEXAMPLE 8 Using a Half-Angle Identity to Find an Exact ValueFind the exact value of cos 15° using the half-angle identity for cosine.TEACHING TIP Have students compare the value of cos 15° in Example 8 to the value in Example 1(a) of Section 7.3, where we used the identity for the cosine of the difference of two angles. Although the expressions look completely different, they are equal, as suggested by a calculator approximation for both, .96592583.Solutioncos 15cos1 30 23 21cos 30 23 2Choose the positive square root.1 2122 232 2Simplify the radicals. (Section R.7) Now try Exercise 51.EXAMPLE 9 Using a Half-Angle Identity to Find an Exact ValueFind the exact value of tan 22.5° using the identity tanSolutionA 21sin A cos A .Since 22.5 tan 22.51 245 , replace A with 45°. tan 45° 2 sin 45 1 cos 452 2 21 2 Now multiply numerator and denominator by 2. Then rationalize the denominator. tan 22.5 2 2 2 2 2 2 1 2 2 2 2 2 2 1Now try Exercise 53.2 222 22LIALMC07_0321227638.QXP2/26/0410:47 AMPage 640640 CHAPTER 7 Trigonometric Identities and Equationss 2EXAMPLE 10Finding Functions of2 3,Given Information about ss s sGiven cos sSolution2with3 2s2 , find cos 2 , sin 2 , and tan 2 . 3 2Since s s 2 2 ,Divide by 2. (Section 1.7)ss 20ands 23 4s3 2 Figure 7terminates in quadrant II. See Figure 7. In quadrant II, the values of cos 2 and s s tan 2 are negative and the value of sin 2 is positive. Now use the appropriate halfangle identities and simplify the radicals. s sin 2 cos s 2 s 2s sin 2 s cos 21 2 12 31 62 36 6 5 6 5 5s26 6 30 630 6tanNotice that it is not necessary to use a half-angle identity for tan 2 once we find s s sin 2 and cos 2 . However, using this identity would provide an excellent check.Now try Exercise 59.EXAMPLE 11Simplifying Expressions Using the Half-Angle IdentitiesSimplify each expression. (a)SolutionA (a) This matches part of the identity for cos 2 .1cos 12x 2(b)1cos 5 sin 5cos Replace A with 12 x to get 1A 21cos A 2cos 12x 2A 2cos12x 2cos 6x. 5 to get(b) Use the third identity for tan 1given earlier with A tan 5 . 2cos 5 sin 5Now try Exercises 67 and 71.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6417.4 Double-Angle Identities and Half-Angle Identities6417.4 Exercises1. cos 2. cos 3. cos x 4. cos x 5. cos 2 sin 2 4 2 5 5 ; sin 5 5 2 4 102 12 6 6Use identities to find values of the sine and cosine functions for each angle measure. See Example 1. 1. 2. , given cos 2 , given cos 2 3 and 5 3 and 4 terminates in quadrant I terminates in quadrant III x x 0 0 0 0 0 014 ; sin 4 42 ; sin x 12 30 ; sin x 6 17 ; 253. x, given cos 2x 4. x, given cos 2x 5. 2 , given sin 6. 2 , given cos 7. 2x, given tan x 8. 2x, given tan x 9. 2 , given sin 10. 2 , given cos5 and 12 2 2 and 3 2 2 and cos 5 12 and sin 13 2 and cos x 5 and sin x 3 5 and cos 7 3 and sin 521 25 119 120 6. cos 2 ; sin 2 169 169 4 3 7. cos 2x ; sin 2x 5 5 8 15 ; sin 2x 8. cos 2x 17 17 39 9. cos 2 ; 49 4 55 sin 2 49 19 10. cos 2 ; 25 3 2 66 11. sin 2 25 2 2 3 3 12. 13. 14. 3 2 2 2 2 15. 16. 2 4 1 1 17. tan 102° 18. tan 68° 2 4 1 1 cos 94.2° 20. sin 59° 19. 4 16 3 21. cos 3x 4 cos x 3 cos x 22. sin 4x 4 sin x cos3 x 4 sin3 x cos x 3 tan x tan3 x 23. tan 3x 1 3 tan2 x 24. cos 4x 8 cos4 x 8 cos2 x 1 25. cos4 x sin4 x cos 2x 4 tan x cos2 x 2 tan x 26. sin 2x 1 tan2 xUse an identity to write each expression as a single trigonometric function value or as a single number. See Example 3. 11. cos2 15° 14. 1 sin2 15° 1° 2 2 tan 15° 1 tan2 15° 1° 15. 2 cos2 67 2 12. tan 34° 18. 2 1 tan2 34° 13. 1 1 16. cos2 1 19. 4 2 sin2 15° 1 22 sin2 228tan 51° 17. 1 tan2 51° 20. 1 sin 29.5° cos 29.5° 81 sin2 47.1° 2Express each function as a trigonometric function of x. See Example 4. 21. cos 3x 22. sin 4x 23. tan 3x 24. cos 4xGraph each expression and use the graph to conjecture an identity. Then verify your conjecture algebraically. 25. cos4 x sin4 x 26. 4 tan x cos2 x 2 tan x 1 tan2 xVerify that each equation is an identity. See Example 2. 27. sin x cos x2sin 2x128. sec 2xsec2 x sec4 x 2 sec2 x sec4 xLIALMC07_0321227638.QXP2/26/0410:47 AMPage 642642 CHAPTER 7 Trigonometric Identities and Equations 41. sin 160° sin 44° 42. sin 225° sin 55° 5 5 43. cos 5x cos x 2 2 1 1 44. cos x cos 9x 2 2 45. 2 sin 3x sin x 46. 2 cos 6.5x cos 1.5x 47. 2 sin 11.5° cos 36.5° 48. 2 cos 98.5° sin 3.5° 49. 2 cos 6x cos 2x 50. 2 cos 6x sin 3x 2 2 2 3 51. 52. 2 2 2 3 53. 54. 2 3 2 2 3 2 3 55. 56. 2 2 10 13 59. 60. 61. 3 4 4 50 20 6 62. 10 50 10 5 63. 10 50 15 10 64. 10 5 65. 7 66. 529. sin 4x 31. 2 cos 2x sin 2x4 sin x cos x cos 2x cot x tan x 4 sin3 x cos x30.1cos 2x sin 2xcot x 8 sin3 x cos x32. sin 4x 34. cos 2x 36. cot x cot x x 24 sin x cos x 1 1 tan x tan x 1 tan2 x tan2 x cos 2x cos x sin2 x x 2233. sin 2x cos 2x 35. tan x 37. sec2 x 2 cot xsin 2x 2 csc 2x 2 cos x138. cot 2 40.x 39. sin2 2tan x sin x 2 tan xsin 2x 2 sin xcos2sin2x 2Write each expression as a sum or difference of trigonometric functions. See Example 6. 41. 2 sin 58° cos 102° 43. 5 cos 3x cos 2x 42. 2 cos 85° sin 140° 44. sin 4x sin 5xWrite each expression as a product of trigonometric functions. See Example 7. 45. cos 4x 48. sin 102° cos 2x sin 95° 46. cos 5x 49. cos 4x cos 8x cos 8x 47. sin 25° 50. sin 9x sin sin 3x 48°Use a half-angle identity to find each exact value. See Examples 8 and 9. 51. sin 67.5° 54. tan 195° 52. sin 195° 55. cos 165° 53. cos 195° 56. sin 165°9 57.Explain how you could use an identity of this section to find the exact value of sin 7.5°. (Hint: 7.5 1 1 30 .) 2 2 the identity tan 2 these answers are the same, without using a calculator. (Hint: If a and a2 b2, then a b.)A 2 1 cos A 1 cos A can be used to find tan 22.5° sin A 1 cos A can be used to find tan 22.5°58. The identity tan A 2322 , and1. Show that 0 and b 0Find each of the following. See Example 10. 59. cos x , given cos x 2 1 , with 0 4 5 , with 8 2 3 , with 90° 5 1 , with 180° 5 2, with 0 3, with x x 270° 180° 2 x x 180° 270° 2x 60. sin , given cos x 2 61. tan 62. cos 63. sin 64. cos 65. tan 66. cot 2 2 , given sin , given sinx , given tan x 2 x , given cot x 2 2 2 , given tan , given tan27 , with 180° 3 5 , with 90° 2LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6437.4 Double-Angle Identities and Half-Angle Identities64367. sin 20° 68. cos 38° 69. tan 73.5° 70. cot 82.5° 71. tan 29.87° 72. tan 79.1° 74. (a) tan 2 cannot be used. is undefined, so itUse an identity to write each expression with a single trigonometric function. See Example 11. 67. 70. 1 1 1 cos 40° 2 cos 165° cos 165° 68. 71. 1 1 cos 76° 2 69. 72. 1 1 1 cos 147° cos 147°sin (b) tan 2 x cos2x xcos 59.74° sin 59.74°sin 158.2° cos 158.2°1 cos A sin AA 73. Use the identity tan A 1 sincos A to derive the equivalent identity tan A 2 2 by multiplying both the numerator and denominator by 1 cos A.75. 78. (b) 80. (b)2 84° 76. 106° 77. 3.9 R b 2 79. (a) cos 2 R b tan (c) 54° 4 50 v 2 sin 2 (a) D 32 approximately 35 ft74. Consider the expression tan2x.(a) Why can't we use the identity for tan A B to express it as a function of x alone? sin (b) Use the identity tan cos to rewrite the expression in terms of sine and cosine. x cot x. (c) Use the result of part (b) to show that tan 2 (Modeling) Mach Number An airplane flying faster than sound sends out sound waves that form a cone, as shown in the figure. The cone intersects the ground to form a hyperbola. As this hyperbola passes over a particular point on the ground, a sonic boom is heard at that point. If is the angle at the vertex of the cone, then 1 , sin 2 m where m is the Mach number for the speed of the plane. (We assume m 1.) The Mach number is the ratio of the speed of the plane and the speed of sound. Thus, a speed of Mach 1.4 means that the plane is flying at 1.4 times the speed of sound. In Exercises 75­78, one of the values or m is given. Find the other value. 3 5 75. m 76. m 77. 78. 30° 60° 2 4 (Modeling) Solve each problem. See Example 5. 79. Railroad Curves In the United States, circular railroad curves are designated by the degree of curvature, the central angle subtended by a chord of 100 ft. See the figure. (Source: Hay, W. W., Railroad Engineering, John Wiley &amp; Sons, 1982.)b 50 50(a) Use the figure to write an expression for cos 2 . (b) Use the result of part (a) and the third half-angle identity for tangent to write an expression for tan 4 . (c) If b 12, what is the measure of angle to the nearest degree? 80. Distance Traveled by a Stone The distance D of an object thrown (or propelled) from height h (feet) at angle with initial velocity v is modeled by the formula v 2 sin cos 2 2Rh Dv sin 2 64h v cos D . 32 See the figure. (Source: Kreighbaum, E. and K. Barthels, Biomechanics, Allyn &amp; Bacon, 1996.) Also see the Chapter 5 Quantitative Reasoning. (a) Find D when h 0; that is, when the object is propelled from the ground. (b) Suppose a car driving over loose gravel kicks up a small stone at a velocity of 36 ft per sec (about 25 mph) and an angle 30°. How far will the stone travel?LIALMC07_0321227638.QXP2/26/0410:47 AMPage 644644 CHAPTER 7 Trigonometric Identities and Equations 81. a 885.6, c 240 82. (a) 885.6,81. Wattage Consumption Refer to Example 5. Use an identity to determine values of c. Check your answer by graphing both expresa, c, and so that W a cos t sions for W on the same coordinate axes. 82. Amperage, Wattage, and Voltage Amperage is a measure of the amount of electricity that is moving through a circuit, whereas voltage is a measure of the force pushing the electricity. The wattage W consumed by an electrical device can be determined by calculating the product of the amperage I and voltage V. (Source: Wilcox, G. and C. Hesselberth, Electricity for Engineering Technology, Allyn &amp; Bacon, 1970.) (a) A household circuit has voltage V 163 sin 120 t when an incandescent lightbulb is turned on with amperage I 1.23 sin 120 t . Graph the wattage W VI consumed by the lightbulb in the window 0, .05 by 50, 300 . (b) Determine the maximum and minimum wattages used by the lightbulb. (c) Use identities to determine values for a, c, and so that W a cos t c. (d) Check your answer by graphing both expressions for W on the same coordinate axes. (e) Use the graph to estimate the average wattage used by the light. For how many watts do you think this incandescent lightbulb is rated?For x = t, W = VI = (163 sin 120 t)(1.23 sin 120 t)3000 ­50.05(b) maximum: 200.49 watts; minimum: 0 watts (c) a 100.245, 240 , c 100.245 (e) 100.245 wattsSummary Exercises on Verifying Trigonometric IdentitiesThese summary exercises provide practice with the various types of trigonometric identities presented in this chapter. Verify that each equation is an identity. 1. tan x 3. tan 2 5. sin t 1 cos t cot csc x 1 sec csc cot x cos t sin t 2. csc cos2 4. sec 6. 8. 10. 1 sin t cos t 2 cos x 1 2 cot 4 x 2 x sin sec x 1 sec t x tan2 2 1 sec t 1 1 tan t 1 2 cot t csc t csc7. sin 2 9. cot 11. 13. sin x cos x sin 12 tan 1 tan2 tan y y tan cos 1 1 2 cos2 1 sin cos cot x cot y 1 cot x cot y tanx tan2 2 2 x tan 211 sec t12. 1tan22 cos cos 1 1 cos2 x cos2 x14. csc4 x 16. cos 2x 18.15. cos x 17.sec2 x sec2 x 1 cos s sin s sec x 1 2 csc s tan xtan2 t 1 tan t csc2 t 2tan t 2 tan 2 sec2 2 cos s sin s sin s cos s1sin s cos s x 2 4 cot cot19. tan 4 21. cot s cos s20. tan 22. tan tantan s sin s2 cos2LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6457.5 Inverse Circular Functions 64523. 