How To Find The Difference Between Sin 5 And Cos 5 In Trigonometric Equations

LIALMC07_0768.QXP /6/0 0:7 AM Page 605 7 Trigonometric Identities and Equations In 8 Michael Faraday discovered that when 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 eamples of applications of the trigonometric functions to electricity and other phenomena in the eamples and eercises in this chapter, including a model of the wattage consumption of a toaster in Section 7., Eample 5. 7. Fundamental Identities 7. Verifying Trigonometric Identities 7. Sum and Difference Identities 7. Double-Angle Identities and Half- Angle Identities Summary Eercises on Verifying Trigonometric Identities 7.5 Inverse Circular Functions 7.6 Trigonometric Equations 7.7 Equations Involving Inverse Trigonometric Functions 605

LIALMC07_0768.QXP /6/0 0:7 AM Page 606 606 CHAPTER 7 Trigonometric Identities and Equations 7. Fundamental Identities Negative-Angle Identities Fundamental Identities Using the Fundamental Identities y (, y) y r y O r (, y) sin( ) = y r Figure = sin Negative-Angle Identities As suggested by the circle shown in Figure, an angle having the point, y on its terminal side has a corresponding angle with the point, y on its terminal side. From the definition of sine, sin y r and sin y r, so sin and sin are negatives of each other, or sin sin. (Section 5.) Figure shows an angle in quadrant II, but the same result holds for in any quadrant. Also, by definition, cos r and cos r, (Section 5.) TEACHING TIP Point out that in trigonometric identities, can be an angle in degrees, a real number, or a variable. so cos cos. We can use these identities for sin and cos to find tan in terms of tan : tan sin cos sin sin cos cos tan tan. Similar reasoning gives the following identities. csc csc, sec sec, cot cot This 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 Identities Reciprocal Identities cot tan Quotient Identities tan sin cos sec cos cot cos sin csc sin (continued)

LIALMC07_0768.QXP /6/0 0:7 AM Page 607 7. Fundamental Identities 607 TEACHING TIP Encourage students to 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 sin cos tan sec Negative-Angle Identities sin() sin cos() cos csc() csc sec() sec cot csc tan() tan cot() cot NOTE The most commonly recognized forms of the fundamental identities are given above. Throughout this chapter you must also recognize alternative forms of these identities. For eample, two other forms of sin cos are sin cos and cos sin. 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 Finding Trigonometric Function Values Given One Value and the Quadrant If tan 5 and is in quadrant II, find each function value. (a) sec (b) sin (c) cot TEACHING TIP Warn students that the given information in Eample, tan 5 5, does not mean that sin 5 and cos. Ask them why these values cannot be correct. Solution (a) Look for an identity that relates tangent and secant. tan sec 5 sec 5 9 sec 9 sec 9 sec sec Pythagorean identity tan 5 Combine terms. Take the negative square root. (Section.) Simplify the radical. (Section R.7) We chose the negative square root since sec is negative in quadrant II.

LIALMC07_0768.QXP /6/0 0:7 AM Page 608 608 CHAPTER 7 Trigonometric Identities and Equations (b) (c) tan cot tan sin cot tan cos cos tan sin sin sec tan 5 sin sin 5 cot 5 5 Quotient identity Multiply by cos. Reciprocal identity From part (a), sec ; tan 5 Reciprocal identity Negative-angle identity tan 5 ; Simplify. (Section R.5) Now try Eercises 5, 7, and 9. CAUTION 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. Any trigonometric function of a number or angle can be epressed in terms of any other function. EXAMPLE Epressing One Function in Terms of Another Epress cos in terms of tan. Solution Since sec is related to both cos and tan by identities, start with tan sec. tan sec tan cos tan cos cos tan cos tan tan Take reciprocals. Reciprocal identity Take square roots. Quotient rule (Section R.7); rewrite. Rationalize the denominator. (Section R.7) Choose the sign or the sign, depending on the quadrant of. Now try Eercise.

LIALMC07_0768.QXP /6/0 0:7 AM Page 609 7. Fundamental Identities 609 We can use a graphing calculator to decide whether two functions are identical. See Figure, which supports the identity sin cos. With an identity, you should see no difference in the two graphs. All other trigonometric functions can easily be epressed in terms of sin and/or cos. We often make such substitutions in an epression to simplify it. Y = Y Figure y = tan + cot y = cos sin EXAMPLE Rewriting an Epression in Terms of Sine and Cosine Write tan cot in terms of sin and cos, and then simplify the epression. Solution tan tan cot sin cot cos cos sin sin cos cos sin cos sin sin cos cos sin cos sin Quotient identities Write each fraction with the LCD. (Section R.5) Add fractions. Pythagorean identity Now try Eercise 55. The graph supports the result in Eample. The graphs of y and y appear to be identical. CAUTION When working with trigonometric epressions and identities, be sure to write the argument of the function. For eample, we would not write sin cos ; an argument such as is necessary in this identity. 7. Eercises..6..65..65 7..75 5. 6. 0 0 7. 5 8. 77 5 9. 05 0. 5 8 Concept Check Fill in the blanks.. If tan.6, then tan.. If cos.65, then cos.. If tan.6, then cot.. If cos.8 and sin.6, then tan. Find sin s. See Eample. 9 5. cos s, s in quadrant I 6. cot s, s in quadrant IV 7. coss 5, tan s 0 8. tan s 7, sec s 0 5 9. sec s, tan s 0 0. csc s 8 5. Why is it unnecessary to give the quadrant of s in Eercise 0?

LIALMC07_0768.QXP /6/0 0:7 AM Page 60 60 CHAPTER 7 Trigonometric Identities and Equations. sin. odd. cos 5. even 6. tan 7. odd 8. 9. 0. f f f f f f. cos 5 ; tan 5 ; cot 5 ; 5 sec 5 ; csc 5. sin 6 ; tan 6; 5 cot 6 ; sec 5; csc 56. sin 7 ; 7 cos 7 ; cot ; 7 sec 7 ; csc 7. sin ; 5 cos ; tan ; 5 cot ; sec 5 5. sin ; cos ; 5 5 tan ; sec 5 ; csc 5 Relating Concepts For individual or collaborative investigation (Eercises 7) A function is called an even function if f f for all in the domain of f. A function is called an odd function if f f for all in the domain of f. Work Eercises 7 in order, to see the connection between the negative-angle identities and even and odd functions.. Complete the statement: sin.. Is the function defined by f sin even or odd?. Complete the statement: cos. 5. Is the function defined by f cos even or odd? 6. Complete the statement: tan. 7. Is the function defined by f tan even or odd? Concept Check For each graph, determine whether f f or f f is true. 8. 9. 0. 6. cos ; tan ; 5 cot ; sec 5 ; csc 5 7. sin 7 ; cos ; tan 7 ; cot 7 ; 7 csc 7 7 Find the remaining five trigonometric functions of. See Eample.. sin, in quadrant II. cos, in quadrant I 5. tan, in quadrant IV. csc 5, in quadrant III 5. cot, sin 0 6. sin, cos 0 5 7. sec, sin 0 8. cos, sin 0

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 7. Fundamental Identities 6 8. sin 5 ; tan 5; cot 5 ; 5 sec ; csc 5 5 9. B 0. D. E. C. A. C 5. A 6. E 7. D 8. B.... 5. 6. 7. 8. sin tan p p sin cos cot sin sin tan sec cot csc csc cos cos sec sin sin 9. cos 50. 5. cot 5. csc 5. cos 5. tan 55. sec cos 56. cot 57. cot tan 58. sin cos 59. sin cos 60. cot tan 6. cos 6. tan 9 9 Concept Check For each epression in Column I, choose the epression from Column II that completes an identity. I II cos 9. sin A. sin cos 0. tan B. cot. cos C. sec. tan D.. E. cos Concept Check For each epression in Column I, choose the epression from Column II that completes an identity. You may have to rewrite one or both epressions. I II. tan cos A. sin cos sin cos 5. sec B. sec 6. sec csc C. sin 7. sin D. csc cot sin 8. cos E. tan 9. A student writes cot csc. Comment on this student s work. 0. Another student makes the following claim: Since sin cos, I should be able to also say that sin cos if I take the square root of both sides. Comment on this student s statement.. Concept Check Suppose that cos. Find sin.. Concept Check Find tan if sec p. Write the first trigonometric function in terms of the second trigonometric function. See Eample.. sin ; cos. cot ; sin 5. tan ; sec 6. cot ; csc 7. csc ; cos 8. sec ; sin p Write each epression in terms of sine and cosine, and simplify it. See Eample. 9. cot sin 50. sec cot sin 5. cos csc 5. cot tan 5. sin csc 5. sec sec 55. cos sec 56. cos sin sin 57. 58. sin cos cot 59. sec cos 60. sec csc cos sin 6. sin csc sin 6. cos sin sin tan cot

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 6 CHAPTER 7 Trigonometric Identities and Equations 6. sec 6. sin 65. 56 60 56 60 ; 66. 8 8 ; 9 9 67. sin 68. It is the negative of sin. 69. cos 70. It is the same function. 7. (a) sin (b) cos (c) 5 sin 7. identity 7. not an identity 7. not an identity 75. identity 76. not an identity 6. sin tan cos 6. 65. Concept Check Let cos 5. Find all possible values for sin. sin cos 66. Concept Check Let csc. Find all possible values for. Relating Concepts For individual or collaborative investigation (Eercises 67 7) In Chapter 6 we graphed functions defined by tan sec y c a f b d sec tan with the assumption that b 0. To see what happens when b 0, work Eercises 67 7 in order. 67. Use a negative-angle identity to write y sin as a function of. 68. How does your answer to Eercise 67 relate to y sin? 69. Use a negative-angle identity to write y cos as a function of. 70. How does your answer to Eercise 69 relate to y cos? 7. Use your results from Eercises 67 70 to rewrite the following with a positive value of b. (a) sin (b) cos (c) 5 sin sec Use a graphing calculator to decide whether each equation is an identity. (Hint: In Eercise 76, graph the function of for a few different values of y (in radians).) 7. cos sin 7. 7. sin cos 75. 76. cos y cos cos y sin s sin s cos cos sin 7. Verifying Trigonometric Identities Verifying Identities by Working with One Side Verifying Identities by Working with Both Sides Recall 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 epressions 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_0768.QXP /6/0 0:7 AM Page 6 7. Verifying Trigonometric Identities 6 Looking Ahead to Calculus Trigonometric identities are used in calculus to simplify trigonometric epressions, determine derivatives of trigonometric functions, and change the form of some integrals. TEACHING TIP There is no substitute for eperience when it comes to verifying identities. Guide students through several eamples, giving hints such as Apply a reciprocal identity, or Use a different form of the Pythagorean identity sin cos. Hints for Verifying Identities. 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. For eample, sin is an alternative form of the identity sin cos. Try to rewrite the more complicated side of the equation so that it is identical to the simpler side.. It is sometimes helpful to epress all trigonometric functions in the equation in terms of sine and cosine and then simplify the result.. Usually, any factoring or indicated algebraic operations should be performed. For eample, the epression sin sin can be factored as sin. The sum or difference of two trigonometric epressions, such as sin cos, can be added or subtracted in the same way as any other rational epression. cos. cos sin sin cos sin 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 eample, to verify the identity tan cos, try to think of an identity that relates tan to cos. In this case, since sec cos and sec tan, the secant function is the best link between the two sides. 6. If an epression contains sin, multiplying both numerator and denominator by sin would give sin, which could be replaced with cos. Similar results for sin, cos, and cos 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 Eamples.

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 6 CHAPTER 7 Trigonometric Identities and Equations For s =, cot + = csc (cos + sin ) The graphs coincide, supporting the conclusion in Eample. EXAMPLE Verifying an Identity (Working with One Side) Verify that the following equation is an identity. Solution We use the fundamental identities from Section 7. 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. Steps Right side of given equation cot s csc scos s sin s csc scos s sin s cos s sin s sin s cos s sin s sin s sin s cot s Left side of given equation Reasons csc s sin s Distributive property (Section R.) cos s sin s sin s cot s; sin s The given equation is an identity since the right side equals the left side. Now try Eercise. tan ( + cot ) = sin The screen supports the conclusion in Eample. EXAMPLE Verifying an Identity (Working with One Side) Verify that the following equation is an identity. tan cot Solution We work with the more complicated left side, as suggested in the second hint. Again, we use the fundamental identities from Section 7.. tan cot tan tan cot tan tan tan sec cos sin sin tan Distributive property cot tan tan tan tan sec sec cos cos sin Since the left side is identical to the right side, the given equation is an identity. Now try Eercise 7.

LIALMC07_0768.QXP /6/0 0:7 AM Page 65 7. Verifying Trigonometric Identities 65 EXAMPLE Verifying an Identity (Working with One Side) Verify that the following equation is an identity. Solution tan t cot t sin t cos t We transform the more complicated left side to match the right side. tan t cos t sin t sec t csc t tan t cot t sin t cos t tan t sin t cos t cot t sin t cos t sec t csc t sin t cos t cot t sin t cos t sin t cos t sin t cos t cos t sin t sin t cos t a b c a b a b a c b c (Section R.5) tan t cos sin t cos t t ; cot t sin t The third hint about writing all trigonometric functions in terms of sine and cosine was used in the third line of the solution. cos t sec t; sin t csc t Now try Eercise. TEACHING TIP Show that the identity in Eample can also be verified by multiplying the numerator and denominator of the left side by sin. EXAMPLE Verifying an Identity (Working with One Side) Verify that the following equation is an identity. Solution We work on the right side, using the last hint in the list given earlier to multiply numerator and denominator on the right by sin. sin cos sin sin cos sin sin cos sin cos cos sin cos sin cos sin sin cos Multiply by. (Section R.) y y y sin cos Lowest terms (Section R.5) (Section R.) Now try Eercise 7. Verifying Identities by Working with Both Sides If both sides of an identity appear to be equally comple, 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_0768.QXP /6/0 0:7 AM Page 66 66 CHAPTER 7 Trigonometric Identities and Equations left right common third epression With all steps reversible, the procedure is as shown in the margin. The left side leads to a common third epression, which leads back to the right side. This procedure is just a shortcut for the procedure used in Eamples : 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. Solution Both sides appear equally comple, so we verify the identity by changing each side into a common third epression. We work first on the left, multiplying numerator and denominator by cos. Multiply by. On the right side of the original equation, begin by factoring. We have shown that sec tan sin sin sec tan cos sec tan sec tan cos sec tan sec tan cos sec cos tan cos sec cos tan cos tan cos tan cos sin cos cos sin cos cos sin sin sin sin sin cos cos sin sin sin sin Distributive property y y y (Section R.) Factor sin. Lowest terms Left side of Common third Right side of given equation epression given equation verifying that the given equation is an identity. sin sin sin sec cos tan sin cos cos sin sec tan sin sin sin, sec tan sin cos Now try Eercise 5.

