5.4 Multiple-Angle Identities

Transcription

1 4 CHAPTER 5 Analytic Trigonometry 5.4 Multiple-Angle Identities What you ll learn about Double-Angle Identities Power-Reducing Identities Half-Angle Identities Solving Trigonometric Equations... and why These identities are useful in calculus courses. Double-Angle Identities The formulas that result from letting v in the angle sum identities are called the double-angle identities. We will state them all and prove one, leaving the rest of the proofs as eercises. (See Eercises 1 4.) Double-Angle Identities sin sin u cos u cos u - sin u cos c cos u sin u tan u tan 1 - tan u There are three identities for cos u. This is not unusual; indeed, there are plenty of other identities one could supply for sin u as well, such as sin u sin 1p/ - u. We list the three identities for cos u because they are all useful in various contets and therefore worth memorizing. EXAMPLE 1 Proving a Double-Angle Identity Prove the identity: sin sin u cos u. sin sin 1u + u = sin u cos u + cos u sin u Sine of a sum 1v = u = sin u cos u Now try Eercise 1. Power-Reducing Identities One immediate use for two of the three formulas for cos u is to derive the powerreducing identities. Some simple-looking functions like y = sin u would be quite difficult to handle in certain calculus contets were it not for the eistence of these identities. Power-Reducing Identities sin 1 - cos u cos 1 + cos u tan 1 - cos u 1 + cos u We will also leave the proofs of these identities as eercises. (See Eercises 37 and 3.)

2 SECTION 5.4 Multiple-Angle Identities 49 EXAMPLE Proving an Identity Prove the identity: cos 4 u - sin 4 cos u. cos 4 u - sin 4 1cos u + sin u1cos u - sin u = 1 # 1cos u - sin u = cos u Pythagorean identity Double-angle identity Now try Eercise 15. EXAMPLE 3 Reducing a Power of 4 Rewrite cos 4 in terms of trigonometric functions with no power greater than 1. cos 4 = 1cos = a 1 + cos b = a 1 + cos + cos b 4 = cos cos 4 a b = cos cos 4 Power-reducing identity Power-reducing identity = cos + cos 4 Now try Eercise 39. Half-Angle Identities The power-reducing identities can be used to etend our stock of special angles whose trigonometric ratios can be found without a calculator. As usual, we are not suggesting that this algebraic procedure is any more practical than using a calculator, but we are suggesting that this sort of eercise helps you to understand how the functions behave. In Eploration 1, for eample, we use a power-reducing formula to find the eact values of sin 1p/ and sin 19p/ without a calculator. EXPLORATION 1 Finding the Sine of Half an Angle Recall the power-reducing formula sin 11 - cos u/. 1. Use the power-reducing formula to show that sin 1p/ = 1-1/4.. Solve for sin 1p/. Do you take the positive or negative square root? Why? 3. Use the power-reducing formula to show that sin 19p/ = 1-1/4. 4. Solve for sin 19p/. Do you take the positive or negative square root? Why? A little alteration of the power-reducing identities results in the half-angle identities, which can be used directly to find trigonometric functions of u/ in terms of trigonometric functions of u. As Eploration 1 suggests, there is an unavoidable ambiguity of sign involved with the square root that must be resolved in particular cases by checking the quadrant in which u/ lies.

3 430 CHAPTER 5 Analytic Trigonometry Did We Miss Two Signs? You might have noticed that all of the half-angle identities have unresolved signs ecept for the last two. The fact that we can omit them on the last two identities for tan u/ is a fortunate consequence of two facts: (1) sin u and tan 1u/ always have the same sign (easily observed from the graphs of the two functions in Figure 5.10), and () 1 cos u is never negative. Half-Angle Identities sin u = 1 - cos u B cos u = 1 + cos u B tan u = f B 1 - cos u 1 + cos u 1 - cos u sin u sin u 1 + cos u y 4 π π 4 FIGURE 5.10 The functions sin u and tan 1u/ always have the same sign. Solving Trigonometric Equations New identities always provide new tools for solving trigonometric equations algebraically. Under the right conditions, they even lead to eact solutions. We assert again that we are not presenting these algebraic solutions for their practical value (as the calculator solutions are certainly sufficient for most applications and unquestionably much quicker to obtain), but rather as ways to observe the behavior of the trigonometric functions and their interwoven tapestry of identities. EXAMPLE 4 Using a Double-Angle Identity Solve algebraically in the interval 30, p: sin = cos. sin = cos s in cos = cos s in cos - cos = 0 cos 1 sin - 1 = 0 cos = 0 or sin - 1 = 0 cos = 0 or sin = 1 [0, π ] by [, ] FIGURE 5.11 The function y = sin - cos for 0 p. The scale on the -ais shows intervals of length p/6. This graph supports the solution found algebraically in Eample 4. The two solutions of cos = 0 are = p/ and = 3p/. The two solutions of sin = 1/ are = p/6 and = 5p/6. Therefore, the solutions of sin = cos are p 6, p, 5p 6, 3p. We can support this result graphically by verifying the four -intercepts of the function y = sin - cos in the interval 30, p (Figure 5.11). Now try Eercise 3.