25. 27.tan x y 1 tan xtan y y tan y 1tan x tan2 x24. 2 cos2 26. 28. csc t csc tx tan x 2 1 1tan x tan tsin x2cos4 x sin4 x cos2 x 2 sin x sin3 x cos xsec t 1 x tan 2 2sin 2x1 x cot 2 2cot x7.5 Inverse Circular FunctionsInverse Functions s Inverse Sine Function s Inverse Cosine Function Remaining Inverse Circular Functions s Inverse Function ValuessInverse Tangent FunctionsInverse Functions We first discussed inverse functions in Section 4.1. We give a quick review here for a pair of inverse functions f and f 1. 1. If a function f is one-to-one, then f has an inverse function f 1. 2. In a one-to-one function, each x-value corresponds to only one y-value and each y-value corresponds to only one x-value. 3. The domain of f is the range of f 1, and the range of f is the domain of f 1. 4. The graphs of f and f 1 are reflections of each other about the line y x. 5. To find f 1 x from f x , follow these steps. Step 1 Replace f x with y and interchange x and y. Step 2 Solve for y. Step 3 Replace y with f 1 x .In the remainder of this section, we use these facts to develop the inverse circular (trigonometric) functions.Looking Ahead to CalculusThe inverse circular functions are used in calculus to solve certain types of related-rates problems and to integrate certain rational functions.Inverse Sine Function From Figure 8 and the horizontal line test, we see that y sin x does not define a one-to-one function. If we restrict the domain to the interval 2 , 2 , which is the part of the graph in Figure 8 shown in color, this restricted function is one-to-one and has an inverse function. The range of y sin x is 1, 1 , so the domain of the inverse function will be 1, 1 , and its range will be 2,2 .yTEACHING TIP Mention that the inter2 2 the graph of the sine function to include all possible values of y. While other intervals could also be used, this interval is an accepted convention that is adopted by scientific calculators and graphing calculators. val , contains enough of­ ­2 ­3 2 ­ 2 1( 2 , 1)(0, 0)23 2 x 2(­ 2 , ­1)­1 ­2y = sin xRestricted domain ­ , 2 2[]Figure 8LIALMC07_0321227638.QXP2/26/0410:47 AMPage 646646 CHAPTER 7 Trigonometric Identities and Equationsy 2­10(1, 2 ) ( 3 , 3 ) 2 ( 2 , 4 ) 2 ( 1 , 6) 2(0, 0)x 1(­1, ­6 2)Reflecting the graph of y sin x on the restricted domain across the line y x gives the graph of the inverse function, shown in Figure 9. Some key points are labeled on the graph. The equation of the inverse of y sin x is found by interchanging x and y to get x sin y. This equation is solved for y by writing y sin 1 x (read &quot;inverse sine of x&quot;). As Figure 9 shows, the domain of y sin 1 x is 1, 1 , while the restricted domain of y sin x, 2 , 2 , is the range of y sin 1 x. An alternative notation for sin 1 x is arcsin x.(­ 2 , ­ 4 ) 2(­1, ­ 2 )y=(­ 3 , ­ 3 ) 2­ 2Inverse Sine Functiony sin1x or yarcsin x means that xsin y, for2y2.sin­1x or y = arcsin xFigure 9We can think of y sin 1 x or y arcsin x as &quot;y is the number in the interlog2 4 by writing it in exval 2 , 2 whose sine is x.&quot; Just as we evaluated y ponential form as 2y 4 (Section 4.3), we can write y sin 1 x as sin y x to evaluate it. We must pay close attention to the domain and range intervals.EXAMPLE 1 Finding Inverse Sine ValuesFind y in each equation. (a) yAlgebraic Solutionarcsin1 2(b) ysin11(c) ysin12Graphing Calculator Solution(a) The graph of the function defined by y arcsin x (Figure 9) includes the point 1 , 6 . Thus, 2 arcsin 1 2 6 .Alternatively, we can think of y arcsin 1 as &quot;y is 2 1 the number in , 2 whose sine is 2 .&quot; Then 2 1 we can write the given equation as sin y 2 . 1 Since sin 6 2 and 6 is in the range of the arcsine function, y 6 . (b) Writing the equation y sin 1 1 in the form sin y 1 shows that y 2 . This can be verified by noticing that the point 1, 2 is on the graph of y sin 1 x. (c) Because 2 is not in the domain of the inverse sine function, sin 1 2 does not exist.To find these values with a graphing calculator, we graph Y1 sin 1 X and locate the points with X-values 1 1. Figure 10(a) shows that when X 1 , 2 and 2 Y 6 .52359878. Similarly, Figure 10(b) shows 1, Y 1.570796. that when X 22 2­11­11­2­2(b) Figure 10(a)Since sin 1 2 does not exist, a calculator will give an error message for this input.Now try Exercises 13 and 23.In Example 1(b), it is tempting to give the value of sin 1 1 as 1. Notice, however, that 32 is not in the range of the inverse sine function. Be certain that the number given for an inverse function value is in the range of the particular inverse function being considered.CAUTION 3 3 2 , since sin 2LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6477.5 Inverse Circular Functions 647TEACHING TIP In problems likeExample 1(a), use phrases such as &quot;y is a value in radians between and whose sine is equal 2 2 1 to to help students understand 2 the meaning of inverse sine. Illustrate the answer graphically by 1 graphing y sin x and y in 2 , , then 2 2 determine the point of intersection. the intervalOur observations about the inverse sine function from Figure 9 lead to the following generalizations.INVERSE SINE FUNCTION y sin 1x or y arcsin xDomain:y1, 1Range:2,2y = sin­1 x x 12 2 4y222­1­1 ­ 0 x 1102 204 212y = sin­1 x­ ­Figure 112· The · ·inverse sine function is increasing and continuous on its domain 1, 1 . Its x-intercept is 0, and its y-intercept is 0. Its graph is symmetric with respect to the origin; it is an odd function.Inverse Cosine Function The function y cos 1 x (or y arccos x) is defined by restricting the domain of the function y cos x to the interval 0, as in Figure 12, and then interchanging the roles of x and y. The graph of y cos 1 x is shown in Figure 13. Again, some key points are shown on the graph.y(­1, )y 1 (0, 1)(­ 3 , 56 ) 2 (­ 2 , 34 ) 2 (­ 1 , 23 ) 2(2 ),0­ 0 2 ­1 2x( , ­1) y = cos x­1( 1 , 3) ( 2 ) 2 ( 2 , 4 ) 2 ( 3 , 6 ) 20,0 1 (1, 0)xRestricted domain [0, ]Figure 12y = cos­1 xor y = arccos xFigure 13Inverse Cosine Functiony cos 1 x or y arccos x means that x cos y, for 0 y .LIALMC07_0321227638.QXP2/26/0410:47 AMPage 648648 CHAPTER 7 Trigonometric Identities and Equations5 4EXAMPLE 2 Finding Inverse Cosine ValuesFind y in each equation. (a) y­1 ­.55 4arccos 1(b) ycos12 21Solution(a) Since the point 1, 0 lies on the graph of y arccos x in Figure 13 on the previous page, the value of y is 0. Alternatively, we can think of y arccos 1 as &quot;y is the number in 0, whose cosine is 1,&quot; or cos y 1. Then y 0, since cos 0 1 and 0 is in the range of the arccosine function. (b) We must find the value of y that satisfies cos y 2 , where y is in the interval 0, , the range of the function y cos 1 x. The only value for y that 3 satisfies these conditions is 4 . Again, this can be verified from the graph in Figure 13.Now try Exercises 15 and 21.2­1 ­.5These screens support the results of Example 2, since 3  2.3561945. 41Our observations about the inverse cosine function lead to the following generalizations.INVERSE COSINE FUNCTION y cos 1 x or y arccos xDomain:y1, 1Range: 0,y = cos­1 xx 12 2yy = cos ­1 x3 4 2 402 2­10x 110­1 0Figure 141· The inverse cosine function is decreasing and continuous on its domain · ·1, 1 . Its x-intercept is 1, and its y-intercept is 2 . Its graph is not symmetric with respect to the y-axis or the origin.Inverse Tangent Function Restricting the domain of the function y tan x to the open interval 2 , 2 yields a one-to-one function. By interchanging the roles of x and y, we obtain the inverse tangent function given by y tan 1 x or y arctan x. Figure 15 shows the graph of the restricted tangent function. Figure 16 gives the graph of y tan 1 x.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6497.5 Inverse Circular Functions 649yy = tan xy( 4 , 1)(0, 0)­ 0 2 2 x ­22 1 ­1 (0, 0)(1, 4 )­1 ­ 2(­ 4 , ­1)Restricted domain ­ , 2 2(­ 3 , ­ 6 ) 3 (­ 3 , ­ 3 ) (­1, ­ 4 ), ( 3 , 6 ) (3 x3 ) 321()y = tan­1 x or y = arctan xFigure 15Figure 16Inverse Tangent Functiony tan 1 x or y arctan x means that x tan y, for2y2.­4 0 24INVERSE TANGENT FUNCTION y tan 1 x or y arctan xDomain: ,y 2 x 1 2Range:2,2­44y = tan­1 x2x 1­ 23 3y4 6y = tan­1 x­2 ­1 ­03 306 40­3312­Figure 172­4 04· The inverse tangent function is increasing and continuous on its domain · · ·, . Its x-intercept is 0, and its y-intercept is 0. Its graph is symmetric with respect to the origin; it is an odd function. The lines y 2 and y 2 are horizontal asymptotes.The first three screens show the graphs of the three remaining inverse circular functions. The last screen shows how they are defined.Figure 18Remaining Inverse Circular Functions The remaining three inverse trigonometric functions are defined similarly; their graphs are shown in Figure 18. All six inverse trigonometric functions with their domains and ranges are given in the table on the next page.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 650650 CHAPTER 7 Trigonometric Identities and EquationsRange Inverse Function y y y y y y sin 1 x cos x tan 1 x cot x sec 1 x csc x1 1 1Domain 1, 1 1, 1 , , , 1 , 1 1, 1, 0,Interval2,2Quadrants of the Unit Circle I and IV I and II I and IV I and II2*0,2,20, ,y ,yI and II I and IV2,20*Inverse Function Values The inverse circular functions are formally defined with real number ranges. However, there are times when it may be convenient to find degree-measured angles equivalent to these real number values. It is also often convenient to think in terms of the unit circle and choose the inverse function values based on the quadrants given in the preceding table.EXAMPLE 3 Finding Inverse Function Values (Degree-Measured Angles)Find the degree measure of in the following. (a)Solutionarctan 1(b)sec 1 2(a) Here must be in 90 , 90 , but since 1 1, leads to alternative statement, tan0, must be in quadrant I. The 45 .2. For sec 1 x, is in quadrant I or II. Because (b) Write the equation as sec 60 , since sec 60 2. Note that 60° 2 is positive, is in quadrant I and the degree equivalent of 3 is in the range of the inverse secant function.Now try Exercises 33 and 39.The inverse trigonometric function keys on a calculator give results in the proper quadrant for the inverse sine, inverse cosine, and inverse tangent functions, according to the definitions of these functions. For example, on a cal30 , tan 1 1 45 , and culator, in degrees, sin 1 .5 30 , sin 1 .5 1 cos .5 120 . Finding cot 1 x, sec 1 x, and csc 1 x with a calculator is not as straightforward, because these functions must be expressed in terms of tan 1 x, cos 1 x, and sin 1 x, respectively. If y sec 1 x, for example, then sec y x, which must be written as a cosine function as follows: If sec y x, then 1 cos y x or cos y 1 , x and y cos11 . x*The inverse secant and inverse cosecant functions are sometimes defined with different ranges. We use intervals that match their reciprocal functions (except for one missing point).LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6517.5 Inverse Circular Functions 651In summary, to find sec 1 x, we find cos 1 x . Similar statements apply to csc 1 x and cot 1 x. There is one additional consideration with cot 1 x. Since we take the inverse tangent of the reciprocal to find inverse cotangent, the calculator gives values of inverse cotangent