LIALMC07_0768.QXP /6/0 0:7 AM Page 67 7. Verifying Trigonometric Identities 67 CAUTION Use the method of Eample 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 Eample 5 is to work on the left as follows. sin sec tan cos cos sec tan sin cos cos sin cos sin cos sin sin Fundamental identities (Section 7.) Add and subtract fractions. (Section R.5) Simplify the comple fraction. (Section R.5) Compare this with the result shown in Eample 5 for the right side to see that the two sides indeed agree. L C An Inductor and a Capacitor Figure EXAMPLE 6 Applying a Pythagorean Identity to Radios Tuners 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. The energy stored in the inductor at time t is given by and the energy stored in the capacitor is given by where F is the frequency of the radio station and k is a constant. The total energy E in the circuit is given by Show that E is a constant function. (Source: Weidner, R. and R. Sells, Elementary Classical Physics, Vol., Allyn & Bacon, 97.) Solution Lt k sin Ft Ct k cos Ft, Et Lt Ct. Et Lt Ct Given equation k sin Ft k cos Ft Substitute. ksin Ft cos Ft Factor. (Section R.) k sin cos (Here Ft). k Since k is constant, Et is a constant function. Now try Eercise 85.

LIALMC07_0768.QXP /6/0 0:7 AM Page 68 68 CHAPTER 7 Trigonometric Identities and Equations 7. Eercises. csc sec or sin cos. csc sec or sin cos. sec s. cot 5. 6. or sec cos 7. 8. or csc sin 9. sin t 0. sec s. cos or cot csc sin... sin cos sin sin sec sec 5. sin 6. tan cot or 7. sin sin 8. tan tan 9. cos 0. cot cot or csc cot.... 5. sin cos sin cos sin cos sin cos sin cos 6. 7. tan 8. sec 9. tan 0. cot t. sec. csc Perform each indicated operation and simplify the result. sec. cot. csc csc cot sec. tan scot s csc s. cos sec csc 5. 6. sin csc sec sin 7. cos cos 8. sin sec sin csc sin cos 9. sin t cos t 0. tan s tan s. cos cos. sin cos Factor each trigonometric epression.. sin. sec 5. sin sin 6. tan cot tan cot 7. sin sin 8. tan tan 9. cos cos 0. cot cot. sin cos. sin cos Each epression simplifies to a constant, a single function, or a power of a function. Use fundamental identities to simplify each epression.. tan cos. cot sin 5. sec r cos r sin tan csc sec 6. cot t tan t 7. 8. cos cot 9. sec 0. csc t.. cot tan tan sin sin csc cos In Eercises 68, verify that each trigonometric equation is an identity. See Eamples 5. cot tan. cos. sin csc sec sin tan 5. cos 6. sec cos sec 7. cos tan 8. sin cot 9. cot s tan s sec s csc s 0. sin tan cos sec cos sin. sin sec tan. sec cos sec csc cos. sin cos sin. 5. cos cos sin sin 6. tan sin tan cos cos sin cot

LIALMC07_0768.QXP /6/0 0:7 AM Page 69 7. Verifying Trigonometric Identities 69 9 9 cos 7. cos 8. tan sec 9. 50. 5. 5. 5. 58. 59. sin 60. sin cos csc cos cos cos 6. 6. 6. 6. 65. sec sin sin sec tan sec tan tan s cos s sin s cot s sec s csc s cos s cos cot csc cos cot cot tan tan 5. tan tan sec tan sec 55. sin sec sin csc sec csc cot 56. cot csc tan sin 57. sec sec tan tan sin sec sec tan tan sin sin cos sin sec s tan s sec s tan s sec s tan s cot t cot t sin t tan t sec t sin cos sin cos cos cos cos cot csc cos 66. sec tan sin sin 67. sec csc cos sin cot tan sin cos 68. sin cos 69. A student claims that the equation cos sin is an identity, since by letting or radians we get 0, a true statement. Comment on this student s reasoning. 70. 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? Eplain. 90 tan t cot t tan t cot t cos sin cos sin cos sec tan tan sec csc tan csc

LIALMC07_0768.QXP /6/0 0:7 AM Page 60 60 CHAPTER 7 Trigonometric Identities and Equations 7. sec tan sin cos 7. csc cot sec tan cos 7. cot sin tan 7. tan sin cos sec 75. identity 76. identity 77. not an identity 78. not an identity 8. It is true when sin 0. 8. (a) I k sin (b) For for all integers n, cos, its maimum value, and I attains a maimum value of k. 85. (a) P 6k cos t (b) P 6k sin t n Concept Check Graph each epression and conjecture an identity. Then verify your conjecture algebraically. 7. sec tan sin 7. csc cot sec cos 7. 7. tan sin cos sin tan Graph the epressions 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.) If the equation looks like an identity, prove it algebraically. See Eample. 5 cos s 75. 76. cot s sec s csc s 5 cot s sin s sec s tan s cot s 77. 78. sin s sin s sec s tan s cot s sin s By substituting a number for s or t, show that the equation is not an identity. 79. sincsc s 80. cos s cos s 8. csc t cot t 8. cos t sin t 8. When is sin cos a true statement? (Modeling) Work each problem. 8. 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 cos, where k is a constant. (Source: Winter, C., Solar Power Plants, Springer-Verlag, 99.) (a) Write I in terms of the sine function. 9(b) Eplain why the maimum value of I occurs when P. y 85. Oscillating Spring The distance or displacement y of a weight attached to an oscillating spring from its natural position is modeled by y cost, where t is time in seconds. Potential energy is the energy of position and is given by P ky, where 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., Allyn & Bacon, 97.) (a) Write an epression for P that involves the cosine function. (b) Use a fundamental identity to write P in terms of sint. 86. Radio Tuners Refer to Eample 6. Let the energy stored in the inductor be given by 0 Lt cos 6,000,000t and the energy in the capacitor be given by Ct sin 6,000,000t,

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 7. Sum and Difference Identities 6 86. (a) The sum of L and C equals. 0 0 6 where 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, 0 6 by,, with Xscl 0 7 and Yscl. Interpret the graph. (b) Make a table of values for L, C, and E starting at t 0, incrementing by 0 7. Interpret your results. (c) Use a fundamental identity to derive a simplified epression for Et. (b) Let Y Lt, Y Ct, and Y Et. Y for all inputs. (c) Et 7. Sum and Difference Identities Cosine Sum and Difference Identities Cofunction Identities Sine and Tangent Sum and Difference Identities Cosine Sum and Difference Identities Several eamples presented earlier should have convinced you by now that cosa B does not equal cos A cos B. For eample, if A and B 0, then cosa B cos 0 cos while cos A cos B cos cos 0 0. We can now derive a formula for cosa 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. 0, (cos(a B), sin(a B)) R (cos A, sin A) S y Q (cos B, sin B O A A B B P (, 0) Figure Point Q is on the unit circle, so by the work with circular functions in Chapter 6, the -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 cosa B,sinA B.

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 6 CHAPTER 7 Trigonometric Identities and Equations Angle 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, cosa B sina B 0 Squaring both sides and clearing parentheses gives (Section.) Since sin cos for any value of, we can rewrite the equation as Subtract ; divide by. Although Figure shows angles A and B in the second and first quadrants, respectively, this result is the same for any values of these angles. To find a similar epression for cosa B, rewrite A B as A B and use the identity for cosa B. cosa B cosa B cos A cos B sin A sin B. cos A B cosa B sin A B cos A cos A cos B cos B sin A sin A sin B sin B. cosa B cos A cos B sin A sin B cosa B cos A cos B sin A sin B. cos A cosb sin A sinb cos A cos B sin Asin B cosa B cos A cos B sin A sin B Cosine difference identity Negative-angle identities (Section 7.) Cosine of a Sum or Difference cos(a B) cos A cos B sin A sin B cos(a B) cos A cos B sin A sin B These identities are important in calculus and useful in certain applications. Although a calculator can be used to find an approimation for cos 5, for eample, the method shown below can be applied to get an eact value, as well as to practice using the sum and difference identities. EXAMPLE Finding Eact Cosine Function Values Find the eact value of each epression. (a) cos 5 (b) cos 5 (c) cos 87 cos 9 sin 87 sin 9 Solution (a) To find cos 5, we write 5 as the sum or difference of two angles with known function values. Since we know the eact trigonometric function values of 5 and 0, we write 5 as 5 0. (We could also use 60 5. ) Then we use the identity for the cosine of the difference of two angles.

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 7. Sum and Difference Identities 6 TEACHING TIP In Eample (b), students may benefit from converting radians to 75 in order to realize that 5 6 0 5 can be used in place of 5. The screen supports the solution in Eample (b) by showing that cos 5 = 6. (b) (c) cos 5 cos5 0 cos 5 cos 5 cos 0 sin 5 sin 0 cos 6 cos 6 6 cos 6 sin Cosine difference identity (Section 5.) Cosine sum identity (Section 6.) cos 87 cos 9 sin 87 sin 9 cos87 9 Cosine sum identity 6 sin cos 80 6 ; (Section 5.) Now try Eercises 7, 9, and. 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 these identities 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 Identities cos(90 ) sin cot(90 ) tan sin(90 ) cos sec(90 ) csc tan(90 ) cot csc(90 ) sec Similar identities can be obtained for a real number domain by replacing 90 with. Substituting 90 for A and for B in the identity for cosa B gives cos90 cos 90 cos sin 90 sin 0 cos sin sin. This result is true for any value of since the identity for cosa B is true for any values of A and B.

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 6 CHAPTER 7 Trigonometric Identities and Equations TEACHING TIP Verify results from Eample using a calculator. EXAMPLE Using Cofunction Identities to Find Find an angle that satisfies each of the following. (a) cot tan 5 (b) sin cos0 (c) csc sec Solution (a) Since tangent and cotangent are cofunctions, tan90 cot. (b) (c) 90 sec 0 90 sin cos0 cos90 cos0 0 csc sec sec sec sec cot tan 5 tan90 tan 5 5 65 Cofunction identity Cofunction identity Combine terms. Cofunction identity Set angle measures equal. Now try Eercises 5 and 7. NOTE Because trigonometric (circular) functions are periodic, the solutions in Eample are not unique. We give only one of infinitely many possibilities. If one of the angles A or B in the identities for cosa B and cosa B is a quadrantal angle, then the identity allows us to write the epression in terms of a single function of A or B. EXAMPLE Reducing cosa B to a Function of a Single Variable Write cos80 as a trigonometric function of. Solution Use the difference identity. Replace A with 80 and B with. cos80 cos 80 cos sin 80 sin cos 0 sin cos (Section 5.) Now try Eercise 9.

LIALMC07_0768.QXP /6/0 0:7 AM Page 65 7. Sum and Difference Identities 65 Sine and Tangent Sum and Difference Identities We can use the cosine sum and difference identities to derive similar identities for sine and tangent. Since sin cos90, we replace with A B to get sina B cos90 A B Cofunction identity cos90 A B cos90 A cos B sin90 A sin B Cosine difference identity sina B sin A cos B cos A sin B. Cofunction identities Now we write sina B as sina B and use the identity for sina B. sina B sina B sin A cosb cos A sinb sina B sin A cos B cos A sin B Sine sum identity Negative-angle identities Sine of a Sum or Difference sin(a B) sin A cos B cos A sin B sin(a B) sin A cos B cos A sin B To derive the identity for tana B, we start with tana B We epress this result in terms of the tangent function by multiplying both numerator and denominator by tana B tana B sina B cosa B sin A cos B cos A sin B cos A cos B sin A sin B. sin A cos B cos A sin B cos A cos B cos A cos B sin A sin B cos A cos B sin A cos B cos A sin B cos A cos B cos A cos B cos A cos B sin A sin B cos A cos B cos A cos B sin A cos A sin B cos B sin A cos A sin B cos B tan A tan B tan A tan B cos A cos B. Fundamental identity (Section 7.) Sum identities Simplify the comple fraction. (Section R.5) Multiply numerators; multiply denominators. tan sin cos Replacing B with B and using the fact that tanb tan B gives the identity for the tangent of the difference of two angles.

LIALMC07_0768.QXP /6/0 0:7 AM Page 66 66 CHAPTER 7 Trigonometric Identities and Equations Tangent of a Sum or Difference tan(a B) tan A tan B tan A tan B tan(a B) tan A tan B tan A tan B EXAMPLE Finding Eact Sine and Tangent Function Values Find the eact value of each epression. (a) sin 75 (b) tan 7 (c) sin 0 cos 60 cos 0 sin 60 Solution (a) (b) (c) sin 75 sin5 0 7 sin 5 cos 0 cos 5 sin 0 6 tan tan tan tan tan tan Tangent sum identity (Section 6.) Sine sum identity (Section 5.) Rationalize the denominator. (Section R.7) Multiply. (Section R.7) Combine terms. Factor out. (Section R.5) Lowest terms sin 0 cos 60 cos 0 sin 60 sin0 60 Sine difference identity sin0 sin 0 Negative-angle identity (Section 5.) Now try Eercises 9,, and 5.

LIALMC07_0768.QXP /6/0 0:7 AM Page 67 7. Sum and Difference Identities 67 EXAMPLE 5 Writing Functions as Epressions Involving Functions of Write each function as an epression involving functions of. (a) sin0 (b) tan5 (c) sin80 Solution (a) Using the identity for sina B, (b) (c) tan5 sin0 sin 0 cos cos 0 sin sin cos sin. tan 5 tan tan tan 5 tan tan sin80 sin 80 cos cos 80 sin 0 cos sin Now try Eercises and 7. EXAMPLE 6 Finding Function Values and the Quadrant of A B Suppose that A and B are angles in standard position, with sin A and cos B 5 5, A,,. Find each of the following. (a) sina B (b) tana B (c) the quadrant of A B Solution (a) The identity for sina 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. sin A cos A 6 5 cos A cos A 9 5 B Fundamental identity sin A 5 Subtract 6 5. cos A 5 Since A is in quadrant II, cos A 0. In the same way, sin B. Now use the formula for sina B 5 5 5 sina B. 0 65 6 65 6 65 (b) To find tana B, first use the values of sine and cosine from part (a) to get tan A and tan B 5. tana B 6 6 5 5 5 5 8 6 6 6 5 5

LIALMC07_0768.QXP /6/0 0:7 AM Page 68 68 CHAPTER 7 Trigonometric Identities and Equations (c) From parts (a) and (b), sina B 6 65 and tana B 6 6, 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 Eercise 5. EXAMPLE 7 Applying the Cosine Difference Identity to Voltage Common household electric current is called alternating current because the current alternates direction within the wires. The voltage V in a typical 5-volt outlet can be epressed by the function Vt 6 sin t, where is the angular 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, 988.) (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 00, 00. (c) Determine a value of so that the graph of Vt 6 cost is the same as the graph of Vt 6 sin t. Solution (a) Each cycle is radians at 60 cycles per sec, so the angular speed is per sec. (b) Vt 6 sin t 6 sin 0t. Because the amplitude of the function is 6 (from Section 6.), 00, 00 is an appropriate interval for the range, as shown in Figure 5. 0 60 radians 00 For = t, V(t) = 6 sin 0t 0.05 00 Figure 5 (c) Using the negative-angle identity for cosine and a cofunction identity, cos cos cos sin. Therefore, if, then Vt 6 cost 6 sin t. Now try Eercise 8.