4 SECTION 5.4 Multiple-Angle Identities 431 EXAMPLE 5 Using Half-Angle Identities Solve sin = sin 1/. The graph of y = sin - sin 1/ in Figure 5.1 suggests that this function is periodic with period p and that the equation sin = sin 1/ has three solutions in 30, p. [ π, π ] Solve Algebraically by [, 1] FIGURE 5.1 The graph of y = sin - sin 1/ suggests that sin = sin 1/ has three solutions in 30, p. (Eample 5) sin = sin sin = a 1 - cos b 1 - cos = 1 - cos Half-angle identity Convert to all cosines. cos - cos = 0 cos 11 - cos = 0 cos = 0 or cos = 1 3p = p or or 0 The rest of the solutions are obtained by periodicity: = np, = p + np, = 3p + np, n = 0, 1,, Á Now try Eercise 43. QUICK REVIEW 5.4 (For help, go to Section 5.1.) Eercise numbers with a gray background indicate problems that the authors have designed to be solved without a calculator. In Eercises 1, find the general solution of the equation. 1. tan - 1 = 0. tan + 1 = cos 11 - sin = sin 11 + cos = 0 5. sin + cos = 0 6. sin - cos = sin - 11 cos + 1 = 0. 1sin + 11 cos - 1 = 0 9. Find the area of the trapezoid Find the height of the isosceles triangle. 3 3

5 43 CHAPTER 5 Analytic Trigonometry SECTION 5.4 EXERCISES In Eercises 1 4, use the appropriate sum or difference identity to prove the double-angle identity. 1. cos cos u - sin u. cos cos u cos 1 - sin u 4. tan u tan 1 - tan u In Eercises 5 10, find all solutions to the equation in the interval 30, p. 5. sin = sin 6. sin = sin 7. cos = sin. cos = cos 9. sin - tan = cos + cos = cos In Eercises 11 14, write the epression as one involving only sin u and cos u. 11. sin u + cos u 1. sin u + cos u 13. sin u + cos 3u 14. sin 3u + cos u In Eercises 15, prove the identity. 15. sin 4 = sin cos 16. cos 6 = cos csc = csc tan 1. cot = cot - tan 19. sin 3 = 1sin 14 cos sin 3 = 1sin 13-4 sin 1. cos 4 = 1 - sin cos. sin 4 = 14 sin cos 1 cos - 1 In Eercises 3 30, solve algebraically for eact solutions in the interval 30, p. Use your grapher only to support your algebraic work. 3. cos + cos = 0 4. cos + sin = 0 5. cos + cos 3 = 0 6. sin + sin 3 = 0 7. sin + sin 4 = 0. cos + cos 4 = 0 In Eercises 9 and 30, use a graphing calculator to find all eact solutions in the interval 30, p. 3Hint: All solutions are rational multiples of p.4 9. sin - cos 3 = sin 3 + cos = 0 In Eercises 31 36, use half-angle identities to find an eact value without a calculator. 31. sin tan cos sin 15p/1 35. tan 17p/1 36. cos 1p/ 37. Prove the power-reducing identities: (a) sin 1 - cos u (b) cos 3. (a) Use the identities in Eercise 37 to prove the powerreducing identity tan 1 - cos u 1 + cos u. (b) Writing to Learn Eplain why the identity in part (a) 1 - cos u does not imply that tan B 1 + cos u. In Eercises 39 4, use the power-reducing identities to prove the identity. 39. sin 4 = cos + cos cos 3 = a 1 cos b11 + cos 1 + cos u 41. sin 3 = a 1 sin b11 - cos 4 4. sin 5 = a 1 sin b13-4 cos + cos 4 In Eercises 43 46, use the half-angle identities to find all solutions in the interval 30, p. Then find the general solution sin = cos a cos = sin a b b 45. tan a 46. sin a b = 1 - cos 1 + cos b = cos - 1 Standardized Test Questions 47. True or False The product of two functions with period p has period p. Justify your answer. 4. True or False The function ƒ1 = cos is a sinusoid. Justify your answer. You should answer these questions without using a calculator. 49. Multiple Choice If ƒ1 = sin and g1 = cos, then ƒ1 = (A) ƒ1. (B) ƒ1) ƒ1. (C) ƒ1) g1. (D) ƒ1 g1. (E) ƒ1 g1 + g1 ƒ Multiple Choice sin.5 = (A) (B) (C) (D). (E). B B 51. Multiple Choice How many numbers between 0 and p satisfy the equation sin = cos? (A) None (B) One (C) Two (D) Three (E) Four 5. Multiple Choice The period of the function sin - cos is p p (A) (B) (C) p. (D) p. (E) 4p. 4.. Eplorations 53. Connecting Trigonometry and Geometry In a regular polygon all sides are the same length and all angles are equal in measure. (a) If the perpendicular distance from the center of the polygon with n sides to the midpoint of a side is R, and if the length of the side of the polygon is, show that = R tan u where p/n is the central angle subtended by one side. (b) If the length of one side of a regular 11-sided polygon is approimately 5.7 and R is a whole number, what is the value of R? θ Regular polygon with n sides R θ