with the same range as inverse tangent, 2,2 , which is not the correct range for inverse cotangent. For inverse cotangent, the proper range must be considered and the results adjusted accordingly.EXAMPLE 4 Finding Inverse Function Values with a Calculator1(a) Find y in radians if y (b) Find in degrees ifSolutioncsc13. .3541 .1 1arccot(a) With the calculator in radian mode, enter csc y .3398369095. See Figure 19.3 as sin1 3to get(b) Set the calculator to degree mode. A calculator gives the inverse tangent value of a negative number as a quadrant IV angle. The restriction on the range of arccotangent implies that must be in quadrant II, so enterFigure 19arccot.3541as tan11 .3541180 .As shown in Figure 19,109.4990544 .Now try Exercises 43 and 49.yEXAMPLE 5 Finding Function Values Using Definitions of the Trigonometric FunctionsEvaluate each expression without using a calculator.13 3(a) sin tanx13 2(b) tan cos15 13Solution3 (a) Let tan 1 3 , so tan 2 2 . The inverse tangent function yields values only in quadrants I and IV, and since 3 is positive, is in quadrant I. Sketch 2 in quadrant I, and label a triangle, as shown in Figure 20. By the Pythagorean theorem, the hypotenuse is 13. The value of sine is the quotient of the side opposite and the hypotenuse, so02 = tan ­1 32 Figure 20 ysin tan12 1313 2sin3 13313 . (Section 5.3) 13A ­50 x5 5 1 (b) Let A cos 1 13 . Then, cos A 13 . Since cos x for a negative value of x is in quadrant II, sketch A in quadrant II, as shown in Figure 21.tan cos1A = cos ­1 ­ 5( 13 )5 13tan A12 5Now try Exercises 63 and 65.Figure 21LIALMC07_0321227638.QXP2/26/0410:47 AMPage 652652 CHAPTER 7 Trigonometric Identities and EquationsTEACHING TIP Point out that theequations sin sin x x, cos cos 1x x, and tan tan 1x x are true wherever they are defined. However, sin 1 sin x x, cos 1 cos x x, 1 and tan tan x x are true only for values of x in the restricted domains of the sine, cosine, and tangent functions.1EXAMPLE 6 Finding Function Values Using IdentitiesEvaluate each expression without using a calculator. (a) cos arctanSolution3arcsin1 3(b) tan 2 arcsin2 5(a) Let A arctan 3 and B arcsin 1 , so tan A 3 and sin B 1 . 3 3 Sketch both A and B in quadrant I, as shown in Figure 22. Now, use the cosine sum identity. cos A B 1 3 cos A cos B cos arctan sin arctan From Figure 22, sin A sin B (Section 7.3) 3 cos arcsin 1 3 1 3(1)ycos arctan3arcsin2 A03x3 sin arcsin1cos arctany3 3cos A sin A1 , 2 3 2 ,cos arcsin sin arcsin1 3 1 3cos B sin B2 3 1 . 32,sin arctan3 B01xSubstitute these values into equation (1) to get cos arctan (b) Let arcsin 2 5 3 arcsin 1 3 1 2 2 3 2 2 3 1 3 2 2 6 3 .22Figure 22B. Then, from the double-angle tangent identity, 2 5 tan 2B 2 tan B . tan2 B(Section 7.4)ytan 2 arcsin15 B02xSince arcsin 2 B, sin B 2 . Sketch a triangle in quadrant I, find the 5 5 length of the third side, and then find tan B. From the triangle in Figure 23, 2 tan B , and 21 2 tan 2 arcsin 5 2 12 21 2 21 4 21221Figure 23414 2121 . 17Now try Exercises 69 and 75.While the work shown in Examples 5 and 6 does not rely on a calculator, we can support our algebraic work with one. By entering cos arctan 3 arcsin 1 3 from Example 6(a) into a calculator, we get the approximation .1827293862, the same approximation as when we enter (the exact value we obtained 6 algebraically). Similarly, we obtain the same approximation when we evaluate tan 2 arcsin 2 and 54 21 17 , 2 2 3supporting our answer in Example 6(b).LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6537.5 Inverse Circular Functions 653EXAMPLE 7 Writing Function Values in Terms of uWrite each trigonometric expression as an algebraic expression in u. (a) sin tan 1 uy(b) cos 2 sin 1 uSolution u2 + 11(a) Letu, u &gt; 0x0tan 1 u, so tan u. Here, u may be positive or negative. Since tan 1 u 2 , sketch in quadrants I and IV and label two triangles, 2 as shown in Figure 24. Since sine is given by the quotient of the side opposite and the hypotenuse, sin tan 1 u sin u u2u2 + 1u, u &lt; 0u 1 uu221 1.The result is positive when u is positive and negative when u is negative.Figure 24sin 1 u, so sin (b) Let 1 2 sin2 . cos 2 sin1u. To find cos 2 , use the identity cos 2 cos 2 1 2 sin2 1 2u2uNow try Exercises 83 and 85.EXAMPLE 8 Finding the Optimal Angle of Elevation of a Shot PutThe optimal angle of elevation a shot-putter should aim for to throw the greatest distance depends on the velocity v of the throw and the initial height h of the shot. See Figure 25. One model for that achieves this greatest distance is arcsin v2 2v 2 64h .(Source: Townend, M. Stewart, Mathematics in Sport, Chichester, Ellis Horwood Limited, 1984.)h DFigure 25Suppose a shot-putter can consistently throw the steel ball with h 6.6 ft and v 42 ft per sec. At what angle should he throw the ball to maximize distance?SolutionTo find this angle, substitute and use a calculator in degree mode. arcsin 422 2 422 64 6.6 41.9h 6.6, v 42Now try Exercise 93.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 654654 CHAPTER 7 Trigonometric Identities and Equations7.5 Exercises1. one-to-one 3. domain 4. 1, ;y 2. range tan1Concept CheckComplete each statement. . of y of y sin x. cos x. arcsin x equals the cos 1 x equals the1. For a function to have an inverse, it must bex or4 y arctan x 5. 6. Sketch the reflection of the graph of f across the line y x. 7. (a) 1, 1 (b) 2 2 2 is not in the ,2. The domain of y 3. The range of y4. The point 4 , 1 lies on the graph of y lies on the graph of . 5. If a function f has an inverse and f 6. How can the graph of f Concept Check1tan x. Therefore, the point 1, then f11.(c) increasing (d) domain. 8. (a) 1, 1 (b) 0, (c) decreasing 4 (d) is not in the range. 3 9. (a) , (b) 2 , ;be sketched if the graph of f is known?In Exercises 7­10, write short answers. sin17. Consider the inverse sine function, defined by yx or yarcsin x.(a) What is its domain? (b) What is its range? (c) Is this function increasing or decreasing? (d) Why is arcsin 2 not defined?28. Consider the inverse cosine function, defined by ycos1x or yarccos x.(c) increasing (d) no 10. (a) , 1 1, 2 (b) 0, (c) 11. cos tan1,0 , 10,(a) What is its domain? (b) What is its range? (c) Is this function increasing or decreasing? 2 1 (d) Arccos 1 2 3 . Why is arccos 2 not equal to 9. Consider the inverse tangent function, defined by y4 31? arctan x.2 1, ;tan x or y2 ,12 ; 0, 1 12. Find a (or 180°).,(a) What is its domain? (b) What is its range? (c) Is this function increasing or decreasing? (d) Is there any real number x for which arctan x is not defined? If so, what is it (or what are they)? 10. Give the domain and range of the three other inverse trigonometric functions, as defined in this section. (a) inverse cosecant function (c) inverse cotangent function16. 41 a 14. 2 3 4 4 6 4 60(b) inverse secant function113. 0 17. 20. 23. 26. 29. 32. 35.4 18. 21.15. 3 211. Concept CheckIs sec1a calculated as cos1 a1or as cos1 1 a ? a is calculated as tan119. 0 22. 25. 5 6 3 4 6 412. Concept Check For positive values of a, cot cot 1 a calculated for negative values of a?1 a.How isFind the exact value of each real number y. Do not use a calculator. See Examples 1 and 2. 13. y 16. y 19. y 22. y 25. y 28. y 31. y sin124.2 30 114. y 17. y 20. y 23. y 3 2 2 2 3 3 26. y 29. y 32. ytan sin11 1 3 2 2 2 2 2 2 215. y 18. y 21. y 24. y 27. y 30. ycos cos11 1 23 4 5 30. 6 27. 33. 36. 45 4528. 31.arctan arctan 0 tan111arcsin sin1arccos 0 cos cot134. 120 37. 12011 2 1 3arccos sec1arcsin csc csc11arccotarcsec1LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6557.5 Inverse Circular Functions 65538. 41. 42. 43. 44. 45. 47. 49. 51. 53.30 39. 120 40. 90 7.6713835 97.671207 113.500970 51.1691219 30.987961 46. 29.506181 .83798122 48. .96012698 2.3154725 50. 2.4605221 1.1900238 52. 1.1082303 y 54.y 2 y = cot ­1 x x ­ 2 ­2 0 ­10 1 2 y = csc­1 x xGive the degree measure of . Do not use a calculator. See Example 3. 33. 36. 39. arctan arcsin sec11 2 2 234. 37. 40.arccos cot csc11 2 3 3 135. 38.arcsin csc13 2 21Use a calculator to give each value in decimal degrees. See Example 4. 41. 43. 45. sin csc1.13349122 .39876459 1.942283342. 44. 46.cos cot1.13348816 1.7670492arccos1arcsin .77900016155.y y = sec­156.x ­1y y = arccsc 2x 0 x 1 ­ 2Use a calculator to give each real number value. (Be sure the calculator is in radian mode.) See Example 4. 47. y 49. y 51. y arctan 1.1111111 cot148. y 50. y 52. yarcsin .81926439 sec1­1 0 1x.921701281.287168457.yarcsin .92837781arccos .44624593y = arcsec 1 x2Graph each inverse function as defined in the text. 53. y 56. y cot12 ­2 0 x 2x54. y 57. ycsc1x55. ysec1xarccsc 2x58. 1.003 is not in the domain of y sin 1 x. 59. The domain of y tan 1 x is , . 60. In both cases, the result is x. In each case, the graph is a straight line bisecting quadrants I and III (i.e., the line y x). 61. It is the graph of y x.101 arcsec x 2158. Explain why attempting to find sin message.1.003 on your calculator will result in an error59. Explain why you are able to find tan 1 1.003 on your calculator. Why is this situation different from the one described in Exercise 58?Relating ConceptsFor individual or collaborative investigation (Exercises 60­62)*x 2 60. Consider the function defined by f x 3x 2 and its inverse f 1 x 3 . 1 1 Simplify f f x and f f x . What do you notice in each case? What would the graph look like in each case?­1010­1062. It does not agree because the range of the inverse tangent function is , not , 2 2 , , as was the case in Exercise 61.1061. Use a graphing calculator to graph y tan tan 1 x in the standard viewing window, using radian mode. How does this compare to the graph you described in Exercise 60? 62. Use a graphing calculator to graph y tan 1 tan x in the standard viewing window, using radian and dot modes. Why does this graph not agree with the graph you found in Exercise 61?­1010 *The authors wish to thank Carol Walker of Hinds Community College for making a suggestion on which these exercises are based.­10LIALMC07_0321227638.QXP2/26/0410:47 AMPage 656656 CHAPTER 7 Trigonometric Identities and Equations63. 66. 69. 71. 74. 77. 78. 79. 80. 81. 82. 84. 86. 88. 90. 92.7 3 5 65 7 68. 12 8 7 15 70. 25 7 4 6 3 72. 73. 2 25 5 63 16 76. 2 75. 65 65 10 3 30 20 48 25 3 39 .894427191 .9682458366 .1234399811 .716386406 83. 1 u2 1 u 1 u u u u2264.15 4 120 67. 16965.5Give the exact value of each expression without using a calculator. See Examples 5 and 6. 63. tan arccos 66. sec sin13 4 1 5 4 3 2 tan tan1 164. sin arccos 67. sin 2 tan 70. tan 2 cos 73. sec sec 3 4 31 11 4 12 5 1 465. cos tan12168. cos 2 sin 71. sin 2 cos 74. csc csc1 11 4 1 5 269. cos 2 arctan 72. cos 2 tan 75. cos tan 77. sin sin1 1112 76. cos sin 78. tan cos5 12 1 23 5 3 2cos15 13111sin3 5Use a calculator to find each value. Give answers as real numbers. 79. cos tan1u2 u2 9 9 5 u285. 87. 89. 91. u1 u2 u 2 2 4 4u2 4.580. sin cos1.2581. tan arcsin .1225101482. cot arccos .58236841 0. SeeWrite each expression as an algebraic (nontrigonometric) expression in u, u Example 7.u2 u2u2 u 3 9283. sin arccos u 86. cot arcsin u 89. tan sin184. tan arccos u 87. sin sec u u2 4 u 2 u2185. cos arcsin u 88. cos tan1 1u 2 90. sec cos3 u9 u2 93. (a) 45 (b) 45° 94. (a) 113 (b) 84 (c) 60 (d) 47u u2 9 u 5 u291. sec arccot92. csc arctan(Modeling) Solve each problem. 