LIALMC07_0768.QXP /6/0 0:7 AM Page 69 7. Sum and Difference Identities 69 7. Eercises. F. A. C. D 5. 6 6 6. 7. 6 6 8. 9. 6 6 0.. 0.. cot. cos 75 5. sin 5 6. cos 7. cos 0 8. tan 9. csc56 5 8 0. cot8. tan. cos. cos. tan 5. 5 00 6. 7. 0 80 6 8. 9. 0.. 6 6... 5. 6. 7. 8. Concept Check Match each epression in Column I with the correct epression in Column II to form an identity. I II. cos y A. cos cos y sin sin y. cos y B. sin sin y cos cos y. sin y C. sin cos y cos sin y. sin y D. sin cos y cos sin y E. F. cos sin y sin cos y cos cos y sin sin y Use identities to find each eact value. (Do not use a calculator.) See Eample. 5. cos 75 6. cos5 7. cos 05 8. cos05 (Hint: 05 60 5) (Hint: 05 60 5) 9. cos 0. cos. cos 0 cos 50 sin 0 sin 50. cos 7 9 Write each function value in terms of the cofunction of a complementary angle. See Eample.. tan 87. sin 5 5. cos 6. sin 5 7. sin 5 8. cot 9 9. sec 6 0. tan 7 8 0 cos 9 7 sin sin 9 9 Use the cofunction identities to fill in each blank with the appropriate trigonometric function name. See Eample.. cot. 6. sin 57. Find an angle that makes each statement true. See Eample. 5. tan cot5 6. sin cos 0 7. 8. sin 5 cos 5 Use identities to find the eact value of each of the following. See Eample. 9. sin 5 0. tan 5. tan. sin. sin. tan 7 7 5. sin 76 cos cos 76 sin 6. sin 0 cos 50 cos 0 sin 50 tan 80 tan 55 tan 80 tan55 7. 8. tan 80 tan 55 tan 80 tan55 sin 7 cot 8 cot 0 tan 0 6

LIALMC07_0768.QXP /6/0 0:7 AM Page 60 60 CHAPTER 7 Trigonometric Identities and Equations 9. sin 0. cos. sin. sin. cos sin. sin 5. cos sin 6. cos sin 7. tan tan tan 8. tan tan tan 9. 50. tan tan 6 6 5. (a) (b) (c) 65 65 6 (d) I 66 86 5. (a) (b) 5 5 86 (c) 66 (d) IV 0 5. (a) 9 5 (b) 9 85 5 (c) 0 0 (d) II 6 5. (a) (b) (c) 6 65 65 6 (d) IV 55. (a) 6 (b) (c) 77 85 85 6 (d) II 8 6 56. (a) 77 (b) (c) 85 85 77 (d) III 6 58. 6 59. 6 60. (a) 6 (b) 6 6. 6. 6 6. 6. 65. 66. 6 Use identities to write each epression as a function of or. See Eamples and 5. 9. cos90 0. cos80. cos. cos. sin5. sin80 5. sin 6. sin 7. tan60 5 6 8. 9. tan 50. tan tan 0 Use the given information to find (a) coss t, (b) sins t, (c) tans t, and (d) the quadrant of s t. See Eample 6. 5. cos s and sin t 5, s and t in quadrant I 5 5. cos s and sin t, s and t in quadrant II 5 5 5. sin s and sin t, s in quadrant II and t in quadrant IV 5. sin s and sin t, s in quadrant I and t in quadrant III 5 55. cos s 8 and cos t, s and t in quadrant III 7 5 56. cos s 5 and sin t, s in quadrant II and t in quadrant I 7 5 Relating Concepts For individual or collaborative investigation (Eercises 57 60) The identities for cosa B and cosa B can be used to find eact values of epressions like cos 95 and cos 55, where the angle is not in the first quadrant. Work Eercises 57 60 in order, to see how this is done. 57. By writing 95 as 80 5, use the identity for cosa B to epress cos 95 as cos 5. 58. Use the identity for cosa B to find cos 5. 59. By the results of Eercises 57 and 58, cos 95. 60. Find each eact value using the method shown in Eercises 57 59. (a) cos 55 (b) cos Find each eact value. Use the technique developed in Relating Concepts Eercises 57 60. 6. sin 65 6. tan 65 6. sin 55 6. tan 85 65. tan 66. sin 6

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 7. Sum and Difference Identities 6 7. sin cos tan 7. tan tan 79. 80. 6 and 6; no 8. (a) 5 lb (c) 0 9 9 67. Use the identity cos90 sin, and replace with 90 A, to derive the identity cos A sin90 A. 68. Eplain how the identities for seca B, csca B, and cota B can be found by using the sum identities given in this section. 69. Why is it not possible to use a method similar to that of Eample 5(c) to find a formula for tan70? 70. Concept Check Show that if A, B, and C are angles of a triangle, then sina B C 0. Graph each epression and use the graph to conjecture an identity. Then verify your conjecture algebraically. tan 7. sin 7. tan Verify that each equation is an identity. 7. 7. 75. 76. 77. 78. sin y sin y sin cos y tan y tan y cos tan cot cos sin sins t tan s tan t cos s cos t sin y tan tan y sin y tan tan y sins t sin t Eercises 79 and 80 refer to Eample 7. 79. How many times does the current oscillate in.05 sec? 80. What are the maimum and minimum voltages in this outlet? Is the voltage always equal to 5 volts? (Modeling) Solve each problem. 8. 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 eerted on the back muscles can be modeled by the equation coss t cos t F.6W sin 90, sin where W is the weight of the person. (Source: Metcalf, H., Topics in Classical Biophysics, Prentice-Hall, 980.) (a) Calculate F when W 70 lb and. (b) Use an identity to show that F is approimately equal to.9w cos. (c) For what value of is F maimum? tan tan y tan tan y sin s sin t cos t 0

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 6 CHAPTER 7 Trigonometric Identities and Equations 8. (a) 08 lb (b) 6. 8. (a) The pressure P is oscillating. For = t, cos[ P(t) =. 0 06t ] 0.9.05 0.05.05 (b) The pressure oscillates and amplitude decreases as r increases. For = r, P(r) = r r cos[.9 ] 0,60 0 0 (c) P a cos ct n 8. (a) For = t, V = V + V = 0 sin 0t + 0 cos 0t 60 0.05 8. Back Stress Refer to Eercise 8. (a) Suppose a 00-lb person bends at the waist so that. Estimate the force eerted on the person s back muscles. (b) Approimate graphically the value of that results in the back muscles eerting a force of 00 lb. 8. 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 r cos r ct 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 maimum 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., 988.) Let ft and c 06 ft per sec. (a) Let a. lb per ft. Graph the sound pressure at distance r 0 ft from its source in the window 0,.05 by.05,.05. Describe P at this distance. (b) Now let a and t 0. Graph the sound pressure in the window 0, 0] by,. 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. 8. Voltage of a Circuit When the two voltages V 0 sin 0t and V 0 cos 0 t are 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, 98.) (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 sin0t. (c) Use identities to verify that your epression for V is valid. 5.9 60 (b) a 50; 5.5 7. Double-Angle Identities and Half-Angle Identities Double-Angle Identities Product-to-Sum and Sum-to-Product Identities Half-Angle Identities TEACHING TIP A common error is to write cos A as cos A. Double-Angle Identities When A B in the identities for the sum of two angles, these identities are called the double-angle identities. For eample, to derive an epression for cos A, we let B A in the identity cosa B cos A cos B sin A sin B. cos A cosa A cos A cos A sin A sin A Cosine sum identity (Section 7.) cos A cos A sin A

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 7. Double-Angle Identities and Half-Angle Identities 6 TEACHING TIP Students might find it helpful to see each formula illustrated with a concrete eample that they can check. For instance, you might show that cos 60 cos 0 cos 0 sin 0. Two other useful forms of this identity can be obtained by substituting either cos A sin A or sin A cos A. Replace cos A with the epression sin A to get cos A cos A sin A sin A sin A cos A sin A, or replace sin A with cos A to get cos A cos A sin A cos A cos A cos A cos A cos A cos A. Fundamental identity (Section 7.) Fundamental identity We find sin A with the identity sina B sin A cos B cos A sin B, letting B A. sin A sina A sin A cos A cos A sin A sin A sin A cos A Using the identity for tana B, we find tan A. tan A tana A tan A tan A tan A tan A tan A tan A tan A Sine sum identity Tangent sum identity Looking Ahead to Calculus The identities cos A sin A and cos A cos A can be rewritten as Double-Angle Identities cos A cos A sin A cos A cos A sin A sin A cos A tan A cos A sin A tan A tan A sin A cos A and cos A cos A. These identities are used to integrate the functions fa sin A and ga cos A. EXAMPLE Finding Function Values of Given Information about Given cos 5 and sin 0, find sin, cos, and tan. Solution To find sin, we must first find the value of sin. sin 5 sin 6 5 sin cos ; Simplify. cos 5 sin 5 Choose the negative square root since sin 0.

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 6 CHAPTER 7 Trigonometric Identities and Equations Using the double-angle identity for sine, sin sin cos Now we find cos, using the first of the double-angle identities for cosine. (Any of the three forms may be used.) The value of tan can be found in either of two ways. We can use the doubleangle identity and the fact that 5 5 5. tan tan tan 9 7 9 Simplify. (Section R.5) Alternatively, we can find tan by finding the quotient of sin and cos. tan sin cos 5 sin 5 ; cos 5 cos cos sin 9 5 6 5 7 5 tan sin cos 5. 5 6 8 5 7 7 7 Now try Eercise 9. EXAMPLE Verifying a Double-Angle Identity Verify that the following equation is an identity. cot sin cos Solution We start by working on the left side, using the hint from Section 7. about writing all functions in terms of sine and cosine. cot sin cos sin sin cos sin cos sin cos cos Quotient identity Double-angle identity cos cos, so cos cos The final step illustrates the importance of being able to recognize alternative forms of identities. Now try Eercise 7.

LIALMC07_0768.QXP /6/0 0:7 AM Page 65 7. Double-Angle Identities and Half-Angle Identities 65 EXAMPLE Simplify each epression. (a) cos 7 sin 7 Solution Simplifying Epressions Using Double-Angle Identities (a) This epression suggests one of the double-angle identities for cosine: cos A cos A sin A. Substituting 7 for A gives (b) If this epression were sin 5 cos 5, we could apply the identity for sin A directly since sin A sin A cos A. We can still apply the identity with A 5 by writing the multiplicative identity element as. (b) sin 5 cos 5 cos 7 sin 7 cos 7 cos. sin 5 cos 5 sin 5 cos 5 Multiply by in the form. sin 5 cos 5 Associative property (Section R.) sin 5 sin A cos A sin A, with A 5 sin 0 sin 0 (Section 5.) Now try Eercises and 5. Identities involving larger multiples of the variable can be derived by repeated use of the double-angle identities and other identities. EXAMPLE Deriving a Multiple-Angle Identity Write sin in terms of sin. Solution sin sin sin cos cos sin sin cos cos cos sin sin sin cos cos sin sin sin sin sin sin sin sin sin sin sin sin sin sin Sine sum identity (Section 7.) Double-angle identities Multiply. cos sin Distributive property (Section R.) Combine terms. Now try Eercise.

LIALMC07_0768.QXP /6/0 0:7 AM Page 66 66 CHAPTER 7 Trigonometric Identities and Equations The net eample applies a multiple-angle identity to answer a question about electric current. 000 For = t, (6 sin 0t) W(t) = 5 EXAMPLE 5 Determining Wattage Consumption If 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 where 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, 998.) Graph the wattage W consumed by a typical toaster with R 5 and V 6 sin 0t in the window 0,.05 by 500, 000. How many oscillations are there? Solution Substituting the given values into the wattage equation gives W V R W V R, 6 sin 0t. 5 To determine the range of W, we note that sin 0t has maimum value, so 6 the epression for W has maimum value 5 77. The minimum value is 0. The graph in Figure 6 shows that there are si oscillations. 0.05 500 Figure 6 Now try Eercise 8. Product-to-Sum and Sum-to-Product Identities Because they make it possible to rewrite a product as a sum, the identities for cosa B and cosa B are used to derive a group of identities useful in calculus. Adding the identities for cosa B and cosa B gives cosa B cos A cos B sin A sin B cosa B cos A cos B sin A sin B cosa B cosa B cos A cos B or cos A cos B cosa B cosa B. Similarly, subtracting cosa B from cosa B gives Looking Ahead to Calculus The product-to-sum identities are used in calculus to find integrals of functions that are products of trigonometric functions. One classic calculus tet includes the following eample: Evaluate cos 5 cos d. The first solution line reads: We may write cos 5 cos cos 8 cos. sin A sin B cosa B cosa B. Using the identities for sina B and sina B in the same way, we get two more identities. Those and the previous ones are now summarized. Product-to-Sum Identities cos A cos B [cos(a B) cos(a B)] sin A sin B [cos(a B) cos(a B)] (continued)

LIALMC07_0768.QXP /6/0 0:7 AM Page 67 7. Double-Angle Identities and Half-Angle Identities 67 sin A cos B [sin(a B) sin(a B)] cos A sin B [sin(a B) sin(a B)] EXAMPLE 6 Using a Product-to-Sum Identity Write cos sin as the sum or difference of two functions. A Solution Use the identity for cos A sin B, with and cos sin sin sin B. sin sin Now try Eercise. 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 Identities sin A sin B sin A B cos A B sin A sin B cos A B sin A B cos A cos B cos A B cos A B cos A cos B sin A B sin A B EXAMPLE 7 Using a Sum-to-Product Identity Write sin sin as a product of two functions. Solution Use the identity for sin A sin B, with and sin sin cos sin cos 6 sin cos sin cos sin A B. sin sin (Section 7.) Now try Eercise 5.

LIALMC07_0768.QXP /6/0 0:7 AM Page 68 68 CHAPTER 7 Trigonometric Identities and Equations Half-Angle Identities From the alternative forms of the identity for cos A, we derive three additional identities for sin A, cos A, and tan A. These are known as half-angle identities. To derive the identity for sin A, start with the following double-angle identity for cosine and solve for sin. cos sin sin cos cos sin Add sin ; subtract cos. Divide by ; take square roots. (Section.) sin A cos A Let A, so A ; substitute. The sign in this identity indicates that the appropriate sign is chosen depending on the quadrant of. For eample, if A A is a quadrant III angle, we choose the negative sign since the sine function is negative in quadrant III. We derive the identity for cos A using the double-angle identity cos cos. cos cos cos cos cos cos Add. Rewrite; divide by. Take square roots. cos A cos A Replace with A. An identity for tan A comes from the identities for sin A and cos A. tan A sin A cos A cos A cos A cos A cos A We derive an alternative identity for tan A using double-angle identities. tan A sin A cos A sin A cos A cos A tan A sin A cos A sin A cos A Multiply by cos A in numerator and denominator. Double-angle identities From this identity for tan A, we can also derive tan A cos A sin A.

