Chapter 5 Solutions to Odd-Numbered Problems

Section 5.1 Solutions

1. All three functions, F,GF,G, and HH, are even.

This is because F(x)=sin(x)sin(x)=(sinx)(sinx)=sin2x=F(x),G(x)=cos(x)cos(x)=cosxcosx=cos2x=G(x)F(x)=sin(x)sin(x)=(sinx)(sinx)=sin2x=F(x),G(x)=cos(x)cos(x)=cosxcosx=cos2x=G(x) and H(x)=tan(x)tan(x)=(tanx)(tanx)=tan2x=H(x)H(x)=tan(x)tan(x)=(tanx)(tanx)=tan2x=H(x).

3. When cost=0cost=0, then sect=10sect=10, which is undefined.

5. sinxsinx

7. secxsecx

9. csctcsct

11. 11

13. sec2xsec2x

15. sin2x+1sin2x+1

17. 1sinx1sinx

19. 1cotx1cotx

21. tanxtanx

23. 4secxtanx4secxtanx

25. ±1cot2x+1±1cot2x+1

27. ±1sin2xsinx±1sin2xsinx

29. Answers will vary. Sample proof:
cosxcos3x=cosx(1cos2x)cosxcos3x=cosx(1cos2x)
=cosxsin2x=cosxsin2x

31. Answers will vary. Sample proof:

1+sin2xcos2x=1cos2x+sin2xcos2x=sec2x+tan2x=tan2x+1+tan2x=1+2tan2x1+sin2xcos2x=1cos2x+sin2xcos2x=sec2x+tan2x=tan2x+1+tan2x=1+2tan2x

33. Answers will vary. Sample proof:

cos2xtan2x=1sin2x(sec2x1)=1sin2xsec2x+1=2sin2xsec2xcos2xtan2x=1sin2x(sec2x1)=1sin2xsec2x+1=2sin2xsec2x

35. False

37. False

39. Proved with negative and Pythagorean identities

41. True

3sin2θ+4cos2θ=3sin2θ+3cos2θ+cos2θ=3(sin2θ+cos2θ)+cos2θ=3+cos2θ3sin2θ+4cos2θ=3sin2θ+3cos2θ+cos2θ=3(sin2θ+cos2θ)+cos2θ=3+cos2θ

Section 5.2 Solutions

1. The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures xx, the second angle measures π2xπ2x. Then sinx=cos(π2x)sinx=cos(π2x). The same holds for the other cofunction identities. The key is that the angles are complementary.

3. sin(x)=sinxsin(x)=sinx, so sinxsinx is odd. cos(x)=cos(0x)=cosxcos(x)=cos(0x)=cosx, so cosxcosx is even.

5. 2+642+64

7. 624624

9. 2323

11. 22sinx22cosx22sinx22cosx

13. 12cosx32sinx12cosx32sinx

15. cscθcscθ

17. cotxcotx

19. tan(x10)tan(x10)

21. sin(ab)=(45)(13)(35)(223)=46215sin(ab)=(45)(13)(35)(223)=46215
cos(a+b)=(35)(13)(45)(223)=38215cos(a+b)=(35)(13)(45)(223)=38215

23. 264264

25. sinxsinx

Graph of y=sin(x) from -2pi to 2pi.

27. cot(π6x)cot(π6x)

Graph of y=cot(pi/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi/6.

29. cot(π4+x)cot(π4+x)

Graph of y=cot(pi/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi/4.

31. sinx2+cosx2sinx2+cosx2

Graph of y = sin(x) / rad2 + cos(x) / rad2 - it looks like the sin curve shifted by pi/4.