93. Angle of Elevation of a Shot Put Refer to Example 8. (a) What is the optimal angle when h 0? (b) Fix h at 6 ft and regard as a function of v. As v gets larger and larger, the graph approaches an asymptote. Find the equation of that asymptote. 94. Angle of Elevation of a Plane Suppose an airplane flying faster than sound goes directly over you. Assume that the plane is flying at a constant altitude. At the instant you feel the sonic boom from the plane, the angle of elevation to the plane is given by 2 arcsin 1 , mwhere m is the Mach number of the plane's speed. (The Mach number is the ratio of the speed of the plane and the speed of sound.) Find to the nearest degree for each value of m. (a) m 1.2 (b) m 1.5 (c) m 2 (d) m 2.5LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6577.5 Inverse Circular Functions 65795. (a) 18 (b) 18 (c) 15 (e) 1.4142151 m (Note: Due to the computational routine, there may be a discrepancy in the last few decimal places.) y=195. Observation of a Painting A painting 1 m high and 3 m from the floor will cut off an angle to an observer, where tan11              1x x2 2 2.3tan ­1(x x2 + 2)10Assume that the observer is x meters from the wall where the painting is displayed and that the eyes of x the observer are 2 m above the ground. (See the figure.) Find the value of for the following values of x. Round to the nearest degree. (a) 1 (b) 2 (c) 3 (d) Derive the formula given above. (Hint: Use the identity for tan . Use right triangles.) (e) Graph the function for with a graphing calculator, and determine the distance that maximizes the angle. (f) The idea in part (e) was first investigated in 1471 by the astronomer Regiomontanus. (Source: Maor, E., Trigonometric Delights, Princeton University Press, 1998.) If the bottom of the picture is a meters above eye level and the top of the picture is b meters above eye level, then the optimum value of x is ab meters. Use this result to find the exact answer to part (e). 96. Landscaping Formula A shrub is planted in a 100-ft-wide space between buildings measuring 75 ft and 150 ft tall. The location of the shrub determines how much sun it receives each day. Show that if is the angle in the figure and x is the distance of the shrub from the taller building, then the value of (in radians) is given by arctan 75 100 x arctan 150 . x0 ­.5 Radian mode(f) 2 97. about 44.7%150 ft 75 ftx 100 ft97. Communications Satellite Coverage The figure shows a stationary communications satellite positioned 20,000 mi above the equator. What percent of the equator can be seen from the satellite? The diameter of Earth is 7927 mi at the equator.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 658658 CHAPTER 7 Trigonometric Identities and Equations7.6 Trigonometric EquationsSolving by Linear Methods s Solving by Factoring s Solving by Quadratic Methods s Solving by Using Trigonometric Identities s Equations with Half-Angles s Equations with Multiple Angles s Applications Looking Ahead to CalculusThere are many instances in calculus where it is necessary to solve trigonometric equations. Examples include solving related-rates problems and optimization problems.Earlier in this chapter, we studied trigonometric equations that were identities. We now consider trigonometric equations that are conditional; that is, equations that are satisfied by some values but not others.Solving by Linear Methods Conditional equations with trigonometric (or circular) functions can usually be solved using algebraic methods and trigonometric identities.EXAMPLE 1 Solving a Trigonometric Equation by Linear MethodsySolve 2 sin= 210° ' = 30°0 x10 over the interval 0 , 360 .SolutionBecause sin is the first power of a trigonometric function, we use the same method as we would to solve the linear equation 2x 1 0. 2 sin 1 2 sin sin 0 1 1 2Subtract 1. (Section 1.1) Divide by 2.1(a)y= 330°0' = 30°x(b)Figure 26To find values of that satisfy sin must be in 2 , we observe that either quadrant III or IV since the sine function is negative only in these two 1 quadrants. Furthermore, the reference angle must be 30° since sin 30 2 . The graphs in Figure 26 show the two possible values of , 210° and 330°. The solution set is 210 , 330 . Alternatively, we could determine the solutions by referring to Figure 11 in Section 6.2 on page 546.Now try Exercise 11.Solving by FactoringEXAMPLE 2 Solving a Trigonometric Equation by FactoringSolve sin x tan xSolutionsin x over the interval 0 , 360 . sin x tan x sin x 0 0 tan x x 45 1 tan x or 1 x 225Subtract sin x. Factor. (Section R.4)sin x tan x sin x tan x sin x x 0 or 0 xsin x 1 or 1800 Zero-factor property (Section 1.4)The solution set is 0 , 45 , 180 , 225 .Now try Exercise 31.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6597.6 Trigonometric Equations 659CAUTION There are four solutions in Example 2. Trying to solve the equation by dividing each side by sin x would lead to just tan x 1, which would give x 45 or x 225 . The other two solutions would not appear. The missing solutions are the ones that make the divisor, sin x, equal 0. For this reason, we avoid dividing by a variable expression.Solving by Quadratic Methods In Section 1.6, we saw that an equation in the form au2 bu c 0, where u is an algebraic expression, is solved by quadratic methods. The expression u may also be a trigonometric function, as in the equation tan2 x tan x 2 0.EXAMPLE 3 Solving a Trigonometric Equation by FactoringSolve tan2 xSolutiontan x20 over the interval 0, 2 . tan2 xThis equation is quadratic in form and can be solved by factoring. tan x tan x 2 2 2 tan x 0 0 0 2Factor. Zero-factor propertytan x tan x 1 tan x 0 1 or or1 tan xThe solutions for tan x 1 over the interval 0, 2 are x 4 and x 54 . 2 over that interval, we use a scientific calculator set in To solve tan x 1.1071487. This is a quadrant IV radian mode. We find that tan 1 2 number, based on the range of the inverse tangent function. (Refer to Figure 11 in Section 6.2 on page 546.) However, since we want solutions over the interval 0, 2 , we must first add to 1.1071487, and then add 2 . x x 1.1071487 1.1071487 2 2.0344439 5.1760366The solutions over the required interval form the solution set 4 , 5 , 4 2.0, 5.2 .      EXAMPLE 4 Solving a Trigonometric Equation Using the Quadratic FormulaFind all solutions of cot x cot xSolutionWe multiply the factors on the left and subtract 1 to get the equation in standard quadratic form. cot 2 x 3 cot x 1 0 (Section 1.4)Since this equation cannot be solved by factoring, we use the quadratic formula, with a 1, b 3, c 1, and cot x as the variable.       Exact values Approximate values to the nearest tenth Now try Exercise 21.31.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 660660 CHAPTER 7 Trigonometric Identities and Equations Quadratic formula with a 1, b 3, c 1(Section 1.4)cot x cot xTEACHING TIP Unlike the cosinefunction, equations of the form y tan x will not contain a second value for x between 2 and3 2943 2 or cot x133.302775638.3027756377 Use a calculator.We cannot find inverse cotangent values directly on a calculator, so we use the 1 fact that cot x tan x , and take reciprocals to get tan x x .3027756377 .2940013018 or tan x or x 3.302775638 1.276795025.2 solutions to y tan x are found using period radians.. Remind students that otherTo find all solutions, we add integer multiples of the period of the tangent function, which is , to each solution found above. Thus, all solutions of the equation are written as .2940013018 n and 1.276795025 n , where n is any integer.*Now try Exercise 43.Solving by Using Trigonometric Identities Recall that squaring both sides of an equation, such as x 4 x 2, will yield all solutions but may also give extraneous values. (In this equation, 0 is a solution, while 3 is extraneous. Verify this.) The same situation may occur when trigonometric equations are solved in this manner.EXAMPLE 5 Solving a Trigonometric Equation by SquaringSolve tan xTEACHING TIP Point out inExample 5 that our first goal is to rewrite the equation in terms of a single trigonometric function.3sec x over the interval 0, 2 .Solution1tan2 x tan x tan2 x2Since the tangent and secant functions are related by the identity sec2 x, square both sides and express sec2 x in terms of tan2 x. tan x 2 2 3 tan x 3 tan x 2 3 3 3 sec x sec2 x 1 2 1 35 6xy2x22xyy2(Section R.3)tan2 x 3 3Pythagorean identity (Section 7.1) Subtract 3 tan2 x.3 tan x tan xDivide by 2 3; rationalize the denominator. (Section R.7)TEACHING TIP Explain that whenever possible, answers should be given in exact form, such as 11 , rather than as decimal 6 approximations.The possible solutions are Left side: tan xand11 6. Now check them. Try 56 first. 5 6 2 3 3 3 3 3 3 2 3 33 5 6tan(Section R.7)Right side:sec xsecNot equal*We usually give solutions of equations as solution sets, except when we ask for all solutions of a trigonometric equation.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6617.6 Trigonometric Equations 661y = tan x + 3 ­ sec x4The check shows that Left side:5 6is not a solution. Now check tan sec 11 6 11 6 2 311 611 6. 3 2 3Equal3 33 3302Right side:­4 Dot mode; radian mode The graph shows that on the interval [0, 2 ), the only x-intercept of the graph of y = tan x + 3 ­ sec x is 5.7595865, which is an approximation for 11 , the sol6 ution found in Example 5.This solution satisfies the equation, sois the solution set.Now try Exercise 41.The methods for solving trigonometric equations illustrated in the examples can be summarized as follows.Solving a Trigonometric Equation1. Decide whether the equation is linear or quadratic in form, so you can determine the solution method. 2. If only one trigonometric function is present, first solve the equation for that function. 3. If more than one trigonometric function is present, rearrange the equation so that one side equals 0. Then try to factor and set each factor equal to 0 to solve. 4. If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval. 5. Try using identities to change the form of the equation. It may be helpful to square both sides of the equation first. If this is done, check for extraneous solutions.Some trigonometric equations involve functions of half-angles or multiple angles.Equations with Half-AnglesEXAMPLE 6 Solving an Equation Using a Half-Angle IdentitySolve 2 sinx 21 (b) give all solutions.(a) over the interval 0, 2 , andSolution(a) Write the interval 0, 2 0 xas the inequality 2 .x The corresponding interval for 2 is0x 2.Divide by 2. (Section 1.7)LIALMC07_0321227638.QXP2/26/0410:47 AMPage 662662 CHAPTER 7 Trigonometric Identities and Equations y = 2 sin x ­ 1 2x To find all values of 2 over the interval 0, x first solve for sin 2 .that satisfy the given equation,42 sin0 2x 2 x 21 1 2Divide by 2.sin­4 The x-intercepts are the solutions found in Example 6. Using Xscl = makes it 3 possible to support the exact solutions by counting the tick marks from 0 on the graph.The two numbers over the interval 0, x 2 x 6 3 or or x 2 x 5 6with sine value 2 are16and5 6, so5 . Multiply by 2. 35 3, 3The solution set over the given interval isTEACHING TIP As a slight variationof the problem in Example 6, x replace with u, solve 2 2 sin u 1 for u, and then multiply the solutions by 2 to find x..