LIALMC07_0768.QXP /6/0 0:7 AM Page 69 7. Double-Angle Identities and Half-Angle Identities 69 Half-Angle Identities TEACHING TIP Point out that the first identity for tan A follows directly from the cosine and sine half-angle identities; however, the other two identities for tan A are more useful. cos A cos A tan A cos A cos A tan A sin A cos A sin A cos A tan A cos A sin A NOTE The last two identities for tan A 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 terminates. For eample, if an angle A, then A which lies in quadrant II. In quadrant II, cos A and tan A 6, are negative, while sin A is positive. TEACHING TIP Have students compare the value of cos 5 in Eample 8 to the value in Eample (a) of Section 7., where we used the identity for the cosine of the difference of two angles. Although the epressions look completely different, they are equal, as suggested by a calculator approimation for both,.965958. EXAMPLE 8 Using a Half-Angle Identity to Find an Eact Value Find the eact value of cos 5 using the half-angle identity for cosine. Solution cos 5 cos 0 cos 0 Choose the positive square root. Simplify the radicals. (Section R.7) Now try Eercise 5. EXAMPLE 9 Using a Half-Angle Identity to Find an Eact Value Find the eact value of tan.5 using the identity tan Solution Since.5 5, replace A with 5. tan.5 tan 5 sin 5 cos 5 Now multiply numerator and denominator by. Then rationalize the denominator. A sin A cos A. tan.5 Now try Eercise 5.

LIALMC07_0768.QXP /6/0 0:7 AM Page 60 60 CHAPTER 7 Trigonometric Identities and Equations s s 0 EXAMPLE 0 Finding Functions of Given Information about s Given cos s with find cos s sin s and tan s, s,,,. Solution Since and Divide by. (Section.7) s, s s s terminates in quadrant II. See Figure 7. In quadrant II, the values of cos and tan s are negative and the value of sin s s is positive. Now use the appropriate halfangle identities and simplify the radicals. Figure 7 sin s 6 6 6 cos s 5 6 0 6 s tan sin s 6 cos s 6 0 5 5 6 Notice that it is not necessary to use a half-angle identity for tan s once we find sin s and cos s. However, using this identity would provide an ecellent check. Now try Eercise 59. EXAMPLE Simplifying Epressions Using the Half-Angle Identities Simplify each epression. cos (a) (b) cos 5 sin 5 Solution (a) This matches part of the identity for cos A. cos A cos A Replace A with to get cos cos cos 6. (b) Use the third identity for tan A given earlier with A 5 to get cos 5 tan 5 sin 5. Now try Eercises 67 and 7.

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 7. Double-Angle Identities and Half-Angle Identities 6 7. Eercises. cos 5 5 ;. cos ;.. 5. cos ; cos 0 6 ; cos 7 5 ; sin 5 sin 5 5 sin sin 0 sin 6 6 6. cos 9 sin 0 69 ; 69 7. cos sin 5 ; 5 8. cos 8 sin 5 7 ; 7 9. cos 9 9 ; sin 55 9 0. cos 9 5 ; sin 66. 5... 5. 6. 7. tan 0 8. tan 68 9. cos 9. 0. sin 59 6. cos cos cos. sin sin cos sin cos. tan tan tan tan. cos 8 cos 8 cos 5. cos sin cos tan cos tan 6. sin tan Use identities to find values of the sine and cosine functions for each angle measure. See Eample.., given cos and terminates in quadrant I 5., given cos and terminates in quadrant III 5., given sin and cos 0 5 6., given cos and sin 0 7., given tan and cos 0 8., given tan 5 and sin 0 9., given sin 5 and cos 0 7 0., given cos and sin 0 5 Use an identity to write each epression as a single trigonometric function value or as a single number. See Eample. tan 5. cos 5 sin 5.. sin 5 tan 5. 5. cos 6. cos 8 67 sin tan 5 tan 7. 8. 9. sin 7. tan 5 tan 0.., given cos 5 and., given cos and sin 9.5 cos 9.5 8 Epress each function as a trigonometric function of. See Eample.. cos. sin. tan. cos Graph each epression and use the graph to conjecture an identity. Then verify your conjecture algebraically. 5. cos sin 6. Verify that each equation is an identity. See Eample. tan cos tan tan 7. 8. sec sec sec sin cos sin sec sec

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 6 CHAPTER 7 Trigonometric Identities and Equations.... 5. 6. 7. 8. 9. 50. 5. 5. 5. 5. 55. 56. 0 59. 60. 6. 6. 6. 6. sin 60 sin sin 5 sin 55 5 cos 5 5 cos cos cos 9 sin sin cos 6.5 cos.5 sin.5 cos 6.5 cos 98.5 sin.5 cos 6 cos cos 6 sin 50 06 0 50 05 0 50 50 0 65. 7 66. 5 5 9 cos 9. sin sin cos cos 0. cot sin cos. cot tan. sin sin cos 8 sin cos sin.. cos tan sin cos sin sin cos tan 5. tan cot csc 6. 7. 8. cot cos sec cos sin sin 9. sin tan sin 0. sin cos tan sin Write each epression as a sum or difference of trigonometric functions. See Eample 6.. sin 58 cos 0. cos 85 sin 0. 5 cos cos. sin sin 5 Write each epression as a product of trigonometric functions. See Eample 7. 5. cos cos 6. cos 5 cos 8 7. sin 5 sin8 8. sin 0 sin 95 9. cos cos 8 50. sin 9 sin Use a half-angle identity to find each eact value. See Eamples 8 and 9. 5. sin 67.5 5. sin 95 5. cos 95 5. tan 95 55. cos 65 56. sin 65 57. Eplain how you could use an identity of this section to find the eact value of sin 7.5. (Hint: 7.5 0.) 58. The identity tan A cos cos A A can be used to find tan.5, and the identity tan A sin cos A A can be used to find tan.5. Show that these answers are the same, without using a calculator. (Hint: If a 0 and b 0 and a b, then a b. ) Find each of the following. See Eample 0. 59. cos given cos with 0,, 60. sin given cos 5 with, 8, 6. tan given sin with 90, 5, 6. cos given sin with 80, 5, 6. sin given tan, with 0, 6. cos given cot, with, 65. tan given tan 7 with 80,, 66. cot given tan 5 with 90,, 80 cot tan cos cot tan 70 70 80

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 7. Double-Angle Identities and Half-Angle Identities 6 67. sin 0 68. cos 8 69. tan 7.5 70. cot 8.5 7. tan 9.87 7. tan 79. 7. (a) tan is undefined, so it cannot be used. (b) tan sin cos 75. 8 76. 06 77..9 78. 79. (a) cos R b R (b) tan (c) 5 b 50 80. (a) D v sin (b) approimately 5 ft Use an identity to write each epression with a single trigonometric function. See Eample. cos 0 cos 76 cos 7 67. 68. 69. cos 7 cos 65 cos 59.7 sin 58. 70. 7. 7. cos 65 sin 59.7 cos 58. 7. Use the identity tan A sin cos A A to derive the equivalent identity tan A sin cos A A by multiplying both the numerator and denominator by cos A. 7. Consider the epression. (a) Why can t we use the identity for tana B to epress it as a function of alone? (b) Use the identity tan sin cos to rewrite the epression in terms of sine and cosine. (c) Use the result of part (b) to show that cot. (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 verte of the cone, then tan sin m, where m is the Mach number for the speed of the plane. (We assume m.) The Mach number is the ratio of the speed of the plane and the speed of sound. Thus, a speed of Mach. means that the plane is flying at. times the speed of sound. In Eercises 75 78, one of the values or m is given. Find the other value. tan 0 75. m 76. m 5 77. 78. 60 (Modeling) Solve each problem. See Eample 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 00 ft. See the figure. (Source: Hay, W. W., Railroad Engineering, John Wiley & Sons, 98.) (a) Use the figure to write an epression for cos. (b) Use the result of part (a) and the third half-angle identity for tangent to write an epression for tan (c) If b, 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 h by the formula what is the measure of angle to the nearest degree? D v D sin cos v cos v sin 6h. See the figure. (Source: Kreighbaum, E. and K. Barthels, Biomechanics, Allyn & Bacon, 996.) 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 6 ft per sec (about 5 mph) and an angle. How far will the stone travel? 0. θ b 50 50 θ θ R

LIALMC07_0768.QXP /6/0 0:7 AM Page 6 6 CHAPTER 7 Trigonometric Identities and Equations 8. a 885.6, c 885.6, 0 8. (a) For = t, W = VI = (6 sin 0t)(. sin 0t) 00 0.05 50 (b) maimum: 00.9 watts; minimum: 0 watts (c) a 00.5,, c 00.5 (e) 00.5 watts 0 8. Wattage Consumption Refer to Eample 5. Use an identity to determine values of a, c, and so that W a cost c. Check your answer by graphing both epressions for W on the same coordinate aes. 8. 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: Wilco, G. and C. Hesselberth, Electricity for Engineering Technology, Allyn & Bacon, 970.) (a) A household circuit has voltage V 6 sin0t when an incandescent lightbulb is turned on with amperage I. sin0t. Graph the wattage W VI consumed by the lightbulb in the window 0,.05 by 50, 00. (b) Determine the maimum and minimum wattages used by the lightbulb. (c) Use identities to determine values for a, c, and so that W a cost c. (d) Check your answer by graphing both epressions for W on the same coordinate aes. (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? Summary Eercises on Verifying Trigonometric Identities These summary eercises provide practice with the various types of trigonometric identities presented in this chapter. Verify that each equation is an identity.. tan cot sec csc.. tan csc cot. csc cos sin csc sec sec sin t sin t 5. 6. cos t cos t sin t cos t sec t tan t 7. sin tan 8. cos tan tan 9. cot 0. sec t tan cos cot t csc t sin cos sec t sin y cot cot y.. tan cos cos y cot cot y cos sin tan.. csc cot cos tan cos cos 5. 6. cos sec cos tan tan sec tan t sin s 7. 8. cos s tan 9. tan 0. tan cos s csc s tan t csc t tan t sin s tan sec sec cot s tan s cos s sin s.. tan cot cos cos s sin s sin s cos s tan cot

LIALMC07_0768.QXP /6/0 0:7 AM Page 65 7.5 Inverse Circular Functions 65. tan y tan y tan y tan y tan. cos tan tan sin 5. cos sin csc t tan 6. sec t tan t cos csc t 7. sin sin sin 8. cot tan cos cot 7.5 Inverse Circular Functions Inverse Functions Inverse Sine Function Inverse Cosine Function Inverse Tangent Function Remaining Inverse Circular Functions Inverse Function Values Inverse Functions We first discussed inverse functions in Section.. We give a quick review here for a pair of inverse functions f and f.. If a function f is one-to-one, then f has an inverse function f.. In a one-to-one function, each -value corresponds to only one y-value and each y-value corresponds to only one -value.. The domain of f is the range of f, and the range of f is the domain of f.. The graphs of f and f are reflections of each other about the line y. 5. To find f from f, follow these steps. Step Replace f with y and interchange and y. Step Solve for y. Step Replace y with f. In the remainder of this section, we use these facts to develop the inverse circular (trigonometric) functions. Looking Ahead to Calculus The 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 does not define a one-to-one function. If we restrict the domain to the interval,, 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 is,, so the domain of the inverse function will be,, and its range will be,. TEACHING TIP Mention that the interval contains enough of, 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. (0, 0) (, ) y = sin y Restricted domain [, ] Figure 8 (, )

LIALMC07_0768.QXP /6/0 0:7 AM Page 66 66 CHAPTER 7 Trigonometric Identities and Equations (, 0 6) (, (0, 0) ) 6 (, ) (, ) y = sin or y = arcsin (, ) y Figure 9 (, ) (, ) (, ) Reflecting the graph of y sin on the restricted domain across the line y 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 is found by interchanging and y to get sin y. This equation is solved for y by writing y sin (read inverse sine of ). As Figure 9 shows, the domain of y sin is,, while the restricted domain of y sin,,, is the range of y sin. An alternative notation for sin is arcsin. Inverse Sine Function y sin or y arcsin means that sin y, for y We can think of y sin or y arcsin as y is the number in the interval, whose sine is. Just as we evaluated y log by writing it in eponential form as y (Section.), we can write y sin as sin y to evaluate it. We must pay close attention to the domain and range intervals.. EXAMPLE Find y in each equation. Finding Inverse Sine Values (a) y arcsin (b) y sin (c) y sin Algebraic Solution (a) The graph of the function defined by y arcsin (Figure 9) includes the point, 6. Thus, Alternatively, we can think of y arcsin as y is the number in, whose sine is. Then we can write the given equation as sin y. Since sin 6 and 6 is in the range of the arcsine function, y arcsin 6. (b) Writing the equation y sin in the form sin y shows that y This can be verified by noticing that the point, graph of y sin. is on the (c) Because is not in the domain of the inverse sine function, sin does not eist. 6.. Graphing Calculator Solution To find these values with a graphing calculator, we graph Y sin X and locate the points with X-values and. Figure 0(a) shows that when X, Y 6.559878. Similarly, Figure 0(b) shows that when X, Y.570796. (a) Figure 0 Since sin does not eist, a calculator will give an error message for this input. (b) Now try Eercises and. CAUTION In Eample (b), it is tempting to give the value of sin as, since sin. Notice, however, that 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.

LIALMC07_0768.QXP /6/0 0:7 AM Page 67 7.5 Inverse Circular Functions 67 TEACHING TIP In problems like Eample (a), use phrases such as y is a value in radians between and whose sine is equal to to help students understand the meaning of inverse sine. Illustrate the answer graphically by graphing y sin and y in the interval then,, determine the point of intersection. Our observations about the inverse sine function from Figure 9 lead to the following generalizations. y 0 0 INVERSE SINE FUNCTION y sin or y arcsin Domain:, y 0 y = sin Range:, y = sin Figure The inverse sine function is increasing and continuous on its domain,. Its -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 (or y arccos ) is defined by restricting the domain of the function y cos to the interval 0, as in Figure, and then interchanging the roles of and y. The graph of y cos is shown in Figure. Again, some key points are shown on the graph. y y (0, ) (, 0) 0 (, ) y = cos Restricted domain [0, ] (, ) (, 5 6 ) (, ) (, ) ( 0, ) 0 (, ) (, ) (, 6) (, 0) y = cos or y = arccos Figure Figure Inverse Cosine Function y cos or y arccos means that cos y, for 0 y.

LIALMC07_0768.QXP /6/0 0:7 AM Page 68 68 CHAPTER 7 Trigonometric Identities and Equations 5.5 5.5 These screens support the results of Eample, since.5695. EXAMPLE Find y in each equation. (a) y arccos Finding Inverse Cosine Values Solution (a) Since the point, 0 lies on the graph of y arccos in Figure on the previous page, the value of y is 0. Alternatively, we can think of y arccos as y is the number in 0, whose cosine is, or cos y. Then y 0, since cos 0 and 0 is in the range of the arccosine function. (b) We must find the value of y that satisfies cos y, where y is in the interval 0,, the range of the function y cos. The only value for y that satisfies these conditions is. Again, this can be verified from the graph in Figure. (b) y cos Now try Eercises 5 and. Our observations about the inverse cosine function lead to the following generalizations. INVERSE COSINE FUNCTION y cos or y arccos Domain:, Range: 0, 0 y 0 0 y y = cos y = cos 0 Figure The inverse cosine function is decreasing and continuous on its domain,. Its -intercept is, and its y-intercept is. Its graph is not symmetric with respect to the y-ais or the origin. Inverse Tangent Function Restricting the domain of the function y tan to the open interval, yields a one-to-one function. By interchanging the roles of and y, we obtain the inverse tangent function given by y tan or y arctan. Figure 5 shows the graph of the restricted tangent function. Figure 6 gives the graph of y tan.