6 SECTION 5.4 Multiple-Angle Identities Connecting Trigonometry and Geometry A rhombus is a quadrilateral with equal sides. The diagonals of a rhombus bisect A D the angles of the rhombus and are perpendicular bisectors of each other. Let ABC = u, d 1 = length of AC, and d = length of BD. B C (a) Show that cos u = d and sin u = d 1. (b) Show that sin d 1d. 55. Group Activity Maimizing Volume The ends of a 10-foot-long water trough are isosceles trapezoids as shown in the figure. Find the value of u that maimizes 1 ft θ θ 1 ft the volume of the trough and 1 ft the maimum volume. 56. Group Activity Tunnel Problem A rectangular tunnel is cut through a mountain to make a road. The upper vertices of the rectangle are on the circle + y = 400, as illustrated in the figure. is an identity but 1 - cos = sin A is not an identity. 63. Hawaiian Sunset Table 5. gives the time of day for sunset in Honolulu, HI, on the first day of each month of 009. Table 5. Sunset in Honolulu, 009 Date Day Time 1:30 + Jan 1 1 1:01 Feb 1 3 1: -9 - Mar :36 6 Apr :46 16 May :57 7 Jun :10 40 Jul :17 47 Aug :10 40 Sep :47 17 Oct :19-11 Nov :55-35 Dec :4-4 Source: y 400 (, y) (a) Show that the cross-sectional area of the end of the tunnel is 400 sin u. (b) Find the dimensions of the rectangular end of the tunnel that maimizes its cross-sectional area. Etending the Ideas In Eercises 57 61, prove the double-angle formulas cot cot u - 1 csc 1 csc u sec u cot u csc u sec u 59. sec 60. sec csc u - - sec u 61. sec sec u csc u csc u - sec u 6. Writing to Learn Eplain why 1 - cos = ƒsin ƒ A The second column gives the date as the day of the year, and the fourth column gives the time as the number of minutes past 1:30. (a) Enter the numbers in column (day) into list L1 and the numbers in column 4 (minutes past 1:30) into list L. Make a scatter plot with -coordinates from L1 and y-coordinates from L. (b) Using sine regression, find the regression curve through the points and store its equation in Y1. Superimpose the graph of the curve on the scatter plot. Is it a good fit? (c) Make a new column showing the residuals (the difference between the actual y-value at each point and the y-value predicted by the regression curve) and store them in list L3. Your calculator might have a list called RESID among the NAMES in the LIST menu, in which case the command RESID : L3 will perform this operation. You could also enter L - Y1(L1) : L3. (d) Make a scatter plot with -coordinates from L1 and y-coordinates from L3. Find the sine regression curve through these points and superimpose it on the scatter plot. (e) Writing to Learn Interpret what the two regressions seem to indicate about the periodic behavior of sunset as a function of time. This is not an unusual phenomenon in astronomical data, and it kept astronomers baffled for centuries.