(b) Since this is a sine function with period 4 , all solutions are given by the expressions 3 4n and 5 3 4n , where n is any integer.Now try Exercise 63.Equations with Multiple AnglesEXAMPLE 7 Solving an Equation with a Double AngleSolve cos 2xSolutioncos x over the interval 0, 2 .cos 2xFirst change cos 2x to a trigonometric function of x. Use the identity 2 cos2 x 1 so the equation involves only cos x. Then factor. cos 2x 2 cos x2cos x cos x 0 0 cos x 1Substitute; double-angle identity(Section 7.4)1 1 1 or2 cos x 2 cos x 2 cos x 12cos x 0 1 2Subtract cos x. Factor.1 cos x0 Zero-factor property 1cos xorcos xOver the required interval, x 2 3 or x 4 3 or x 0.The solution set is 0, 23 , 43 .Now try Exercise 65.LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6637.6 Trigonometric Equations 663CAUTION In the solution of Example 7, cos 2x cannot be changed to cos x by dividing by 2 since 2 is not a factor of cos 2x.cos 2x 2cos xThe only way to change cos 2x to a trigonometric function of x is by using one of the identities for cos 2x.EXAMPLE 8 Solving an Equation Using a Multiple-Angle IdentitySolve 4 sin cosSolution3 over the interval 0 , 360 . sin 2 is useful here.2 2 sin 2 (Section 7.4)The identity 2 sin cos 4 sin cos 2 2 sin cos 2 sin 2 sin 2 2 3 3 4 33 2 sin cosDivide by 2.TEACHING TIP Students should beaware that y sin ax may have as many as 2a solutions from 0° to 360°.From the given interval 0 360 , the interval for 2 is 0 List all solutions over this interval. 2 or 60 , 120 , 420 , 480 30 , 60 , 210 , 240Divide by 2.2720 .The final two solutions for 2 were found by adding 360° to 60° and 120°, respectively, giving the solution set 30 , 60 , 210 , 240 .Now try Exercise 83.Applications Music is closely related to mathematics.EXAMPLE 9 Describing a Musical Tone from a Graphy = .004 sin(300 x).006A basic component of music is a pure tone. The graph in Figure 27 models the sinusoidal pressure y P in pounds per square foot from a pure tone at time x t in seconds. (a) The frequency of a pure tone is often measured in hertz. One hertz is equal to one cycle per second and is abbreviated Hz. What is the frequency f in hertz of the pure tone shown in the graph? (b) The time for the tone to produce one complete cycle is called the period. Approximate the period T in seconds of the pure tone. (c) An equation for the graph is y .004 sin 300 x . Use a calculator to estimate all solutions to the equation that make y .004 over the interval 0, .02 .0.04­.006Figure 27LIALMC07_0321227638.QXP2/26/0410:47 AMPage 664664 CHAPTER 7 Trigonometric Identities and Equations SolutionY2 = .004.007(a) From the graph in Figure 27 on the previous page, we see that there are 6 6 cycles in .04 sec. This is equivalent to .04 150 cycles per sec. The pure tone has a frequency of f 150 Hz. (b) Six periods cover a time of .04 sec. One period would be equal to 1 T .04 150 or .006 sec. 60.02(c) If we reproduce the graph in Figure 27 on a calculator as Y1 and also graph a second function as Y2 .004, we can determine that the approximate values of x at the points of intersection of the graphs over the interval 0, .02 are .0017, .0083, and .015.­.007Y1 = .004 sin(300 X)Figure 28The first value is shown in Figure 28.Now try Exercise 87.A piano string can vibrate at more than one frequency when it is struck. It produces a complex wave that can mathematically be modeled by a sum of several pure tones. If a piano key with a frequency of f1 is played, then the corresponding string will not only vibrate at f1 but it will also vibrate at the higher frequencies of 2f1, 3f1, 4f1, . . ., nf 1 , . . . . f1 is called the fundamental frequency of the string, and higher frequencies are called the upper harmonics. The human ear will hear the sum of these frequencies as one complex tone. (Source: Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag, 1975.)EXAMPLE 10Analyzing Pressures of Upper HarmonicsSuppose that the A key above middle C is played. Its fundamental frequency is f1 440 Hz, and its associated pressure is expressed as P1 The string will also vibrate at f2 880, f3 1320, f4 1760, f5 2200, . . . Hz. .002 sin 880 t .The corresponding pressures of these upper harmonics are P2P = P1 + P2 + P3 + P4 + P5.005.002 sin 1760 t , 2 andP3.002 sin 2640 t , 3 P5 .002 sin 4400 t . 5P4 The graph of.002 sin 3520 t , 40.01PP 1P 2P 3P 4P, 5shown in Figure 29, is &quot;saw-toothed.&quot;­.005Figure 29(a) What is the maximum value of P? (b) At what values of x does this maximum occur over the interval 0, .01 ?LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6657.6 Trigonometric Equations 665P = P1 + P2 + P3 + P4 + P5.005Solution(a) A graphing calculator shows that the maximum value of P is approximately .00317. See Figure 30..010(b) The maximum occurs at x .000188, .00246, .00474, .00701, and .00928. Figure 30 shows how the second value is found; the others are found similarly.Now try Exercise 89. 1. Solve the linear equation for cot x. 2. Solve the linear equation for sin x. 3. Solve the quadratic equation for sec x by factoring. 4. Solve the quadratic equation for cos x by the zero-factor property. 5. Solve the quadratic equation for sin x using the quadratic formula. 6. Solve the quadratic equation for tan x using the quadratic formula. 7. Use an identity to rewrite as an equation with one trigonometric function.­.005Figure 307.6 Exercises8. Use an identity to rewrite as an equation with one trigonometric function. 4 , , 9. 10. 135°, 180°, 3 3 3 7 , 225°, 270° 11. 4 4 12. 14. 17. 18. 19. 21. 22. 23. 24. 25. 26. 27. 28. 30. 31. 32. 33. 34. 35. 36. 2 5 , 3 3 4 , 13. 6 , 5 6Concept Check Refer to the summary box on solving a trigonometric equation. Decide on the appropriate technique to begin the solution of each equation. Do not solve the equation. 1. 2 cot x 3. 5 sec2 x 5. 9 sin x 7. tan x21 5 sin x cot x1 12. sin x 4. 2 cos2 x 6. tan x 8. cos2 x223 cos x 1 2 1 06 sec x 0 Answer each question.4 tan x sin2 xConcept Check15. 0 16. 02 5 5 , , 3 4 39. Suppose you are solving a trigonometric equation for solutions over the interval 0, 2 , and your work leads to 2x 23 , 2 , 83 . What are the corresponding values of x? 10. Suppose you are solving a trigonometric equation for solutions over the interval 0°, 360° , and your work leads to 1 45°, 60°, 75°, 90°. What are the corre3 sponding values of ? Solve each equation for exact solutions over the interval 0, 2 . See Examples 1 ­4. 11. 2 cot x 13. 2 sin x 15. tan2 x 17. cot x 19. cos x 21. 2 sin2 x23 7 11 , , , 4 4 6 6 20. 7 3 11 , , 6 2 6 0, 2 4 , 3 3 3 11 , , , 6 2 2 61 3 3 1 4 0112. sin x 14. 2 sec x 16. sec2 x 1 0 1 0 18. csc x 20. 2 cos x 22. 2 cos2 x22 1 23 sec x 1 2 0 1 0 3 cos x cos x 33 cot x 1 3 sin x2 csc x30°, 210°, 240°, 300° 0°, 45°, 225° 90°, 210°, 330° 60°, 135°, 240°, 315° 45°, 135°, 225°, 315° 0°, 180° 29. 45°, 225° 90°, 270° 0°, 30°, 150°, 180° 0°, 90°, 180°, 270° 0°, 45°, 135°, 180°, 225°, 315° 45°, 135°, 225°, 315° 53.6°, 126.4°, 187.9°, 352.1° 78.0°, 282.0°2 cos xSolve each equation for exact solutions over the interval 0°, 360° . See Examples 2 ­ 5. 23. cot 25. 2 sin 27. tan 29. csc 33. sec23 2 sin 1 cot 2 cot sin tan23024. tan 26. tan 28. cos21 cos 1 sin2 cos sin2 21 10 3 cotcsc 0 0 tan2 2 tan 1 03 cos 0 0 4 cos30. sin231. 2 tan2232. sin2 cos2 34. cos235. 9 sin6 sin36. 4 cos1LIALMC07_0321227638.QXP2/26/0410:47 AMPage 666666 CHAPTER 7 Trigonometric Identities and Equations 37. 149.6°, 329.6°, 106.3°, 286.3° 38. 38.4°, 218.4°, 104.8°, 284.8° 39. 0 40. 68.5°, 291.5° 41. 57.7°, 159.2° 42. 114.3°, 335.7° 43. .9 2n , 2.3 2n , 3.6 2n , 5.8 2n , where n is any integer 44. 5 3 3 2n , 2n ,37. tan2 39. sin2 41. cot4 tan 2 sin 2 csc2 3 30 038. 3 cot 2 40. 2 cos2 42. 2 sin3 cot 2 cos 1 2 cos1 10 0Determine all solutions of each equation in radians (for x) or degrees (for ) to the nearest tenth as appropriate. See Example 4. 43. 3 sin2 x 45. 4 cos x 47. 5 sec2 49. 3 2 tan tan22sin x 1 0 6 sec 11044. 2 cos2 x 46. 2 cos x 48. 3 sin2 50. sec22cos x 5 cos x sin 2 tan1 2 2 4 02n , where n is any integer2 45. 2n , 2n , 3 3 5 4 2n , 2n , where n 3 3 2 is any integer 46. 2n , 3 4 2n , where n is any integer 3 47. 33.6° 360° n, 326.4° 360° n, where n is any integer 48. 90° 360° n, 221.8° 360° n, 318.2° 360° n, where n is any integer 49. 45° 180° n, 108.4° 180° n, where n is any integer 50. 135° 360° n, 315° 360° n, 71.6° 360° n, 251.6° 360° n, where n is any integer 51. .6806, 1.4159 52. 0, .3760 55. 56. 57. 58. 59. 60. 61. 62. 63. 11 13 23 , , , 12 12 12 12 2 4 5 , , 3 3 3 3 , , 7 11 , 2 6 6 0, 3 , , , 2 4 5 , , , 3 3 3The following equations cannot be solved by algebraic methods. Use a graphing calculator to find all solutions over the interval 0, 2 . Express solutions to four decimal places. 51. x 2 sin x x3 cos x 0 52. x 3 cos2 x 1 x 2 19 53.Explain what is wrong with the following solution. Solve tan 2 2 over the interval 0, 2 . tan 2 tan 2 2 tan 4 The solutions are or 2 2 2 1 5 49 54.x The equation cot csc 2 1 0 has no solution over the interval 0, 2 . Using this information, what can be said about the graph of4 x 2and 54 .ycotx 2cscx 21over this interval? Confirm your answer by actually graphing the function over the interval. Solve each equation for exact solutions over the interval 0, 2 . See Examples 6 ­ 8. 55. cos 2x 58. sin 3x 61. 0 1 3 2 56. cos 2x 59. 3 tan 3x 62. 2 3 sin 2x sin 2x x 2 2 0 1 2 3 3 57. sin 3x 60. cot 3x x 63. sin 2 66. cos 2x 69. sin x 2 1 3 2 cos x cos x 2 sin 0 x 27 13 19 25 31 , , , , 18 18 18 18 18 18 7 13 19 25 31 , , , , 18 18 18 18 18 18 3 5 11 13 , , , 8 8 8 8 5 13 17 , , 12 12 12 12 , 3 2 2 ,2 cos 2x 0 x 2 4 cos x 264. tan 4x 67. 8 sec2 70. sec x 265. sin x 68. sin2LIALMC07_0321227638.QXP2/26/0410:47 AMPage 6677.6 Trigonometric Equations 66764. 65. 66. 69.0, 0, 0,3 5 3 7 , , , , 4 2 4 4 2 4 , , 3 , , 5 3 67. 0 68. 0Solve each equation in Exercises 71 ­78 for exact solutions over the interval 0°, 360° . In Exercises 79 ­86, give all exact solutions. See Examples 6 ­ 8. 71. 2 sin 3 1 2 cos 2 2 cos2 sin 2 sin 2 4 sin 2 0 1 0 72. 75. 2 78. cos 81. csc2 2 2 cos 2 3 sin 3 cos 2 2 sec 8 sin cos 3 73. cos 76. 2 79. 1 82. cos 85. 2 cos 222 3 cos sin1 3 cos 2 sin2 2 1 cos 22 4 , 3 3 70. 074. sin22 1277. 2 sin280. sin 2 83. 2 86. sin71. 15°, 45°, 135°, 165°, 255°, 285° 72. 75°, 105°, 255°, 285° 73. 0° 74. 180° 75. 120°, 240° 76. 300° 77. 30°, 150°, 270° 78. 60°, 90°, 270°, 300° 79. 0° 360° n, 30° 360° n, 150° 360° n, 180° 360° n, where n is any integer 80. 45° 360° n, 90° 360° n, 225° 360° n, 270° 360° n, where n is any integer 81. 60° 360° n, 300° 360° n, where n is any integer 82. 70.5° 360° n, 289.5° 360° n, where n is any integer 83. 11.8° 360° n, 78.2° 360° n, 191.8° 360° n, 258.2° 360° n, where n is any integer 84. 22.5° 360° n, 112.5° 360° n, 202.5° 360° n, 292.5° 360° n, where n is any integer 85. 30° 360° n, 90° 360° n, 150° 360° n, 210° 360° n, 270° 360° n, 330° 360° n, where n is any integer 86. 0° 360° n, 60° 360° n, 180° 360° n, 300° 360° n, where n is any integer 87. (a) .00164 and .00355 (b) .00164, .00355 (c) outward 88. (a) 3 beats per sec For x = t, P(t) = .005 sin 440 t + .005 sin 446 t.0184. 4 cos 2(Modeling) Solve each problem. See Examples 9 and 10. 