LIALMC07_0768.QXP /6/0 0:7 AM Page 69 7.5 Inverse Circular Functions 69 y (0, 0) 0 (, ) y = tan (, ) Restricted domain (, ) y (, ) (0, 0) (, ( 6), ) (, ) 6 (, (, ) ) y = tan or y = arctan Figure 5 Figure 6 Inverse Tangent Function y tan or y arctan means that tan y, for y. 0 INVERSE TANGENT FUNCTION y tan or y arctan Domain:, Range:, y 6 0 0 6 y = tan 0 y y = tan 0 Figure 7 The inverse tangent function is increasing and continuous on its domain,. Its -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 and y 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 8 Remaining Inverse Circular Functions The remaining three inverse trigonometric functions are defined similarly; their graphs are shown in Figure 8. All si inverse trigonometric functions with their domains and ranges are given in the table on the net page.

LIALMC07_0768.QXP /6/0 0:7 AM Page 650 650 CHAPTER 7 Trigonometric Identities and Equations Range Inverse Quadrants of the Function Domain Interval Unit Circle y sin, y cos, 0, y tan, y cot, 0, y sec,, y csc,, 0,, y,,, I and IV I and II I and IV I and II * I and II, y 0* I and IV 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 Finding Inverse Function Values (Degree-Measured Angles) Find the degree measure of in the following. (a) arctan Solution (a) Here must be in 90,90, but since 0, must be in quadrant I. The alternative statement, tan, leads to (b) Write the equation as sec. For sec, is in quadrant I or II. Because is positive, is in quadrant I and since sec 60. Note that 60 the degree equivalent of is in the range of the inverse secant function. (b) 60, sec 5. Now try Eercises and 9. 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 eample, on a calculator, in degrees, sin.5 0, sin.5 0, tan 5, and cos.5 0. Finding cot, sec, and csc with a calculator is not as straightforward, because these functions must be epressed in terms of tan, cos, and sin, respectively. If y sec, for eample, then sec y, which must be written as a cosine function as follows: If sec y, then or cos y and y cos cos y,. *The inverse secant and inverse cosecant functions are sometimes defined with different ranges. We use intervals that match their reciprocal functions (ecept for one missing point).

LIALMC07_0768.QXP /6/0 0:7 AM Page 65 7.5 Inverse Circular Functions 65 In summary, to find we find cos sec,. Similar statements apply to csc and cot. There is one additional consideration with cot. 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,,, which is not the correct range for inverse cotangent. For inverse cotangent, the proper range must be considered and the results adjusted accordingly. EXAMPLE Finding Inverse Function Values with a Calculator (a) Find y in radians if y csc. (b) Find in degrees if arccot.5. Figure 9 Solution (a) With the calculator in radian mode, enter csc as sin to get y.9869095. See Figure 9. (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 enter arccot.5 as tan.5 80. As shown in Figure 9, 09.9905. Now try Eercises and 9. 0 y = tan 5 Figure 0 y 0 A A = cos 5 ( ) Figure EXAMPLE 5 Finding Function Values Using Definitions of the Trigonometric Functions Evaluate each epression without using a calculator. (a) sintan Solution (a) Let so tan. The inverse tangent function yields values only in quadrants I and IV, and since is positive, is in quadrant I. Sketch in quadrant I, and label a triangle, as shown in Figure 0. By the Pythagorean theorem, the hypotenuse is. The value of sine is the quotient of the side opposite and the hypotenuse, so tan, (b) tancos 5 sintan sin. (Section 5.) (b) Let A Then, cos A 5 cos. 5. Since cos for a negative value of is in quadrant II, sketch A in quadrant II, as shown in Figure. tancos 5 tan A 5 Now try Eercises 6 and 65.

LIALMC07_0768.QXP /6/0 0:7 AM Page 65 65 CHAPTER 7 Trigonometric Identities and Equations TEACHING TIP Point out that the equations sinsin, coscos, and tantan are true wherever they are defined. However, sin sin, cos cos, and tan tan are true only for values of in the restricted domains of the sine, cosine, and tangent functions. 0 0 y y y A B Figure EXAMPLE 6 Finding Function Values Using Identities Evaluate each epression without using a calculator. (a) cosarctan arcsin (b) tan arcsin 5 Solution (a) Let A arctan and B arcsin, so tan A and sin B. Sketch both A and B in quadrant I, as shown in Figure. Now, use the cosine sum identity. From Figure, cosarctan cos A, sinarctan sin A, Substitute these values into equation () to get cosarctan arcsin (b) Let arcsin 5 B. Then, from the double-angle tangent identity, tan arcsin 5 tan B tan B tan B. cosa B cos A cos B sin A sin B (Section 7.) (Section 7.) cosarctan arcsin cosarctan cosarcsin sinarctan sinarcsin () cosarcsin cos B sinarcsin sin B.. 6, 0 B 5 Figure Since arcsin 5 B, sin B 5. Sketch a triangle in quadrant I, find the length of the third side, and then find tan B. From the triangle in Figure, tan B and, tan arcsin 5 7. Now try Eercises 69 and 75. While the work shown in Eamples 5 and 6 does not rely on a calculator, we can support our algebraic work with one. By entering cosarctan arcsin from Eample 6(a) into a calculator, we get the approimation.87986, the same approimation as when we enter 6 (the eact value we obtained algebraically). Similarly, we obtain the same approimation when we evaluate tan arcsin 5 and, supporting our answer in Eample 6(b). 7

LIALMC07_0768.QXP /6/0 0:7 AM Page 65 7.5 Inverse Circular Functions 65 EXAMPLE 7 Writing Function Values in Terms of u Write each trigonometric epression as an algebraic epression in u. (a) sintan u (b) cos sin u y Solution 0 u + u + u, u > 0 u, u < 0 (a) Let so tan u. Here, u may be positive or negative. Since tan u, sketch in quadrants I and IV and label two triangles, as shown in Figure. Since sine is given by the quotient of the side opposite and the hypotenuse, tan u, sintan u sin u u uu u. Figure The result is positive when u is positive and negative when u is negative. (b) Let so sin u. To find cos, use the identity cos sin. sin u, cos sin u cos sin u Now try Eercises 8 and 85. EXAMPLE 8 Finding the Optimal Angle of Elevation of a Shot Put The 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 5. One model for that achieves this greatest distance is arcsin v v 6h. (Source: Townend, M. Stewart, Mathematics in Sport, Chichester, Ellis Horwood Limited, 98.) h D Figure 5 Suppose a shot-putter can consistently throw the steel ball with h 6.6 ft and v ft per sec. At what angle should he throw the ball to maimize distance? Solution To find this angle, substitute and use a calculator in degree mode. arcsin 66.6.9 h 6.6, v Now try Eercise 9.

LIALMC07_0768.QXP /6/0 0:7 AM Page 65 65 CHAPTER 7 Trigonometric Identities and Equations 7.5 Eercises. one-to-one. range. domain., ; y tan or y arctan 5. 6. Sketch the reflection of the graph of f across the line y. 7. (a), (b), (c) increasing (d) is not in the domain. 8. (a), (b) 0, (c) decreasing (d) is not in the range. 9. (a), (b), (c) increasing (d) no 0. (a),, ;,0 0, (b),, ; 0,, (c), ; 0,. cos. Find a tan (or 80 ). a. 0. 5. 6. 7. 8. 9. 0 0..... 5. 6 6. 7. 8. 9. 6 0. 6. 6.. 5. 0 5. 60 6. 5 7. 0 5 5 Concept Check Complete each statement.. For a function to have an inverse, it must be.. The domain of y arcsin equals the of y sin.. The range of y cos equals the of y cos.. The point, lies on the graph of y tan. Therefore, the point lies on the graph of. 5. If a function f has an inverse and f, then f. 6. How can the graph of be sketched if the graph of f is known? f Concept Check In Eercises 7 0, write short answers. 7. Consider the inverse sine function, defined by y sin or y arcsin. (a) What is its domain? (b) What is its range? (c) Is this function increasing or decreasing? (d) Why is arcsin not defined? 8. Consider the inverse cosine function, defined by y cos or y arccos. (a) What is its domain? (b) What is its range? (c) Is this function increasing or decreasing? (d) Arccos. Why is arccos not equal to? 9. Consider the inverse tangent function, defined by y tan or y arctan. (a) What is its domain? (b) What is its range? (c) Is this function increasing or decreasing? (d) Is there any real number for which arctan is not defined? If so, what is it (or what are they)? 0. Give the domain and range of the three other inverse trigonometric functions, as defined in this section. (a) inverse cosecant function (b) inverse secant function (c) inverse cotangent function. Concept Check Is sec a calculated as cos a or as cos? a. Concept Check For positive values of a, cot a is calculated as tan a. How is cot a calculated for negative values of a? Find the eact value of each real number y. Do not use a calculator. See Eamples and.. y sin 0. y tan 5. y cos 6. y arctan 7. y sin 8. y cos 9. y arctan 0 0. y arcsin. y arccos 0. y tan. y sin. 5. y arccos 6. y arcsin 7. y cot 8. y sec 9. y csc 0. y arccot. y arcsec. y csc y cos

LIALMC07_0768.QXP /6/0 0:7 AM Page 655 7.5 Inverse Circular Functions 655 8. 0 9. 0 0. 90. 7.6785. 97.6707..500970. 5.699 5. 0.98796 6. 9.5068 7..8798 8..960698 9..575 50..605 5..9008 5..080 5. 5. y 55. y 56. 57. y y = cot 0 0 y = sec y 0 y = arcsec 58..00 is not in the domain of y sin. 59. The domain of y tan is,. 60. In both cases, the result is. In each case, the graph is a straight line bisecting quadrants I and III (i.e., the line y ). 6. It is the graph of y. 0 0 0 0 6. It does not agree because the range of the inverse tangent function is, not,,, as was the case in Eercise 6. 0 0 0 y y = csc y = arccsc Give the degree measure of. Do not use a calculator. See Eample. arctan.. 5. 6. 7. 8. 9. 0. arcsin sec Use a calculator to give each value in decimal degrees. See Eample. sin.9.... 5. 6. csc.98 Use a calculator to give each real number value. (Be sure the calculator is in radian mode.) See Eample. 7. y arctan. 8. y arcsin.8969 9. y cot.9708 50. y sec.8768 5. y arcsin.98778 5. y arccos.659 Graph each inverse function as defined in the tet. 5. y cot 5. y csc 55. y sec 56. y arccsc 57. y arcsec arccos.987659 arccos cot csc 58. Eplain why attempting to find sin.00 on your calculator will result in an error message. 59. Eplain why you are able to find tan.00 on your calculator. Why is this situation different from the one described in Eercise 58? Relating Concepts cos.886 arcsin.7790006 cot.76709 For individual or collaborative investigation (Eercises 60 6)* arcsin csc 60. Consider the function defined by f and its inverse f. Simplify f f and f f. What do you notice in each case? What would the graph look like in each case? 6. Use a graphing calculator to graph y tantan in the standard viewing window, using radian mode. How does this compare to the graph you described in Eercise 60? 6. Use a graphing calculator to graph y tan tan in the standard viewing window, using radian and dot modes. Why does this graph not agree with the graph you found in Eercise 6? 0 0 0 *The authors wish to thank Carol Walker of Hinds Community College for making a suggestion on which these eercises are based.

LIALMC07_0768.QXP /6/0 0:7 AM Page 656 656 CHAPTER 7 Trigonometric Identities and Equations 6. 7 5 5 6. 65. 5 66. 56 0 7 67. 68. 69 8 69. 7 70. 5 5 7 6 7. 5 7. 5 7. 7. 6 75. 76. 6 65 65 77. 78. 0 0 0 8 5 9 79..8979 80..9685866 8..998 8..768606 8. u 8. u u 85. u 86. u u 87. u u uu 9 u 88. 89. u 9 90. u 5 u 9. u u 9. 9 u 9 u 9. (a) 5 (b) 9. (a) (b) 8 (c) 60 (d) 7 5 Give the eact value of each epression without using a calculator. See Eamples 5 and 6. 6. tanarccos 6. sinarccos 65. costan 66. 67. sin tan 68. cos sin 69. cos arctan 70. tan cos 7. sin cos 5 7. cos tan 7. secsec 7. csccsc 5 75. 76. cossin 5 5 costan tan cos 77. sinsin 78. tancos 5 sin tan secsin 5 Use a calculator to find each value. Give answers as real numbers. 79. costan.5 80. sincos.5 8. tanarcsin.50 8. cotarccos.5868 Write each epression as an algebraic (nontrigonometric) epression in u, u 0. See Eample 7. 8. sinarccos u 8. tanarccos u 85. cosarcsin u 86. cotarcsin u 87. sinsec 88. costan u 89. tansin 90. seccos u u 5 u 9. secarccot 9. cscarctan u u u 5 9 u u u (Modeling) Solve each problem. 9. Angle of Elevation of a Shot Put Refer to Eample 8. (a) What is the optimal angle when h 0? (b) Fi 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. 9. 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 arcsin m where 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. (b) m.5 (c) m (d) m.5,

LIALMC07_0768.QXP /6/0 0:7 AM Page 657 7.5 Inverse Circular Functions 657 95. (a) 8 (b) 8 (c) 5 (e).5 m (Note: Due to the computational routine, there may be a discrepancy in the last few decimal places.).5 y = tan ( ) + 0 0 Radian mode (f) 97. about.7% 95. Observation of a Painting A painting m high and m from the floor will cut off an angle to an observer, where tan Assume that the observer is meters from the wall where the painting is displayed and that the eyes of the observer are m above the ground. (See the figure.) Find the value of (a) (b) (c). for the following values of. Round to the nearest degree. (d) Derive the formula given above. (Hint: Use the identity for. Use right triangles.) (e) Graph the function for with a graphing calculator, and determine the distance that maimizes the angle. (f) The idea in part (e) was first investigated in 7 by the astronomer Regiomontanus. (Source: Maor, E., Trigonometric Delights, Princeton University Press, 998.) 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 is ab meters. Use this result to find the eact answer to part (e). 96. Landscaping Formula A shrub is planted in a 00-ft-wide space between buildings measuring 75 ft and 50 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 is the distance of the shrub from the taller building, then the value of (in radians) is given by. 00 arctan 50 arctan 75 θ α tan 50 ft 75 ft 00 ft 97. Communications Satellite Coverage The figure shows a stationary communications satellite positioned 0,000 mi above the equator. What percent of the equator can be seen from the satellite? The diameter of Earth is 797 mi at the equator.