87. Pressure on the Eardrum No musical instrument can generate a true pure tone. A pure tone has a unique, constant frequency and amplitude that sounds rather dull and uninteresting. The pressures caused by pure tones on the eardrum are sinusoidal. The change in pressure P in pounds per square foot on a person's eardrum from a pure tone at time t in seconds can be modeled using the equation P A sin 2 f t ,where f is the frequency in cycles per second, and is the phase angle. When P is positive, there is an increase in pressure and the eardrum is pushed inward; when P is negative, there is a decrease in pressure and the eardrum is pushed outward. (Source: Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag, 1975.) A graph of the tone middle C is shown in the figure. (a) Determine algebraically the values of t for which P 0 over 0, .005 . (b) From the graph and your answer in part (a), determine the interval for which P 0 over 0, .005 . (c) Would an eardrum hearing this tone be vibrating outward or inward when P 0?For x = t, P(t) = .004 sin 2 (261.63)t +.005[7]0.005­.00588. Hearing Beats in Music Musicians sometimes tune instruments by playing the same tone on two different instruments and listening for a phenomenon known as beats. Beats occur when two tones vary in frequency by only a few hertz. When the two instruments are in tune, the beats disappear. The ear hears beats because the pressure slowly rises and falls as a result of this slight variation in the frequency. This phenomenon can be seen using a graphing calculator. (Source: Pierce, J., The Science of Musical Sound, Scientific American Books, 1992.) (a) Consider two tones with frequencies of 220 and 223 Hz and pressures P1 .005 sin 440 t and P2 .005 sin 446 t, respectively. Graph the pressure P P1 P2 felt by an eardrum over the 1-sec interval .15, 1.15 . How many beats are there in 1 sec? (b) Repeat part (a) with frequencies of 220 and 216 Hz. (c) Determine a simple way to find the number of beats per second if the frequency of each tone is given..151.15­.01LIALMC07_0321227638.QXP2/26/0410:47 AMPage 668668 CHAPTER 7 Trigonometric Identities and Equations 88. (b) 4 beats per sec For x = t, P(t) = .005 sin 440 t + .005 sin 432 t.0189. Pressure of a Plucked String If a string with a fundamental frequency of 110 Hz is plucked in the middle, it will vibrate at the odd harmonics of 110, 330, 550, . . . Hz but not at the even harmonics of 220, 440, 660, . . . Hz. The resulting pressure P caused by the string can be modeled by the equation P .003 sin 220 t .003 sin 660 t 3 .003 sin 1100 t 5 .003 sin 1540 t. 7.151.15­.01(Source: Benade, A., Fundamentals of Musical Acoustics, Dover Publications, 1990; Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag, 1975.) (a) Graph P in the window 0, .03 by .005, .005 . (b) Use the graph to describe the shape of the sound wave that is produced. (c) See Exercise 87. At lower frequencies, the inner ear will hear a tone only when the eardrum is moving outward. Determine the times over the interval 0, .03 when this will occur. 90. Hearing Difference Tones Small speakers like those found in older radios and telephones often cannot vibrate slower than 200 Hz--yet 35 keys on a piano have frequencies below 200 Hz. When a musical instrument creates a tone of 110 Hz, it also creates tones at 220, 330, 440, 550, 660, . . . Hz. A small speaker cannot reproduce the 110-Hz vibration but it can reproduce the higher frequencies, which are called the upper harmonics. The low tones can still be heard because the speaker produces difference tones of the upper harmonics. The difference between consecutive frequencies is 110 Hz, and this difference tone will be heard by a listener. We can model this phenomenon using a graphing calculator. (Source: Benade, A., Fundamentals of Musical Acoustics, Dover Publications, 1990.) (a) In the window 0, .03 by pressure 1, 1 , graph the upper harmonics represented by the(c) The number of beats is equal to the absolute value of the difference in the frequencies of the two tones. 89. (a) For x = t, P(t) = .003 sin 220 t +.003 3 .003 5 .003 7sin 660 t + sin 1100 t + sin 1540 t.0050.03­.005(b) The graph is periodic, and the wave has &quot;jagged square&quot; tops and bottoms. (c) This will occur when t is in one of these intervals: .0045, .0091 , .0136, .0182 , .0227, .0273 . 90. (a) For x = t, P(t) =1 sin[2 2 1 sin[2 3 1 sin[2 41 1 1 sin 2 220 t sin 2 330 t sin 2 440 t . 2 3 4 (b) Estimate all t-coordinates where P is maximum. (c) What does a person hear in addition to the frequencies of 220, 330, and 440 Hz? (d) Graph the pressure produced by a speaker that can vibrate at 110 Hz and above. P 91. Daylight Hours in New Orleans The seasonal variation in length of daylight can be modeled by a sine function. For example, the daily number of hours of daylight in New Orleans is given by h 35 3 7 2 x sin , 3 365(220)t] + (330)t] + (440)t]1where x is the number of days after March 21 (disregarding leap year). (Source: Bushaw, Donald et al., A Sourcebook of Applications of School Mathematics. Copyright © 1980 by The Mathematical Association of America.).030(a) On what date will there be about 14 hr of daylight? (b) What date has the least number of hours of daylight? (c) When will there be about 10 hr of daylight? (Modeling) Alternating Electric Current The study of alternating electric current requires the solutions of equations of the form i I max sin 2 ft,­1(b) .0007576, .009847, .01894, .02803 (c) 110 Hzfor time t in seconds, where i is instantaneous current in amperes, I max is maximum current in amperes, and f is the number of cycles per second. (Source: Hannon, R. H., BasicLIALMC07_0321227638.QXP2/26/0410:48 AMPage 6697.6 Trigonometric Equations 66990. (d) For x = t, P(t) = sin[2 (110)t] +1 2 1 3 1 4Technical Mathematics with Calculus, W. B. Saunders Company, 1978.) Find the smallest positive value of t, given the following data. 92. i 94. i 40, I max I max, f 100, f 60 60 93. i 95. i 50, I max 1 I max, f 2 100, f 60 120sin[2 (220)t] + sin[2 (330)t] + sin[2 (440)t]2(Modeling) Solve each problem. 96. Accident Reconstruction The model0.03.342D cosh cos216D 2 V 02­291. (a) 91.3 days after March 21, on June 20 (b) 273.8 days after March 21, on December 19 (c) 228.7 days after March 21, on November 4, and again after 318.8 days, on February 2 92. .001 sec 93. .0007 sec 94. .004 sec 95. .0014 sec 1 96. 14° 97. (a) sec 4 1 (b) sec (c) .21 sec 6 1 98. (a) 2 sec (b) 3 sec 3 99. (a) One such value is (b) One such value is 4 . 3 .is used to reconstruct accidents in which a vehicle vaults into the air after hitting an obstruction. V0 is velocity in feet per second of the vehicle when it hits, D is distance (in feet) from the obstruction to the landing point, and h is the difference in height (in feet) between landing point and takeoff point. Angle is the takeoff angle, the angle between the horizontal and the path of the vehicle. Find to the nearest degree if V0 60, D 80, and h 2. 97. Electromotive Force In an electric circuit, let V cos 2 tmodel the electromotive force in volts at t seconds. Find the smallest positive value of t where 0 t 1 for each value of V. 2 (a) V 0 (b) V .5 (c) V .25 98. Voltage Induced by a Coil of Wire induces a voltage modeled by e 20 sin A coil of wire rotating in a magnetic field t 42,where t is time in seconds. Find the smallest positive time to produce each voltage. (a) 0 (b) 10 3 99. Movement of a Particle A particle moves along a straight line. The distance of the particle from the origin at time t is modeled by st 2 2 3 sin t 2 cos t. 3 2 2Find a value of t that satisfies each equation. (a) s t (b) s t9 100. Explain what is wrong with the following solution for all x over the interval 0, 2 of the equation sin2 x sin x 0.sin2 x sin x sin x 1 sin x x The solution set is . 0 0 1 2Divide by sin x. Add 1.2LIALMC07_0321227638.QXP2/26/0410:48 AMPage 670670 CHAPTER 7 Trigonometric Identities and Equations7.7 Equations Involving Inverse Trigonometric FunctionsSolving for x in Terms of y Using Inverse FunctionssSolving Inverse Trigonometric EquationsUntil now, the equations in this chapter have involved trigonometric functions of angles or real numbers. Now we examine equations involving inverse trigonometric functions.Solving for x in Terms of y Using Inverse FunctionsEXAMPLE 1 Solving an Equation for a Variable Using Inverse NotationSolve y3 cos 2x for x.Solution We want cos 2x alone on one side of the equation so we can solve for 2x, and then for x.y y 3 2x x3 cos 2x cos 2x arccos y 3Divide by 3. Definition of arccosine (Section 7.5) Multiply by 2 . Now try Exercise 7.11 y arccos 2 3Solving Inverse Trigonometric EquationsEXAMPLE 2 Solving an Equation Involving an Inverse Trigonometric FunctionSolve 2 arcsin xSolution.First solve for arcsin x, and then for x. 2 arcsin x arcsin x x x 2 sin 1 2Divide by 2. Definition of arcsine (Section 7.5)(Section 6.2)Verify that the solution satisfies the given equation. The solution set is 1 .Now try Exercise 23.LIALMC07_0321227638.QXP2/26/0410:48 AMPage 6717.7 Equations Involving Inverse Trigonometric Functions 671EXAMPLE 3 Solving an Equation Involving Inverse Trigonometric FunctionsySolve cosSolution1xsin111 . 2 u. Then sin u cos1Let sin1 21 2and for u in quadrant I, the equationbecomes2 u01xxu x. Alternative form1 2. 3 23cos u See Figure 31. Since x cos u, xSketch a triangle and label it using the facts that u is in quadrant I and sin uFigure 31, and the solution set is3 2. Check.Now try Exercise 29.EXAMPLE 4 Solving an Inverse Trigonometric Equation Using an IdentitySolve arcsin xSolutionarccos x6.Isolate one inverse function on one side of the equation. arccos x arcsin x 6 arccos x x u x sin u cos cos u sin by definition.Substitute. (2)arcsin x6Add arccos x.(1)sin arccos x Let u6Definition of arcsinearccos x, so 0 sin u sin u 6 666Sine sum identity (Section 7.3)Substitute this result into equation (2) to get sin u cosy6cos u sin6x.(3)From equation (1) and by the definition of the arcsine function, 21 uarccos x arccos x6 3 .2Subtract 6 . (Section 1.7)1 ­xx2x2 30Figure 32Since 0 arccos x , we must have 0 arccos x 3 . Thus, x 0, and we can sketch the triangle in Figure 32. From this triangle we find that sin u 1 x 2.LIALMC07_0321227638.QXP2/26/0410:48 AMPage 672672 CHAPTER 7 Trigonometric Identities and EquationsNow substitute into equation (3) using sin u 3 cos 6 x. 2 , and cos u sin u cos 1 6 x2 1 cos u sin 3 2 x2 3 3 31 x 3 1 x 6 1 2 x x22 21x 2, sin61 2,x x 2x x x x2 2 2(3)Multiply by 2. Subtract x. Square both sides. (Section 1.6) Distributive property (Section R.1) Add 3x 2.3x3 x x To check, replace x with arcsin3 24x3 4 3 2Solve for x; Choose the positive square root because x 0. Quotient rule (Section R.7)in the original equation: arccos3 23 23 2 3 6 6,as required. The solution set is.Now try Exercise 31.7.7 Exercises1. C 5. x 6. x 7. x 8. x 9. x 2. A 3. C 4. C y arccos 5 arcsin 4y 1 arccot 2y 3 arcsec 12y 1 y arctan 2 3Concept Check A. arctan 1Answer each question. x 2 2 B. arccos 0 x 2 2 C. arcsin 0 x 3 31. Which one of the following equations has solution 0? 2. Which one of the following equations has solution 4 ? A. arcsin x B. arccos x C. arctan x3. Which one of the following equations has solution 34 ? A. arctan 1 x B. arcsin 2 2 x C. arccos6?2 2 1 2x4. Which one of the following equations has solution A. arctan 3 3 x B. arccos 1 2 xC. arcsinxLIALMC07_0321227638.QXP2/26/0410:48 AMPage 6737.7 Equations Involving Inverse Trigonometric Functions 67310. x 11. x 12. x 13. x 14. x 15. x 16. x 17. x 18. x 19. x 20. x 23. 25. 27. 30. 33. 35. 3 5 3 4 1 2 22 arcsiny 3 y 4 arccos 6 3 arcsin y y 1 arccos 5 2 y 1 arccot 5 3 3 arccos y 1 1 arctan y 2 arcsin y 2 arccot y 1 y 4 arcsin 2 arccos 2 3 y 3 24. 0 3 12 5 3 6 29. 32. 4 5 3 2 2 4Solve each equation for x. See Example 1. 5. y 8. 6y 11. y 14. y 17. y 20. y 5 cos x 1 sec x 2 6 cos x 4 2 3 cos x 6. 4y 9. y 12. y 15. y 18. y sin x 3 tan 2x sin cos x cot x x 3 3 1 7. 2y 10. y 13. y 16. y 19. y cot 3x 3 sin x 22 cos 5x tan 2x 2 sin x 1 43 cot 5x sin x 49 21. 9 22.Refer to Exercise 17. A student attempting to solve this equation wrote as the first step y sin x 2 , inserting parentheses as shown. Explain why this is incorrect. Explain why the equation sin required.)1xcos12 cannot have a solution. (No work isSolve each equation for exact solutions. See Examples 2 and 3. 23. 4 cos 3126. 28.y 4 y 3 arctan sin124. 4 2 3 44 tan1y25. 2 arccos 27. arcsin x 29. cos126. arccos y 28. arctan x 30. cot13 arccos tan16 5 1331. 0 34. 0 1 2 36.x3 5x4 35 5Solve each equation for exact solutions. See Example 4. 31. sin62137. 0 38. 0 39. Y = arcsin X ­ arccos X ­2 ­2xtan114 3 2 6 2 arccos x32. sin1xtan132 3 3 2 2 0 333. arccos x 35. arcsin 2x2 arcsin arccos x tan134. arccos x 36. arcsin 2x 38. sin12 arcsin arcsin x tan1­837. cos621xxxx40. Y1 = arcsin X ­ arccos X Y2 =2 ­239. Provide graphical support for the solution in Example 4 by showing that the graph of y arcsin x6has x-intercept3 2.8660254.40. Provide graphical support for the solution in Example 4 by showing that the x-coordinate of the point of intersection of the graphs of Y1 arcsin X arccos X and Y26is3 2.8660254.­841. 4.4622037 42. 2.2824135The following equations cannot be solved by algebraic methods. Use a graphing calculator to find all solutions over the interval 0, 6 . Express solutions to as many decimal places as your calculator displays. 41. arctan x3x2042.sin1.2x3xLIALMC07_0321227638.QXP2/26/0410:48 AMPage 674674 CHAPTER 7 Trigonometric Identities and Equations(Modeling) Solve each problem. 43. (a) A .00506, .484; P .00506 sin 440 t .484 43. Tone Heard by a Listener When two sources located at different positions pro(b) The two graphs are the same. duce the same pure tone, the human ear will often hear one sound that is equal to the For x = t, sum of the individual tones. Since the sources are at different locations, they will P(t) = .00506 sin(440 t + .484) have different phase angles . If two speakers located at different positions produce P1(t) + P2(t) = .0012 sin(440 t + .052) + pure tones P1 A 1 sin 2 ft A 2 sin 2 ft 1 and P2 2 , where 1, 4 .004 sin(440 t + .61) , then the resulting tone heard by a listener can be written as 2.0064P.01A sin 2 ft A, where A 1 cos arctan1A 2 cos1 122A 1 sin2 21A 2 sin220and­.006A 1 sin A 1 cosA 2 sin A 2 cos.(Source: Fletcher, N. and T. Rossing, The Physics of Musical Instruments, Second Edition, Springer-Verlag, 1998.) (a) Calculate A and if A 1 .0012, 1 .052, A 2 .004, and 2 .61. Also find an expression for P A sin 2 ft if f 220. (b) Graph Y1 P and Y2 P1 P2 on the same coordinate axes over the interval 0, .01 . Are the two graphs the same?44. (a) A .0035, .470; P .0035 sin 600 t .47 (b) The two graphs are the same. For x = t, P(t) = .0035 sin(600 t + .47) P1(t) + P2(t) = .0025 sin 600 t + .001 sin 600 t +.006(7(6)) +45. Depth of Field44. Tone Heard by a Listener Repeat Exercise 43, with A 1 A 2 .001, 2 6 , and f 300..0025,17,0.01When a large-view camera is used to take a picture of an object that is not parallel to the film, the lens board should be tilted so that the planes containing the subject, the lens board, and the film intersect in a line. This gives the best &quot;depth of field.&quot; See the figure. (Source: Bushaw, Donald et al., A Sourcebook of Applications of School Mathematics. Copyright © 1980 by The Mathematical Association of America.)Subject Lens y x Film z­.00645. (a) tan x (b) tan (c) (d) x tan arctan arctanx ; tan z y x tan x y x y tan x 2 ux zy46. (a) x (b)ysin u,2(a) Write two equations, one relating , x, and z, and the other relating , x, y, and z. (b) Eliminate z from the equations in part (a) to get one equation relating , , x, and y. (c) Solve the equation from part (b) for . (d) Solve the equation from part (b) for . 46. Programming Language for Inverse Functions In Visual Basic, a widely used programming language for PCs, the only inverse trigonometric function available is arctangent. The other inverse trigonometric functions can be expressed in terms of arctangent as follows. y (a) Let u arcsin x. Solve the equation for x in terms of u. (b) Use the result of part (a) to label the three sides of the triangle in the figure in terms of x. (c) Use the triangle from part (b) to write an equation for tan u in terms of x. (d) Solve the equation from part (c) for u.1 u 0x x1 ­ x2(c) tan ux 1 x2 1 x2u0xLIALMC07_0321227638.QXP2/26/0410:48 AMPage 6757.7 Equations Involving Inverse Trigonometric Functions 675x 1 x2 1 x2 e 1 47. (a) t arcsin 2 f E max (b) .00068 sec 48. (b) (i) approximately .94 or 4.26 (ii) approximately .60 or 6.64 (c) (i) approximately .54 (ii) approximately .61 3 49. (a) t arcsin 3y 4 (b) .27 sec 50. y = sec ­1 x (d) u arctan47. Alternating Electric Current In the study of alternating electric current, instantaneous voltage is modeled by e E max sin 2 ft,where f is the number of cycles per second, E max is the maximum voltage, and t is time in seconds. (a) Solve the equation for t. (b) Find the smallest positive value of t if E max calculator. 48. Viewing Angle of an Observer While visiting a museum, Marsha Langlois views a painting that is 3 ft high and hangs 6 ft above the ground. See the figure. Assume her eyes are 5 ft above the ground, and let x be the distance from the spot where she is standing to the wall displaying the painting. (a) Show that , the viewing angle subtended by the painting, is given by tan112, e5, and f100. Use a3 ft6 ft5 ft­4 0 Radian mode4x ft4 xtan11 . x(b) Find the value of x for each value of . (i) 6 (ii) 8(c) Find the value of for each value of x. (i) x 4 (ii) x 3 49. Movement of an Arm In the exercises for Section 6.3 we found the equation y 4 t 1 , sin 3 3where t is time (in seconds) and y is the angle formed by a rhythmically moving arm. (a) Solve the equation for t. (b) At what time(s) does the arm form an angle of .3 radian? 50. The function y sec 1 x is not found on graphing calculators. However, with some models it can be graphed as y 2 x 0 x 0 2 tan1x21.(This formula appears as Y1 in the screen here.) Use the formula to obtain the graph of y sec 1 x in the window 4, 4 by 0, .LIALMC07_0321227638.QXP2/26/0410:48 AMPage 676676 CHAPTER 7 Trigonometric Identities and EquationsChapter 7 SummaryNEW SYMBOLSsin 1 x (arcsin x) inverse sine of x cos 1 x (arccos x) inverse cosine of x tan 1 x (arctan x) inverse tangent of x cot 1 x (arccot x) inverse cotangent of x sec 1 x (arcsec x) inverse secant of x csc 1 x (arccsc x) inverse cosecant of xQUICK REVIEWCONCEPTS EXAMPLES7.1 Fundamental IdentitiesReciprocal Identities cot 1 tan sec 1 cos csc 1 sinIf is in quadrant IV and sin . sin csc 1 sin 1 1 1 9 25 16 25 sin 13 53 5,find csc , cos , and 5 3Quotient Identities tan sin cos cot cos sin sin2 3 5 sec22cos2 cos2 cos2Pythagorean Identities sin2 cos2 1 1 cot 2 tan2 csc2 116 25 4 5 3 5cos is positive in quadrant IV.Negative-Angle Identities sin csc sin csc cos sec cos sec tan cot tan cotcos sin7.2 Verifying Trigonometric IdentitiesSee the box titled Hints for Verifying Identities on page 613.7.3 Sum and Difference IdentitiesCofunction Identities cos 90° sin 90° tan 90° sin cos cot cot 90° sec 90° csc 90° tan csc secFind a value of such that tan tan cot 90° 90° Find the exact value of cos sin A sin B sin A sin B cos A sin B cos A sin B 2 6 4 cos 15° cos 30° 3 2 2 2 4 15° . 45°cot 78°. cot 78° cot 78° 78° 12°Sum and Difference Identities cos A cos A sin A sin A B B B B cos A cos B cos A cos B sin A cos B sin A cos Bcos 30° cos 45° 1 2 6sin 30° sin 45° 2 2 2 4LIALMC07_0321227638.QXP2/26/0410:48 AMPage 677CHAPTER 7Summary 677CONCEPTSEXAMPLESSum and Difference Identities tan A tan A B B tan A tan B 1 tan A tan B tan A tan B 1 tan A tan BWrite tan4in terms of tan . 4 tan 4 tan 1 tan 4 tan 1 1 tan tantan4tan17.4 Double-Angle Identities and Half-Angle IdentitiesDouble-Angle Identities cos 2A cos 2A cos2 A 2 cos2 A sin2 A 1 cos 2A sin 2A 2 tan A 1 tan2 A 1 2 sin2 A 2 sin A cos AGiven cos5 13and sin0, find sin 2 .Sketch a triangle in quadrant II and use it to find sin : 12 y sin 13 . sin 2 2 sin 2 12 13 cos 5 13 120 16913 12 ­5xtan 2AProduct-to-Sum Identities cos A cos B sin A sin B sin A cos B cos A sin B 1 cos A 2 1 cos A 2 1 sin A 2 1 sin A 2 B B B B cos A cos A sin A sin A B B B BWrite sin sinsin 2 as the difference of two functions. sin 2 1 cos 2 1 cos 2 1 cos 2 1 cos 3 2 3 3 2 cos 1 cos 2 1 cos 2 cos 2Sum-to-Product Identities sin A sin A cos A cos A sin B sin B cos B cos B 2 sin 2 cos 2 cos 2 sin A 2 A 2 A 2 A 2 B B B B cos sin cos sin A 2 A 2 A 2 A 2 B B B BWrite cos coscos 3 as a product of two functions. cos 3 2 cos 2 cos 4 2 3 2 cos cos 2 2 3 22 cos 2 cos 2 cos 2 cosHalf-Angle Identities cos tan A 2 1 cos A 2 sin A 2 1 cos A 2Find the exact value of tan 67.5°. We choose the last form with A tan 67.5° tan 12 2135°. 12 2 2 2A 2 A tan 2A sin A 1 cos A tan 1 cos A 2 1 cos A 1 cos A The sign is chosen based on the sin A quadrant of A. 2135° 22 21 2 2cos 135° sin 135° 2 2 2or21Rationalize the denominator; simplify.LIALMC07_0321227638.QXP2/26/0410:48 AMPage 678678 CHAPTER 7 Trigonometric Identities and EquationsCONCEPTSEXAMPLES7.5 Inverse Circular FunctionsRange Inverse Function y y y y y y sin 1 x cos x tan x cot x sec x csc 1 xy1 1 1 1Evaluate y cos 1 0. Write y cos 1 0 as cos y cos20. Then y2,becauseDomain 1, 1 1, 1 , , , 1 , 1 1, 1,Interval2, 2Quadrants of the Unit Circle I and IV I and II I and IV I and II20 and2is in the range of cos 1 x.0,2, 20, 0,2, 2,y ,yI and II I and IVy0Evaluate sin tan13 4.12(1, 2 )0 x 1(­1, ) y = cos ­1 x (1, 0)­1 0 13 Let u tan 1 3 . Then tan u 4 4 . Since tan tive in quadrant IV, sketch a triangle as shown.yx is nega-­1 ­(0, 2 )x(­1, ­ 2 )2y = sin­1 xy 24 u 5x­3y = tan­1 x­2 ­1 ­0(1, 4 ) x1 2We want sin tan sin u3 5.13 4sin u. From the triangle,(­1, ­ 4 ) 2See page 649 for graphs of the other inverse circular (trigonometric) functions.7.6 Trigonometric EquationsSolving a Trigonometric Equation 1. Decide whether the equation is linear or quadratic in form, so you can determine the solution method. 2. If only one trigonometric function is present, first solve the equation for that function. 3. If more than one trigonometric function is present, rearrange the equation so that one side equals 0. Then try to factor and set each factor equal to 0 to solve. 4. If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval. 5. Try using identities to change the form of the equation. It may be helpful to square both sides of the equation first. If this is done, check for extraneous solutions.Solve tan 3 2 Use a linear method. tan3 over the interval 0°, 360° . 3 tan 2 3 60° 3Another solution over 0°, 360° is 60° 180° 240°.The solution set is 60°, 240° .LIALMC07_0321227638.QXP2/26/0410:48 AMPage 679CHAPTER 7Review Exercises 679Chapter 7 Review Exercises2. A 3. C 4. F 5. D cos2 6. E 7. 