LIALMC07_0768.QXP /6/0 0:7 AM Page 658 658 CHAPTER 7 Trigonometric Identities and Equations 7.6 Trigonometric Equations Solving by Linear Methods Solving by Factoring Solving by Quadratic Methods Solving by Using Trigonometric Identities Equations with Half-Angles Equations with Multiple Angles Applications Looking Ahead to Calculus There are many instances in calculus where it is necessary to solve trigonometric equations. Eamples 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. y EXAMPLE Solving a Trigonometric Equation by Linear Methods ' = 0 0 = 0 Solve sin 0 over the interval 0, 60. Solution Because sin is the first power of a trigonometric function, we use the same method as we would to solve the linear equation 0. (a) sin 0 sin Subtract. (Section.) y sin Divide by. = 0 0 ' = 0 (b) Figure 6 To find values of that satisfy sin, we observe that must be in either quadrant III or IV since the sine function is negative only in these two quadrants. Furthermore, the reference angle must be 0 since sin 0. The graphs in Figure 6 show the two possible values of, 0 and 0. The solution set is 0, 0. Alternatively, we could determine the solutions by referring to Figure in Section 6. on page 56. Solving by Factoring Now try Eercise. EXAMPLE Solving a Trigonometric Equation by Factoring Solve sin tan sin over the interval 0, 60. Solution sin tan sin sin tan sin 0 sin tan 0 Subtract sin. Factor. (Section R.) sin 0 or tan 0 Zero-factor property (Section.) tan 0 or 80 5 or 5 The solution set is 0,5, 80, 5. Now try Eercise.

LIALMC07_0768.QXP /6/0 0:7 AM Page 659 7.6 Trigonometric Equations 659 CAUTION There are four solutions in Eample. Trying to solve the equation by dividing each side by sin would lead to just tan, which would give 5 or 5. The other two solutions would not appear. The missing solutions are the ones that make the divisor, sin, equal 0. For this reason, we avoid dividing by a variable epression. Solving by Quadratic Methods In Section.6, we saw that an equation in the form au bu c 0, where u is an algebraic epression, is solved by quadratic methods. The epression u may also be a trigonometric function, as in the equation tan tan 0. EXAMPLE Solving a Trigonometric Equation by Factoring Solve tan tan 0 over the interval 0,. Solution This equation is quadratic in form and can be solved by factoring. tan tan 0 tan tan 0 Factor. tan 0 or tan 0 Zero-factor property tan or tan The solutions for tan over the interval 0, are and 5. To solve tan over that interval, we use a scientific calculator set in radian mode. We find that tan.0787. This is a quadrant IV number, based on the range of the inverse tangent function. (Refer to Figure in Section 6. on page 56.) However, since we want solutions over the interval 0,, we must first add to.0787, and then add..0787.09.0787 5.76066 The solutions over the required interval form the solution set, 5,.0, 5.. Eact values Approimate values to the nearest tenth Now try Eercise. EXAMPLE Solving a Trigonometric Equation Using the Quadratic Formula Find all solutions of cot cot. Solution We multiply the factors on the left and subtract to get the equation in standard quadratic form. cot cot 0 (Section.) Since this equation cannot be solved by factoring, we use the quadratic formula, with a, b, c, and cot as the variable.

LIALMC07_0768.QXP /6/0 0:7 AM Page 660 660 CHAPTER 7 Trigonometric Identities and Equations TEACHING TIP Unlike the cosine function, equations of the form y tan will not contain a second value for between Remind students that other. solutions to y tan are found using period radians. and cot 9 Quadratic formula with a, b, c (Section.) Use a calculator. We cannot find inverse cotangent values directly on a calculator, so we use the fact that cot and take reciprocals to get tan, cot.077568 or cot.0775677 tan.0775677 or tan.077568.90008 or.7679505. To 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.90008 n and.7679505 n, where n is any integer.* Now try Eercise. Solving by Using Trigonometric Identities Recall that squaring both sides of an equation, such as, will yield all solutions but may also give etraneous values. (In this equation, 0 is a solution, while is etraneous. Verify this.) The same situation may occur when trigonometric equations are solved in this manner. TEACHING TIP Point out in Eample 5 that our first goal is to rewrite the equation in terms of a single trigonometric function. TEACHING TIP Eplain that whenever possible, answers should be given in eact form, such as rather than as decimal 6, approimations. EXAMPLE 5 Solving a Trigonometric Equation by Squaring Solve tan sec over the interval 0,. Solution Since the tangent and secant functions are related by the identity tan sec, square both sides and epress sec in terms of tan. tan tan sec tan tan tan y y y (Section R.) Pythagorean identity (Section 7.) Subtract tan. Divide by ; rationalize the denominator. (Section R.7) The possible solutions are and Now check them. Try first. Left side: Right side: tan sec tan tan 5 6 6. tan tan 5 6 sec sec 5 6 5 6 (Section R.7) Not equal *We usually give solutions of equations as solution sets, ecept when we ask for all solutions of a trigonometric equation.

LIALMC07_0768.QXP /6/0 0:7 AM Page 66 7.6 Trigonometric Equations 66 0 y = tan + sec Dot mode; radian mode The graph shows that on the interval [0, ), the only -intercept of the graph of y = tan + sec is 5.7595865, which is an approimation for, the solution found in Eample 6 5. The check shows that Left side: Right side: 5 This solution satisfies the equation, so 6 is not a solution. Now check sec 6 6 is the solution set. Equal Now try Eercise. The methods for solving trigonometric equations illustrated in the eamples can be summarized as follows. Solving a Trigonometric Equation. Decide whether the equation is linear or quadratic in form, so you can determine the solution method.. If only one trigonometric function is present, first solve the equation for that function.. 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.. 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 etraneous solutions. 6. tan 6 Some trigonometric equations involve functions of half-angles or multiple angles. Equations with Half-Angles EXAMPLE 6 Solving an Equation Using a Half-Angle Identity Solve sin (a) over the interval 0,, and (b) give all solutions. Solution (a) Write the interval 0, as the inequality 0. The corresponding interval for is 0. Divide by. (Section.7)

LIALMC07_0768.QXP /6/0 0:7 AM Page 66 66 CHAPTER 7 Trigonometric Identities and Equations 0 y = sin The -intercepts are the solutions found in Eample 6. Using Xscl = makes it possible to support the eact solutions by counting the tick marks from 0 on the graph. TEACHING TIP As a slight variation of the problem in Eample 6, replace with u, solve sin u for u, and then multiply the solutions by to find. To find all values of over the interval 0, that satisfy the given equation, first solve for sin. Divide by. The two numbers over the interval 0, with sine value are and so Multiply by. The solution set over the given interval is. (b) Since this is a sine function with period, all solutions are given by the epressions n 6 and sin sin or 5 n, 5 6 or 5., 5 where n is any integer. 6 5 6, Now try Eercise 6. Equations with Multiple Angles EXAMPLE 7 Solving an Equation with a Double Angle Solve cos cos over the interval 0,. Solution First change cos to a trigonometric function of. Use the identity cos cos so the equation involves only cos. Then factor. cos cos cos cos cos cos 0 cos cos 0 cos 0 or cos 0 Substitute; double-angle identity (Section 7.) Subtract cos. Factor. Zero-factor property cos or cos Over the required interval, The solution set is 0,,. or or 0. Now try Eercise 65.

LIALMC07_0768.QXP /6/0 0:7 AM Page 66 7.6 Trigonometric Equations 66 CAUTION In the solution of Eample 7, cos cannot be changed to cos by dividing by since is not a factor of cos. cos cos The only way to change cos to a trigonometric function of is by using one of the identities for cos. TEACHING TIP Students should be aware that y sin a may have as many as a solutions from 0 to 60. EXAMPLE 8 Solving an Equation Using a Multiple-Angle Identity Solve sin cos over the interval 0, 60. Solution The identity sin cos sin is useful here. sin cos sin cos sin sin From the given interval 0 List all solutions over this interval. 60, sin cos sin (Section 7.) Divide by. 60, 0, 0, 80 0, 60, 0, 0 the interval for is 0 70. or Divide by. The final two solutions for were found by adding 60 to 60 and 0, respectively, giving the solution set 0,60, 0, 0. Now try Eercise 8. Applications Music is closely related to mathematics. y =.00 sin(00).006 0.0.006 Figure 7 EXAMPLE 9 Describing a Musical Tone from a Graph A basic component of music is a pure tone. The graph in Figure 7 models the sinusoidal pressure y P in pounds per square foot from a pure tone at time 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. Approimate the period T in seconds of the pure tone. (c) An equation for the graph is y.00 sin00. Use a calculator to estimate all solutions to the equation that make y.00 over the interval 0,.0.

LIALMC07_0768.QXP /6/0 0:7 AM Page 66 66 CHAPTER 7 Trigonometric Identities and Equations Y =.00.007 0.0.007 Y =.00 sin(00x) Figure 8 Solution (a) From the graph in Figure 7 on the previous page, we see that there are 6 6 cycles in.0 sec. This is equivalent to.0 50 cycles per sec. The pure tone has a frequency of f 50 Hz. (b) Si periods cover a time of.0 sec. One period would be equal to T.0 6 50 or.006 sec. (c) If we reproduce the graph in Figure 7 on a calculator as Y and also graph a second function as Y.00, we can determine that the approimate values of at the points of intersection of the graphs over the interval 0,.0 are.007,.008, and.05. The first value is shown in Figure 8. Now try Eercise 87. A piano string can vibrate at more than one frequency when it is struck. It produces a comple wave that can mathematically be modeled by a sum of several pure tones. If a piano key with a frequency of f is played, then the corresponding string will not only vibrate at f but it will also vibrate at the higher frequencies of f, f, f,..., nf,... f 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 comple tone. (Source: Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag, 975.) P = P + P + P + P + P 5.005 0.0.005 Figure 9 EXAMPLE 0 Analyzing Pressures of Upper Harmonics Suppose that the A key above middle C is played. Its fundamental frequency is f 0 Hz, and its associated pressure is epressed as The string will also vibrate at The corresponding pressures of these upper harmonics are The graph of P.00 sin880t. f 880, f 0, f 760, f 5 00,... Hz. P.00 sin760t, P.00 sin60t, P.00 sin50t, and P 5.00 5 sin00t. P P P P P P 5, shown in Figure 9, is saw-toothed. (a) What is the maimum value of P? (b) At what values of does this maimum occur over the interval 0,.0?

LIALMC07_0768.QXP /6/0 0:7 AM Page 665 7.6 Trigonometric Equations 665 P = P + P + P + P + P 5.005 0.0.005 Figure 0 7.6 Eercises Solution (a) A graphing calculator shows that the maimum value of P is approimately.007. See Figure 0. (b) The maimum occurs at.00088,.006,.007,.0070, and.0098. Figure 0 shows how the second value is found; the others are found similarly. Now try Eercise 89.. Solve the linear equation for cot.. Solve the linear equation for sin.. Solve the quadratic equation for sec by factoring.. Solve the quadratic equation for cos by the zero-factor property. 5. Solve the quadratic equation for sin using the quadratic formula. 6. Solve the quadratic equation for tan using the quadratic formula. 7. Use an identity to rewrite as an equation with one trigonometric function. 8. Use an identity to rewrite as an equation with one trigonometric function. 9. 0. 5, 80,,, 5, 70., 7.. 6 6, 5. 5. 0 6. 0, 5 7.,, 5, 5 8. 9. 0...,, 7 6, 6 7 6,, 6 0,, 6,.. 5. 6. 7. 8. 0, 80 9. 0, 0, 0, 00 0,5, 5 90, 0, 0 60, 5, 0, 5 5, 5, 5, 5 0. 90, 70.... 5. 6. 78.0, 8.0,, 6 5, 5 0,0, 50, 80 0,90, 80, 70 0,5, 5, 80, 5, 5 5, 5, 5, 5 5.6, 6., 87.9, 5. Concept Check Refer to the summary bo on solving a trigonometric equation. Decide on the appropriate technique to begin the solution of each equation. Do not solve the equation.. cot. sin. 5 sec 6 sec. cos cos 5. 9 sin 5 sin 6. tan tan 0 7. tan cot 0 8. cos sin Concept Check Answer each question. 9. Suppose you are solving a trigonometric equation for solutions over the interval 0,, and your work leads to,,. What are the corresponding values of? 0. Suppose you are solving a trigonometric equation for solutions over the interval 0, 60, and your work leads to 5, 60, 75, 90. What are the corresponding values of? Solve each equation for eact solutions over the interval 0,. See Eamples.. cot. sin. sin. sec sec 5. tan 0 6. sec 7. cot cot 0 8. csc csc 0 9. cos cos 0 0. cos cos 0. sin sin. cos cos Solve each equation for eact solutions over the interval 0, 60. See Eamples 5.. cot sin 0. tan cos 0 5. sin csc 6. tan cot 7. tan cot 0 8. cos sin 9. csc cot 0 0. sin cos cos. tan sin tan 0. sin cos 0. sec tan tan. cos sin 0 5. 9 sin 6 sin 6. cos cos 8

LIALMC07_0768.QXP /6/0 0:7 AM Page 666 666 CHAPTER 7 Trigonometric Identities and Equations 7. 8. 9.6, 9.6, 06., 86. 8., 8., 0.8, 8.8 9. 0 0. 68.5, 9.5. 57.7, 59..., 5.7..9 n,. n,.6 n, 5.8 n, where n is any integer., n 5, where n is any integer n 5. where n n, n, is any integer 6. n, where n is any integer n, 7..6 60 n, 6. 60 n, where n is any integer 8. 90 60 n,.8 60 n, 8. 60 n, where n is any integer 9. 5 80 n, 08. 80 n, where n is any integer 50. 5 60 n, 5 60 n, 7.6 60 n, 5.6 60 n, where n is any integer 5..6806,.59 5. 0,.760 55.,,, 56.,,, 5 57., 7 6, 6 58. 59. 60. 6. 6. n, 0,,,,, 5 8, 7 8, 8, 9 8, 5 8 8, 8, 7 8, 8, 9 8, 5 8 8, 8, 5 8, 8, 8, 5,, 7 6., 5 n, n, 9 9 7. tan tan 0 8. cot cot 0 9. sin sin 0 0. cos cos 0. cot csc. sin cos Determine all solutions of each equation in radians (for ) or degrees (for ) to the nearest tenth as appropriate. See Eample.. sin sin 0. cos cos 5. cos 0 6. cos 5 cos 0 7. 5 sec 6 sec 8. sin sin tan 9. 50. sec tan The following equations cannot be solved by algebraic methods. Use a graphing calculator to find all solutions over the interval 0,. Epress solutions to four decimal places. 5. 5. cos sin cos 0 5. Eplain what is wrong with the following solution. Solve tan over the interval 0,. or The solutions are and. 5. The equation cot csc 0 has no solution over the interval 0,. Using this information, what can be said about the graph of over this interval? Confirm your answer by actually graphing the function over the interval. Solve each equation for eact solutions over the interval 0,. See Eamples 6 8. 55. cos 56. cos 57. sin 58. sin 0 59. tan 60. cot 6. cos 6. sin 6. sin sin 6. tan 0 65. sin sin 66. cos cos 0 67. 8 68. sin 69. sin cos sec 0 70. sec cos 5 tan tan tan 5 y cot csc tan