1 8. sin 1 cos 1 9. 10. cos2 sin 4 4 ; tan x ; 11. sin x 5 3 4 3 cot x ; 12. cot x 4 5 41 41 csc x ; sec x 5 4 13. E 14. B 15. J 16. A 17. I 18. C 19. H 20. D 4 3 15 ; 21. G 22. B 23. 20 4 15 3 4 3 15 ; ;I 20 4 15 3 4 9 11 12 11 3 ; ; 24. 50 50 4 9 11 1 ; IV 25. 2 12 11 3 26. 6 3 1. BConcept Check For each expression in Column I, choose the expression from Column II that completes an identity. I 1. sec x 3. tan x 5. tan2 x 2. csc x 4. cot x 6. sec2 x 1 A. sin x C. E. sin x cos x 1 cos2 x II B. D. F. 1 cos x 1 cot 2 x cos x sin xUse identities to write each expression in terms of sin 7. sec2 9. tan2 1 tan2 cot 2 8. cot secand cos , and simplify.10. csccot x , given cos x3 511. Use the trigonometric identities to find sin x, tan x, and cot x is in quadrant IV.5 12. Given tan x 4 , where csc x, and sec x. 2andx, use the trigonometric identities to find cot x,Concept Check For each expression in Column I, use an identity to choose an expression from Column II with the same value. I 13. cos 210° 15. tan 35° 14. sin 35° 16. sin 35° A. sin C. 1 35° cos 150° 2 35° sin2 150° cos 15° sin 60° J. cot 125° y , and the quadrant of II B. cos 55° D. 2 sin 150° cos 150° sin 150° sin 60°17. cos 35° 19. sin 75° 21. cos 300°18. cos 75° 20. sin 300° 22. cos 55°E. cos 150° cos 60° F. cot G. cos2 150° I. cos 35° y , tan xH. sin 15° cos 60°For each of the following, find sin x x y. 23. sin x 24. sin x 1 , cos y 4 1 , cos y 10y , cos x4 , x and y in quadrant III 5 4 , x in quadrant I, y in quadrant IV 5Find each of the following. 25. cos 2 , given cos 1 , with 90° 2 1 , with 3 2 y 180°26. sin y, given cos 2yLIALMC07_0321227638.QXP2/26/0410:48 AMPage 680680 CHAPTER 7 Trigonometric Identities and Equations sin 2x sin x x cot cos 2x cos x 2 1 cos 2x tan x sin 2x 2 46. 47. 4 3 3 3 49. 50. 2 4 6 2 3 52. 53. 3 3 4 60° 55. 60° 56. 0° 60.67924514° 41.33444556° 36.4895081° 12.51631252° 73.26220613° 7.673567973° 3 3 1 64. 65. 2 4 67. 4 68. 0 69. 7 427. 28. 45. 48. 51. 54. 57. 58. 59. 60. 61. 62. 63. 66.Graph each expression and use the graph to conjecture an identity. Then verify your conjecture algebraically. 27. sin 2x cos 2x sin x cos x 28. 1 cos 2x sin 2xVerify that each equation is an identity. 29. sin2 x 31. sin2 y cos2 y cos2 x 2 cos A csc 2 2 cos2 tan2 x 2 tan 1 tan tan A csc A cos2 x 30. 2 cos3 x 32. 34. 36. sin 2x sin x 2 tan B sin 2B 2 cot x tan 2x cos x 2 sec x sec2 B csc2 x 2 sec A 1 1 sec 2 sec 2 tan tan2 1 12cos2 x sin2 x sec xsin2 x 2 2 cos x33. 2 cos A 35. 1 37. tan tan2 sin 2sec A 2 tan 238. csc A sin 2A 40. 2 cos2 42. sec2 44. sin3 1 sin 1cos 2A sec A39. 2 tan x csc 2x 41. tan cos2 43. 2 cos3 x10 3 70. 71. 10 2 294 125 6 73. 92 74. 76. 1 , u2 75. 1 u9 72. 7cos2 1 tan2cos xcos2 x sin2 x sec xcos2 sinGive the exact real number value of y. Do not use a calculator. 45. y 48. y 51. y sin13 2 22 2 1 246. y 49. y 52. yarccos cos11 2 2 2 2 3 347. y 50. y 53. ytan13 3 3 177. .463647609, 3.605240263arcsin sec1arctan arccotarccscGive the degree measure of . Do not use a calculator. 54. arccos 1 2 55. arcsin 3 2 56. tan 1 0Use a calculator to give the degree measure of . 57. 60. arctan 1.7804675 cot 4.5046388158. 61.sin1.6604532059. 62.cos 1 .80396577 csc 1 7.4890096arcsec 3.4723155Evaluate the following without using a calculator. 63. cos arccos 66. arcsec sec 69. sin arccos 3 4 1 64. sin arcsin 67. tan13 2 465. arccos cos 68. cos13 4tancos 0170. cos arctan 371. cos csc2LIALMC07_0321227638.QXP2/26/0410:48 AMPage 681CHAPTER 7Review Exercises 68178..7297276562,2,72. sec 2 sin11 373. tan arcsin3 5arccos5 72.411864997 79. 80. , , 3 5 7 , , 4 4 4 3 5 7 9 11 , , , , , 8 8 8 8 8 81. 3 5 , , , 8 8 8Write each of the following as an algebraic (nontrigonometric) expression in u. 74. cos arctan u 1 u2 75. tan arcsec u2 u 14 8Solve each equation for solutions over the interval 0, 2 . 76. sin2 x 79. tan x 1 cot x 77. 2 tan x 80. sec 2x413 15 , 8 81 4078. 3 sin2 x 81. tan2 2x5 sin x 1 0207 9 11 13 15 , , , , 8 8 8 8 8 82. 0 2n , where n is any integer 83. 5 3 84. 6 3 2n , 2n ,Give all solutions for each equation. 82. sec x 2 cos x 2 83. cos 2x cos x 0 84. 4 sin x cos x 32n , where n is any integer 2n , 3 2n ,Solve each equation for solutions over the interval 0°, 360° . If necessary, express solutions to the nearest tenth of a degree. 85. sin2 87. sin 2 89. 3 cos2 3 sin cos 2 2 cos 2 1 1 0 0 86. 2 tan2 88. 2 sin 2 90. 5 cot 2 tan 1 cot 2 0 14 7 2n , 2n , where n 6 3 is any integer 85. {270°} 86. {45°, 153.4°, 225°, 333.4°} 87. {45°, 90°, 225°, 270°} 88. {15°, 75°, 195°, 255°} 89. {70.5°, 180°, 289.5°} 90. {53.5°, 118.4°, 233.5°, 298.4°} 91. x arcsin 2y y 92. x 2 arccos 3 93. x 94. x 95. 0 98. t 1 arctan 2y 3 arcsin 96. 50 3 7 arccos d 5y 4 5 97. 550 450 1 2 3 2 3Solve each equation in Exercises 91 ­ 97 for x. 91. 4y 94. 5y 2 sin x 4 sin x 3 92. y 95. 11 12 50 t for t in terms of d. 3 cos x 2 93. 2y tan 3x 2 2 74 x arctan 3 296. arccos xarcsin97. arccos x 98. Solve darctan 1 550450 cos(Modeling) Solve each problem. 99. Viewing Angle of an Observer A 10-ft-wide chalkboard is situated 5 ft from the left wall of a classroom. See the figure. A student sitting next to the wall x feet from the front of the classroom has a viewing angle of radians. (a) Show that the value of defined by f x arctan 15 x is given by the function 5 . x5 1099. (b) 8.6602567 ft; There may be a discrepancy in the final digits. f (x) = arctan 15 ­ arctan 5 x x1x( )()200arctan­1(b) Graph f x with a graphing calculator to estimate the value of x that maximizes the viewing angle.LIALMC07_0321227638.QXP2/26/0410:48 AMPage 682682 CHAPTER 7 Trigonometric Identities and Equations 100. 48.8° 101. The light beam is completely underwater. 102. (a) 42.2° (b) 90° (c) 48.0° 103. y = csc ­1 x2100. Snell's Law Recall Snell's law from Exercises 102 and 103 of Section 5.3: c1 c2 sin sin1 2,Air2­442 Radian mode­where c 1 is the speed of light in one medium, Water c 2 is the speed of light in a second medium, 1 and 1 and 2 are the angles shown in the figure. Suppose a light is shining up through water into the air as in the figure. As 1 increases, 2 approaches 90°, at which point c1 no light will emerge from the water. Assume the ratio c2 in this case is .752. For what value of 1 does 2 90°? This value of 1 is called the critical angle for water. 101. Snell's Law Refer to Exercise 100. What happens when critical angle?1104. (a), (b)2 ­1 1is greater than the102. British Nautical Mile The British nautical mile is defined as the length of a minute of arc of a meridian. Since Earth is flat at its poles, the nautical mile, in feet, is given by L 6077 31 cos 2 ,A nautical mile is the length on any of the meridians cut by a central angle of measure 1 minute.In both cases, sin 1 .4 .41151685.where is the latitude in degrees. See the figure. (Source: Bushaw, Donald et al., A Sourcebook of Applications of School Mathematics. Copyright © 1980 by The Mathematical Association of America.) (a) Find the latitude between 0° and 90° at which the nautical mile is 6074 ft. (b) At what latitude between 0° and 180° is the nautical mile 6108 ft? (c) In the United States, the nautical mile is defined everywhere as 6080.2 ft. At what latitude between 0° and 90° does this agree with the British nautical mile? 103. The function y csc 1 x is not found on graphing calculators. However, with some models it can be graphed as y x 0 x 0 2 tan1x21.(This formula appears as Y1 in the screen here.) Use the formula to obtain the graph of y csc 1 x in the window 4, 4 by 2, 2 . 104. (a) Use the graph of y sin 1 x to approximate sin 1 .4. (b) Use the inverse sine key of a graphing calculator to approximate sin1.4.Chapter 7 Test1. sin x cos x 6 5 61 ; 61 11. Given tan x 2. Express tan x25 3 6, 2x2 , use trigonometric identities to find sin x and cos x. y , if sin x1 3,61 2. 61sec2 x in terms of sin x and cos x, and simplify. cos y2 5,3. Find sin x y , cos x y , and tan x quadrant III, and y is in quadrant II. 4. Use a half-angle identity to find sinx is in22.5° .LIALMC07_0321227638.QXP2/26/0410:48 AMPage 683CHAPTER 7 Test6833. sin x cos x tan x 4. y y 2y2 42 ; 15 21 4 2 ; 15 2 2 4 21 8 2 422Graph each expression and use the graph to conjecture an identity. Then verify your conjecture algebraically. 5. sec x sin x tan x 6. cot x 2 cot xVerify that each equation is an identity. 7. sec2 B 1 1 sin2 B 8. cos 2A cot A tan A csc A sec A alone.2 5. sec x sin x tan x cos x x 6. cot cot x csc x 2 9. (a) sin (b) sin 10. (a) V 163 cos 1 sec 240 , 2 2 2 t9. Use an identity to write each expression as a trigonometric function of (a) cos 270° (b) sin(b) 163 volts; 11. 1, 1 ;y 2 (0, 0) ­110. Voltage The voltage in common household current is expressed as V 163 sin t, where is the angular speed (in radians per second) of the generator at the electrical plant and t is time (in seconds). (a) Use an identity to express V in terms of cosine. (b) If 120 , what is the maximum voltage? Give the smallest positive value of t when the maximum voltage occurs. 11. Graph y sin 1 x, and indicate the coordinates of three points on the graph. Give the domain and range.(1, 2 )y = sin 1 ­­1x x12. Find the exact value of y for each equation. (a) y arccos tan1(­1, ­ 2 )12. (a) (c) 13. 14. 16. 17. 18. 2 321 2(b) y (d) ysin13 2 2 1 3(c) y(b)0 2 3arcsec3 2 0 (d) 3 5 4 2 (a) (b) 3 9 u 1 u2 15. {90°, 270°} 1 u2 {18.4°, 135°, 198.4°, 315°} 0, 2 4 , 3 313. Find each exact value. (a) cos arcsin (b) sin 2 cos114. Write tan arcsin u as an algebraic (nontrigonometric) expression in u. 15. Solve sin22cos21 for solutions over the interval 0°, 360° .16. Solve csc 2 cot 4 for solutions over the interval 0°, 360° . Express approximate solutions to the nearest tenth of a degree. 17. Solve cos x 18. Solve 2 3 sin cos 2x for solutions over the interval 0, 2 . x 2 3, giving all solutions in radians.4 2 4n , 4n , where 3 3 n is any integer 1 arccos y 19. (a) x 3 4 (b) 5 20. 5 11 17 sec, sec, sec 6 6 619. Solve each equation for x. (a) y cos 3x (b) arcsin x arctan 4 320. (Modeling) Movement of a Runner's Arm A runner's arm swings rhythmically according to the model y 8 cos t 1 3 ,where y represents the angle between the actual position of the upper arm and the downward vertical position and t represents time in seconds. At what times over the interval 0, 3 is the angle y equal to 0?LIALMC07_0321227638.QXP2/26/0410:48 AMPage 684684 CHAPTER 7 Trigonometric Identities and EquationsChapter 7 Quantitative ReasoningHow can we determine the amount of oil in a submerged storage tank?The level of oil in a storage tank buried in the ground can be found in much the same way a dipstick is used to determine the oil level in an automobile crankcase. The person in the figure on the left has lowered a calibrated rod into an oil storage tank. When the rod is removed, the reading on the rod can be used with the dimensions of the storage tank to calculate the amount of oil in the tank. Suppose the ends of the cylindrical storage tank in the figure are circles of radius 3 ft and the cylinder is 20 ft long. Determine the volume of oil in the tank if the rod shows a depth of 2 ft. (Hint: The volume will be 20 times the area of the shaded segment of the circle shown in the figure on the right.)3 ft 2 ft20 9 arctan88165 ft3`

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