LIALMC07_0768.QXP /6/0 0:7 AM Page 667 7.6 Trigonometric Equations 667 6. 0,,,,, 5,, 7 Solve each equation in Eercises 7 78 for eact solutions over the interval 0, 60. In Eercises 79 86, give all eact solutions. See Eamples 6 8. 65. 0,,, 5 7. sin 0 7. cos 7. cos 66. 0, 67. 0 68. 0, 69. 70. 0 7. 7. 5,5, 5, 65, 55, 85 75, 05, 55, 85 7. 0 7. 80 75. 0, 0 76. 00 77. 78. 79. 0, 50, 70 60,90, 70, 00 0 60 n, 0 60 n, 50 60 n, 80 60 n, where n is any integer 80. 5 60 n, 90 60 n, 5 60 n, 70 60 n, where n is any integer 8. 60 60 n, 00 60 n, where n is any integer 8. 70.5 60 n, 89.5 60 n, where n is any integer 8..8 60 n, 78. 60 n, 9.8 60 n, 58. 60 n, where n is any integer 8..5 60 n,.5 60 n, 0.5 60 n, 9.5 60 n, where n is any integer 85. 0 60 n, 90 60 n, 50 60 n, 0 60 n, 70 60 n, 0 60 n, where n is any integer 86. 0 60 n, 60 60 n, 80 60 n, 00 60 n, where n is any integer 87. (a).006 and.0055 (b).006,.0055 (c) outward 88. (a) beats per sec For = t, P(t) =.005 sin 0t +.005 sin 6t.0.5.5.0 7. sin 75. sin 76. cos 77. sin cos 78. cos cos 79. sin cos 80. sin cos 8. csc 8. cos sin sec 8. sin sin 8. cos 8 sin cos 85. cos cos 86. sin sin 0 (Modeling) Solve each problem. See Eamples 9 and 0. 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 sinft, 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, 975.) A graph of the tone middle C is shown in the figure. (a) Determine algebraically the values of t for For = t, which P 0 over 0,.005. P(t) =.00 sin[(6.6)t + 7 ] (b) From the graph and your answer in part (a),.005 determine the interval for which P 0 over 0,.005. (c) Would an eardrum hearing this tone be 0.005 vibrating outward or inward when P 0? 88. 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, 99.) (a) Consider two tones with frequencies of 0 and Hz and pressures P.005 sin 0t and P.005 sin 6t, respectively. Graph the pressure P P P felt by an eardrum over the -sec interval.5,.5. How many beats are there in sec? (b) Repeat part (a) with frequencies of 0 and 6 Hz. (c) Determine a simple way to find the number of beats per second if the frequency of each tone is given..005

LIALMC07_0768.QXP /6/0 0:7 AM Page 668 668 CHAPTER 7 Trigonometric Identities and Equations 88. (b) beats per sec For = t, P(t) =.005 sin 0t +.005 sin t.0.5.5 (c) The number of beats is equal to the absolute value of the difference in the frequencies of the two tones. 89. (a) (b) The graph is periodic, and the wave has jagged square tops and bottoms. (c) This will occur when t is in one of these intervals:.005,.009,.06,.08,.07,.07. 90. (a) For = t, P(t) = sin[(0)t] + sin[(0)t] + sin[(0)t] 0.0.0 For = t, P(t) =.00 sin 0t +.005 0.0.005.00 sin 660t +.00 sin 00t + 5.00 sin 50t 7 (b).0007576,.00987,.089,.080 (c) 0 Hz 89. Pressure of a Plucked String If a string with a fundamental frequency of 0 Hz is plucked in the middle, it will vibrate at the odd harmonics of 0, 0, 550,... Hz but not at the even harmonics of 0, 0, 660,... Hz. The resulting pressure P caused by the string can be modeled by the equation P.00 sin 0t.00 sin 660t.00 sin 00t 5 (Source: Benade, A., Fundamentals of Musical Acoustics, Dover Publications, 990; Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag, 975.) (a) Graph P in the window 0,.0 by.005,.005. (b) Use the graph to describe the shape of the sound wave that is produced. (c) See Eercise 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,.0 when this will occur. 90. Hearing Difference Tones Small speakers like those found in older radios and telephones often cannot vibrate slower than 00 Hz yet 5 keys on a piano have frequencies below 00 Hz. When a musical instrument creates a tone of 0 Hz, it also creates tones at 0, 0, 0, 550, 660,... Hz. A small speaker cannot reproduce the 0-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 0 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, 990.) (a) In the window 0,.0 by,, graph the upper harmonics represented by the pressure P sin0t sin0t sin0t. (b) Estimate all t-coordinates where P is maimum. (c) What does a person hear in addition to the frequencies of 0, 0, and 0 Hz? (d) Graph the pressure produced by a speaker that can vibrate at 0 Hz and above. 9. Daylight Hours in New Orleans The seasonal variation in length of daylight can be modeled by a sine function. For eample, the daily number of hours of daylight in New Orleans is given by h 5 7 sin 65, where is the number of days after March (disregarding leap year). (Source: Bushaw, Donald et al., A Sourcebook of Applications of School Mathematics. Copyright 980 by The Mathematical Association of America.) (a) On what date will there be about hr of daylight? (b) What date has the least number of hours of daylight? (c) When will there be about 0 hr of daylight? (Modeling) Alternating Electric Current The study of alternating electric current requires the solutions of equations of the form i I ma sin ft,.00 sin 50t. 7 for time t in seconds, where i is instantaneous current in amperes, is maimum current in amperes, and f is the number of cycles per second. (Source: Hannon, R. H., Basic I ma

LIALMC07_0768.QXP /6/0 0:8 AM Page 669 7.6 Trigonometric Equations 669 90. (d) For = t, P(t) = sin[(0)t] + sin[(0)t] + sin[(0)t] + sin[(0)t] Technical Mathematics with Calculus, W. B. Saunders Company, 978.) Find the smallest positive value of t, given the following data. 9. i 0, I ma 00, f 60 9. 9. i I ma, f 60 95. (Modeling) Solve each problem. 96. Accident Reconstruction The model i 50, I ma 00, f 0 i f 60 I ma, 0.0 9. (a) 9. days after March, on June 0 (b) 7.8 days after March, on December 9 (c) 8.7 days after March, on November, and again after 8.8 days, on February 9..00 sec 9..0007 sec 9..00 sec 95..00 sec 96. 97. (a) sec (b) sec 6 (c). sec 98. (a) sec (b) sec 99. (a) One such value is (b) One such value is.. 9 is used to reconstruct accidents in which a vehicle vaults into the air after hitting an obstruction. V 0 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 V 0 60, D 80, and h. 97. Electromotive Force In an electric circuit, let model the electromotive force in volts at t seconds. Find the smallest positive value of t where 0 t for each value of V. (a) V 0 (b) V.5 (c) V.5 98. Voltage Induced by a Coil of Wire A coil of wire rotating in a magnetic field induces a voltage modeled by where t is time in seconds. Find the smallest positive time to produce each voltage. (a) 0 (b) 0 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 Find a value of t that satisfies each equation. (a) st 00. Eplain what is wrong with the following solution for all over the interval 0, of the equation sin sin 0. The solution set is..d cos h cos V cos t e 0 sint st sin t cos t. (b) sin sin 0 sin 0 sin 6D, st V 0 Divide by sin. Add.

LIALMC07_0768.QXP /6/0 0:8 AM Page 670 670 CHAPTER 7 Trigonometric Identities and Equations 7.7 Equations Involving Inverse Trigonometric Functions Solving for in Terms of y Using Inverse Functions Solving Inverse Trigonometric Equations Until now, the equations in this chapter have involved trigonometric functions of angles or real numbers. Now we eamine equations involving inverse trigonometric functions. Solving for in Terms of y Using Inverse Functions EXAMPLE Solve y cos for. Solving an Equation for a Variable Using Inverse Notation Solution We want cos alone on one side of the equation so we can solve for, and then for. y y cos cos Divide by. arccos y Definition of arccosine (Section 7.5) arccos y Multiply by. Now try Eercise 7. Solving Inverse Trigonometric Equations EXAMPLE Solving an Equation Involving an Inverse Trigonometric Function Solve arcsin. Solution First solve for arcsin, and then for. arcsin arcsin sin Divide by. Definition of arcsine (Section 7.5) (Section 6.) Verify that the solution satisfies the given equation. The solution set is. Now try Eercise.

LIALMC07_0768.QXP /6/0 0:8 AM Page 67 7.7 Equations Involving Inverse Trigonometric Functions 67 EXAMPLE Solving an Equation Involving Inverse Trigonometric Functions y Solve cos sin. 0 u Figure Solution Let sin u. Then sin u and for u in quadrant I, the equation becomes cos u Alternative form Sketch a triangle and label it using the facts that u is in quadrant I and sin u See Figure. Since cos u,, and the solution set is. Check. cos u.. Now try Eercise 9. 0 y u Figure EXAMPLE Solution Solving an Inverse Trigonometric Equation Using an Identity Isolate one inverse function on one side of the equation. arcsin arccos sinarccos Let u arccos, so 0 u by definition. sinu sinu u cos cos u sin 6sin 6 6 Sine sum identity (Section 7.) Substitute this result into equation () to get From equation () and by the definition of the arcsine function, arcsin arccos 6 6 sin u cos arccos. () Add arccos. () Definition of arcsine Substitute. () Subtract 6. (Section.7) Since 0 arccos, we must have 0 arccos. Thus, 0, and we can sketch the triangle in Figure. From this triangle we find that sin u. arccos 6 6 Solve arcsin arccos 6. 6 6 cos u sin 6.

LIALMC07_0768.QXP /6/0 0:8 AM Page 67 67 CHAPTER 7 Trigonometric Identities and Equations Now substitute into equation () using sin u sin cos 6 6,,, and cos u. sin u cos 6 cos u sin 6 () Multiply by. Subtract. Square both sides. (Section.6) Distributive property (Section R.) Add. Solve for ; Choose the positive square root because 0. To check, replace with arcsin as required. The solution set is in the original equation: arccos. Quotient rule (Section R.7) 6 6, Now try Eercise. 7.7 Eercises. C. A. C. C 5. 6. 7. 8. arccos y 5 arcsin y arccot y arcsec y 9. arctan y Concept Check Answer each question.. Which one of the following equations has solution 0? A. arctan B. arccos 0 C. arcsin 0. Which one of the following equations has solution? A. arcsin B. arccos C. arctan. Which one of the following equations has solution? A. B. C. arccos arctan arcsin. Which one of the following equations has solution 6? A. arctan B. arccos C. arcsin

LIALMC07_0768.QXP /6/0 0:8 AM Page 67 7.7 Equations Involving Inverse Trigonometric Functions 67 0.... arcsin y arccos y 6 arcsiny 5 arccos y Solve each equation for. See Eample. 5. y 5 cos 6. y sin 7. 8. 6y 9. y tan 0. sec. y 6 cos. y sin. y cot y sin y cos 5. 5. 5 arccot y arccos y 6. arctan y 7. arcsin y 8. arccot y 9. arcsin y 0. arccos y.. 0 5. 6. 6 5 7. 8. 9. 0.. 0. 5.. 0 5. 6. 5 5 7. 0 8. 0 9. Y = arcsin X arccos X 6 8 5 0. Y = arcsin X arccos X Y = 6 9 9. y cot 5 5. y cos 6. y tan 7. y sin 8. y cot 9. y sin 0. y cos. Refer to Eercise 7. A student attempting to solve this equation wrote as the first step y sin, inserting parentheses as shown. Eplain why this is incorrect.. Eplain why the equation sin cos cannot have a solution. (No work is required.) Solve each equation for eact solutions. See Eamples and.. y. cos 5. arccos y 6. arccosy 6 7. arcsin arctan 8. arctan arccos 5 9. cos 0. cot tan sin 5 Solve each equation for eact solutions. See Eample.. sin tan.. arccos arcsin. 5. arcsin arccos 6. arcsin arcsin 6 7. cos tan 8. sin tan 0 9. Provide graphical support for the solution in Eample by showing that the graph of y arcsin arccos 6 has -intercept.86605. 0. Provide graphical support for the solution in Eample by showing that the -coordinate of the point of intersection of the graphs of Y arcsin X arccos X and Y 6 is.86605. tan y sin tan arccos arcsin..607..85 8 The following equations cannot be solved by algebraic methods. Use a graphing calculator to find all solutions over the interval 0, 6. Epress solutions to as many decimal places as your calculator displays.. arctan 0. sin.

LIALMC07_0768.QXP /6/0 0:8 AM Page 67 67 CHAPTER 7 Trigonometric Identities and Equations. (a) A.00506, ; (Modeling) Solve each problem. P.00506 sin0t.8. Tone Heard by a Listener When two sources located at different positions produce (b) The two graphs are the same. For = t, P(t) =.00506 sin(0t +.8) the same pure tone, the human ear will often hear one sound that is equal to the sum of the individual tones. Since the sources are at different locations, they will have different phase angles. If two speakers located at different positions produce P (t) + P (t) =.00 sin(0t +.05) + pure tones P A sinft and P A sinft, where,.00 sin(0t +.6), then the resulting tone heard by a listener can be written as.006 P A sinft, where.8 0.0.006.70. (a) A.005, ; P.005 sin600t.7 (b) The two graphs are the same. For = t, P(t) =.005 sin(600t +.7) P (t) + P (t) =.005 sin ( 600t + ) +.00 sin ( 600t + 6 ).006 0.0.006 7 A A cos A cos A sin A sin and. A cos A cos (Source: Fletcher, N. and T. Rossing, The Physics of Musical Instruments, Second Edition, Springer-Verlag, 998.) (a) Calculate A and if A.00,.05, A.00, and.6. Also find an epression for P A sinft if f 0. (b) Graph Y P and Y P P on the same coordinate aes over the interval 0,.0. Are the two graphs the same?. Tone Heard by a Listener Repeat Eercise, with A.005, 7, A.00, 6, and f 00. 5. Depth of Field When 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 depth of field. See the figure. (Source: Bushaw, Donald et al., A Sourcebook of Applications of School Mathematics. Copyright 980 by The Mathematical Association of America.) arctan A sin A sin Subject y Lens 5. (a) tan ; tan y z z (b) y tan tan (c) y (d) arctan tan arctan y tan 0 (c) tan u u 6. (a) sin u, u (b) y (a) Write two equations, one relating,, and z, and the other relating,, y, and z. (b) Eliminate z from the equations in part (a) to get one equation relating,,, and y. (c) Solve the equation from part (b) for. (d) Solve the equation from part (b) for. 6. 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 epressed in terms of arctangent as follows. y (a) Let u arcsin. Solve the equation for 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. (c) Use the triangle from part (b) to write an equation u 0 for tan u in terms of. (d) Solve the equation from part (c) for u. Film z

LIALMC07_0768.QXP /6/0 0:8 AM Page 675 7.7 Equations Involving Inverse Trigonometric Functions 675 (d) u arctan 7. (a) t f arcsin e E ma (b).00068 sec 8. (b) (i) approimately.9 or.6 (ii) approimately.60 or 6.6 (c) (i) approimately.5 (ii) approimately.6 9. (a) t arcsin y (b).7 sec 50. y = sec 0 Radian mode 7. Alternating Electric Current In the study of alternating electric current, instantaneous voltage is modeled by e E ma sin ft, where f is the number of cycles per second, is the maimum voltage, and t is time in seconds. (a) Solve the equation for t. (b) Find the smallest positive value of t if E ma, e 5, and f 00. Use a calculator. 8. Viewing Angle of an Observer While visiting a museum, Marsha Langlois views a painting that is ft ft high and hangs 6 ft above the ground. θ See the figure. Assume her eyes are 5 ft above the ground, and let be the 6 ft distance from the spot where she is 5 ft standing to the wall displaying the painting. ft (a) Show that, the viewing angle subtended by the painting, is given by tan tan. E ma (b) Find the value of for each value of. (i) (ii) 6 8 (c) Find the value of for each value of. (i) (ii) 9. Movement of an Arm In the eercises for Section 6. we found the equation y t sin, where 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. radian? 50. The function y sec is not found on graphing calculators. However, with some models it can be graphed as y. 0 0 tan (This formula appears as Y in the screen here.) Use the formula to obtain the graph of y sec in the window, by 0,.

LIALMC07_0768.QXP /6/0 0:8 AM Page 676 676 CHAPTER 7 Trigonometric Identities and Equations Chapter 7 Summary NEW SYMBOLS sin (arcsin ) cos (arccos ) tan (arctan ) inverse sine of inverse cosine of inverse tangent of cot (arccot ) sec (arcsec ) csc (arccsc ) inverse cotangent of inverse secant of inverse cosecant of QUICK REVIEW CONCEPTS EXAMPLES 7. Fundamental Identities Reciprocal Identities cot tan Quotient Identities Pythagorean Identities sin Negative-Angle Identities sin sin csc csc tan sin cos cos sec cos cot csc cos cos sec sec cot cos sin csc sin tan sec tan tan cot cot If is in quadrant IV and sin 5, find csc, cos, and sin. sin 5 cos cos csc sin 5 5 cos 9 5 6 5 cos 6 5 5 sin sin 5 cos is positive in quadrant IV. 7. Verifying Trigonometric Identities See the bo titled Hints for Verifying Identities on page 6. 7. Sum and Difference Identities Cofunction Identities cos90 sin cot90 tan sin90 cos sec90 csc tan90 cot csc90 sec Sum and Difference Identities cosa B cos A cos B sin A sin B cosa B cos A cos B sin A sin B sina B sin A cos B cos A sin B sina B sin A cos B cos A sin B Find a value of such that tan cot 78. cot90 cot 78 90 tan cot 78 78 Find the eact value of cos5. cos5 cos0 5 cos 0 cos 5 sin 0 sin 5 6 6

LIALMC07_0768.QXP /6/0 0:8 AM Page 677 CHAPTER 7 Summary 677 CONCEPTS EXAMPLES Sum and Difference Identities Write in terms of tan. tan A tan B tan tan tan tana B tan A tan B tan tan tan A tan B tan tana B tan A tan B tan tan tan 7. Double-Angle Identities and Half-Angle Identities Double-Angle Identities cos A cos A sin A cos A cos A tan A cos A sin A sin A sin A cos A tan A tan A Given cos 5 and sin 0, find sin. Sketch a triangle in quadrant II and use it to find sin : sin. y sin sin cos 5 0 69 5 Product-to-Sum Identities cos A cos B cosa B cosa B Write sin sin as the difference of two functions. sin sin cos cos sin A sin B cosa B cosa B cos cos sin A cos B sina B sina B cos cos cos A sin B sina B sina B cos cos Sum-to-Product Identities sin A sin B sin A B cos A B sin A sin B cos A B sin A B cos A cos B cos A B cos A B cos A cos B sin A B sin A B Half-Angle Identities cos A cos A tan A cos A cos A tan A cos A sin A sin A cos A tan A sin A cos A The sign is chosen based on the quadrant of A. Write cos cos as a product of two functions. cos cos cos cos cos cos cos cos Find the eact value of tan 67.5. We choose the last form with A 5. tan 67.5 tan 5 cos cos cos 5 sin 5 or Rationalize the denominator; simplify.

LIALMC07_0768.QXP /6/0 0:8 AM Page 678 678 CHAPTER 7 Trigonometric Identities and Equations CONCEPTS EXAMPLES 7.5 Inverse Circular Functions (, ) y 0 (, ) y = sin y = tan 0 (, ) See page 69 for graphs of the other inverse circular (trigonometric) functions. y (, ) (, ) ( 0, ) Range Inverse Quadrants Function Domain Interval of the Unit Circle y sin y cos y tan y cot y sec y csc,,,,,,,, 0,, y,, y 0, 0,, 0, 0 y y = cos I and IV I and II I and IV I and II I and II I and IV (, 0) Evaluate y cos 0. Write y cos 0 as cos y 0. Then y, because cos 0 and is in the range of cos. Evaluate Let u. Then tan u tan. Since tan is negative in quadrant IV, sketch a triangle as shown. We want sin u 5. sintan. sintan sin u y 5 u. From the triangle, 7.6 Trigonometric Equations Solving a Trigonometric Equation. Decide whether the equation is linear or quadratic in form, so you can determine the solution method.. If only one trigonometric function is present, first solve the equation for that function.. 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.. 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 etraneous solutions. Solve tan over the interval 0, 60. Use a linear method. tan tan Another solution over 0, 60 is The solution set is 60,0. 60 60 80 0.

LIALMC07_0768.QXP /6/0 0:8 AM Page 679 CHAPTER 7 Review Eercises 679 Chapter 7 Review Eercises. B. A. C. F 5. D 6. E 7. 8. cos sin 9. cos 0. cos sin. sin tan 5 ; ; cot 9 IV 5. ; 6 6.. cot 5 ; csc sec 5 ;. E. B 5. J 6. A 7. I 8. C 9. H 0. D. G. B. 5 ; 0 5 5 ; I 0 5 ; 9. ; ; 50 50 Concept Check For each epression in Column I, choose the epression from Column II that completes an identity. I II. sec. csc A. B. sin cos. tan. cot sin C. D. cos cot 5. tan 6. sec E. cos F. cos sin Use identities to write each epression in terms of sin and cos, and simplify. 7. sec tan 8. 9. tan cot 0. csc cot. Use the trigonometric identities to find sin, tan, and cot, given cos 5 and is in quadrant IV.. Given tan 5, where, use the trigonometric identities to find cot, csc, and sec. Concept Check For each epression in Column I, use an identity to choose an epression from Column II with the same value. I. cos 0. sin 5 5. tan5 6. sin 5 7. cos 5 8. cos 75 9. sin 75 0. sin 00. cos 00. cos55 cot sec II A. sin5 B. cos 55 cos 50 C. D. sin 50 cos 50 E. F. G. H. cos 50 cos 60 sin 50 sin 60 cot5 cos 50 sin 50 sin 5 cos 60 cos 5 sin 60 I. cos5 J. cot 5 For each of the following, find sin y, cos y, tan y, and the quadrant of y.. sin cos y and y in quadrant III, 5,. sin cos y in quadrant I, y in quadrant IV 0, 5, Find each of the following. 5. cos, given cos with 90, 6. sin y, given cos y, with y 80

LIALMC07_0768.QXP /6/0 0:8 AM Page 680 680 CHAPTER 7 Trigonometric Identities and Equations sin sin 7. cos cos cot 8. cos tan sin 5. 6. 7. 8. 9. 50. 6 5. 5. 5. 5. 60 55. 60 56. 0 57. 60.6795 58..556 59. 6.89508 60..565 6. 7.606 6. 7.6756797 6. 6. 65. 66. 67. 7 68. 0 69. 70. 0 9 7. 7. 0 7 7. 9 56 9 7. u 75. u 76., 77..667609,.60506 Graph each epression and use the graph to conjecture an identity. Then verify your conjecture algebraically. sin sin 7. 8. cos cos Verify that each equation is an identity. 9. sin sin y cos y cos 0. sin.. cos cos. cos A sec A cos A tan A. csc A 5. tan tan csc 6. 7. tan sin cos 8. csc A sin A sec A cos A sec A 9. tan csc tan 0. cos sin cos cos cos sin sec sin sin sec tan B sin B sec B cot tan csc cos tan tan.. sec sec tan cos tan cos tan tan sec. cos cos cos sin. sin sin cos sin sec Give the eact real number value of y. Do not use a calculator. 5. 6. y arccos y sin 7. y tan 8. y arcsin 9. y cos 50. y arctan 5. y sec 5. y arccsc 5. y arccot Give the degree measure of. Do not use a calculator. 5. 55. 56. arccos Use a calculator to give the degree measure of. arctan.780675 57. 58. 59. cot.50688 arcsin sin.66050 arcsec.755 60. 6. 6. tan 0 cos.8096577 csc 7.890096 Evaluate the following without using a calculator. 6. cosarccos 6. sinarcsin 65. 66. arcsecsec 67. tan tan 68. cos cos 0 69. sinarccos 70. cosarctan 7. coscsc arccoscos

LIALMC07_0768.QXP /6/0 0:8 AM Page 68 78..7977656,.86997 79.,, 5, 7 80. 8. 8, 8, 5 8, 5 8 8, 8, 9 8, 8, 8, 5 8 8. 0 n, where n is any integer 8.,, n, where n is any integer n 8.,, 6 n n,, where n 6 n n is any integer 85. {70 } 86. {5, 5., 5,. } 87. {5, 90, 5, 70 } 88. {5, 75, 95, 55 } 89. {70.5, 80, 89.5 } 90. {5.5, 8.,.5, 98. } 9. arcsin y 7 5 7 9. 9. 9. arcsin 5y 95. 0 96. 5 97. 7 98. 99. (b) 8.660567 ft; There may be a discrepancy in the final digits. f() = arctan( ) arctan( ) 5 5 0 0 arccos y arctan y t 50, 8, 8, 5 8, 7 8, 9 8, 8, n arccos d 550 50 7. 7. sec sin CHAPTER 7 Review Eercises 68 Write each of the following as an algebraic (nontrigonometric) epression in u. u 7. 75. tanarcsec u cosarctan u u Solve each equation for solutions over the interval 0,. 76. sin 77. tan 0 78. sin 5 sin 0 79. tan cot 80. sec 8. tan 0 Give all solutions for each equation. 8. sec 8. cos cos 0 8. sin cos cos Solve each equation for solutions over the interval 0, 60. If necessary, epress solutions to the nearest tenth of a degree. 85. sin sin 0 86. tan tan 87. sin cos 88. sin 89. cos cos 0 90. 5 cot cot 0 Solve each equation in Eercises 9 97 for. 9. y sin 9. y cos 9. y tan 9. 5y sin 95. 96. arccos arcsin arctan 7 97. arccos arctan 98. Solve d 550 50 cos for t in terms of d. 50 t (Modeling) Solve each problem. 99. Viewing Angle of an Observer A 0-ft-wide chalkboard is situated 5 ft from the left wall of a classroom. See the figure. A student sitting net to the wall feet from the front of the classroom has a viewing angle of radians. (a) Show that the value of is given by the function defined by f arctan 5 arctan 5. tanarcsin 5 arccos 5 7 5 0 (b) Graph f with a graphing calculator to estimate the value of that maimizes the viewing angle.

LIALMC07_0768.QXP /6/0 0:8 AM Page 68 68 CHAPTER 7 Trigonometric Identities and Equations 00. 8.8 0. The light beam is completely underwater. 0. (a). (b) 90 (c) 8.0 0. y = csc 0. (a), (b) Radian mode In both cases, sin..5685. 00. Snell s Law Recall Snell s law from Eercises 0 and 0 of Section 5.: where c is the speed of light in one medium, Water c is the speed of light in a second medium, and and are the angles shown in the figure. Suppose a light is shining up through water into the air as in the figure. As increases, approaches 90, at which point c no light will emerge from the water. Assume the ratio in this case is.75. For c what value of does 90? This value of is called the critical angle for water. 0. Snell s Law Refer to Eercise 00. What happens when is greater than the critical angle? 0. 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 cos, c sin, c sin where is the latitude in degrees. See the figure. (Source: Bushaw, Donald et al., A Sourcebook of Applications of School Mathematics. Copyright 980 by The Mathematical Association of America.) (a) Find the latitude between 0 and 90 at which the nautical mile is 607 ft. (b) At what latitude between 0 and 80 is the nautical mile 608 ft? (c) In the United States, the nautical mile is defined everywhere as 6080. ft. At what latitude between 0 and 90 does this agree with the British nautical mile? 0. The function y csc is not found on graphing calculators. However, with some models it can be graphed as y 0 0 tan. (This formula appears as Y in the screen here.) Use the formula to obtain the graph of y csc in the window, by,. 0. (a) Use the graph of y sin to approimate sin.. (b) Use the inverse sine key of a graphing calculator to approimate sin.. Air A nautical mile is the length on any of the meridians cut by a central angle of measure minute. Chapter 7 Test. sin 56 6 ; cos 66 6.. Given tan 5 6,, use trigonometric identities to find sin and cos.. Epress tan sec in terms of sin and cos, and simplify.. Find sin y, cos y, and tan y, if sin, cos y 5, is in quadrant III, and y is in quadrant II.. Use a half-angle identity to find sin.5.

LIALMC07_0768.QXP /6/0 0:8 AM Page 68 CHAPTER 7 Test 68. sin y ; 5 cos y ; 5 tan y 8. 5. sec sin tan cos 6. cot cot csc 9. (a) sin (b) 0. (a) V 6 cos (b) 6 volts; sec 0., ;,. (a) (b) (c) 0 (d). (a) 5 (b) 9. u u u 5. {90, 70 } 6. {8., 5, 98., 5 } 7. (, ) 0,, (0, 0) y 8.,, where n n n is any integer 9. (a) arccos y (b) 5 y = sin (, ) 5 7 0. sec, sec, sec 6 6 6 sin t Graph each epression and use the graph to conjecture an identity. Then verify your conjecture algebraically. 5. sec sin tan 6. cot cot Verify that each equation is an identity. cot A tan A 7. sec B 8. cos A sin B csc A sec A 9. Use an identity to write each epression as a trigonometric function of alone. (a) cos70 (b) 0. Voltage The voltage in common household current is epressed as V 6 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 epress V in terms of cosine. (b) If, what is the maimum voltage? Give the smallest positive value of t when the maimum voltage occurs. 0. Graph y sin, and indicate the coordinates of three points on the graph. Give the domain and range.. Find the eact value of y for each equation. (a) y arccos (b) y sin (c) y tan 0 (d) y arcsec. Find each eact value. (a) cosarcsin (b) sin cos. Write tanarcsin u as an algebraic (nontrigonometric) epression in u. 5. Solve sin cos for solutions over the interval 0, 60. 6. Solve csc cot for solutions over the interval 0, 60. Epress approimate solutions to the nearest tenth of a degree. 7. Solve cos cos for solutions over the interval 0,. 8. Solve sin, giving all solutions in radians. 9. Solve each equation for. (a) y cos (b) arcsin arctan 0. (Modeling) Movement of a Runner s Arm A runner s arm swings rhythmically according to the model y sin 8 cos t, 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, is the angle y equal to 0?

LIALMC07_0768.QXP /6/0 0:8 AM Page 68 68 CHAPTER 7 Trigonometric Identities and Equations Chapter 7 Quantitative Reasoning How 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 ft and the cylinder is 0 ft long. Determine the volume of oil in the tank if the rod shows a depth of ft. (Hint: The volume will be 0 times the area of the shaded segment of the circle shown in the figure on the right.) ft ft 0 9 arctan8 8 65 ft