 

33. They are the same.

35. They are the different, try g(x)=sin(9x)cos(3x)sin(6x)g(x)=sin(9x)cos(3x)sin(6x).

37. They are the same.

39. They are the different, try g(θ)=2tanθ1tan2θg(θ)=2tanθ1tan2θ.

41. They are different, try g(x)=tanxtan(2x)1+tanxtan(2x)g(x)=tanxtan(2x)1+tanxtan(2x).

43. 3122, or 0.25883122, or 0.2588

45. 1+3221+322, or 0.9659

47. tan(x+π4)=tanx+tan(π4)1tanxtan(π4)=tanx+11tanx(1)=tanx+11tanxtan(x+π4)=tanx+tan(π4)1tanxtan(π4)=tanx+11tanx(1)=tanx+11tanx

 

49. cos(a+b)cosacosb=cosacosbcosacosbsinasinbcosacosb=1tanatanbcos(a+b)cosacosb=cosacosbcosacosbsinasinbcosacosb=1tanatanb

51. cos(x+h)cosxh=cosxcoshsinxsinhcosxh=cosx(cosh1)sinxsinhh=cosxcosh1hsinxsinhhcos(x+h)cosxh=cosxcoshsinxsinhcosxh=cosx(cosh1)sinxsinhh=cosxcosh1hsinxsinhh

53. True

55. True. Note that sin(α+β)=sin(πγ)sin(α+β)=sin(πγ) and expand the right hand side.

Section 5.3 Solutions

1. Use the Pythagorean identities and isolate the squared term.

3. 1cosxsinx,sinx1+cosx1cosxsinx,sinx1+cosx, multiplying the top and bottom by 1cosx1cosx and 1+cosx1+cosx, respectively.

5. a) 37323732 b) 31323132 c) 37313731

7. a) 3232 b) 1212 c) 33

9. cosθ=255,sinθ=55,tanθ=12,cscθ=5,secθ=52,cotθ=2cosθ=255,sinθ=55,tanθ=12,cscθ=5,secθ=52,cotθ=2

11. 2sin(π2)=22sin(π2)=2

13. 222222

15. 232232

17. 2+32+3

19. 1212

21. a) 3131331313 b) 2131321313 c) 3232

23. a) 104104 b) 6464 c) 153153

25. 120169,119169,120119120169,119169,120119

27. 21313,31313,2321313,31313,23

29. cos(74)cos(74)

31. cos(18x)cos(18x)

33. 3sin(10x)3sin(10x)

35. 2sin(x)cos(x)=2(sin(x)cos(x))=sin(2x)2sin(x)cos(x)=2(sin(x)cos(x))=sin(2x)

37. sin(2θ)1+cos(2θ)tan2θ=2sin(θ)cos(θ)1+cos2θsin2θtan2θ=2sin(θ)cos(θ)2cos2θtan2θ=sin(θ)cosθtan2θ=cot(θ)tan2θ=tanθsin(2θ)1+cos(2θ)tan2θ=2sin(θ)cos(θ)1+cos2θsin2θtan2θ=2sin(θ)cos(θ)2cos2θtan2θ=sin(θ)cosθtan2θ=cot(θ)tan2θ=tanθ

39. 1+cos(12x)21+cos(12x)2

41. 3+cos(12x)4cos(6x)83+cos(12x)4cos(6x)8

43. 2+cos(2x)2cos(4x)cos(6x)322+cos(2x)2cos(4x)cos(6x)32

45. 3+cos(4x)4cos(2x)3+cos(4x)+4cos(2x)3+cos(4x)4cos(2x)3+cos(4x)+4cos(2x)

47. 1cos(4x)81cos(4x)8

49. 3+cos(4x)4cos(2x)4(cos(2x)+1)3+cos(4x)4cos(2x)4(cos(2x)+1)

51. (1+cos(4x))sinx2(1+cos(4x))sinx2

53. 4sinxcosx(cos2xsin2x)4sinxcosx(cos2xsin2x)

55. 2tanx1+tan2x=2sinxcosx1+sin2xcos2x=2sinxcosxcos2x+sin2xcos2x=2tanx1+tan2x=2sinxcosx1+sin2xcos2x=2sinxcosxcos2x+sin2xcos2x=
2sinxcosx.cos2x1=2sinxcosx=sin(2x)2sinxcosx.cos2x1=2sinxcosx=sin(2x)

57. 2sinxcosx2cos2x1=sin(2x)cos(2x)=tan(2x)2sinxcosx2cos2x1=sin(2x)cos(2x)=tan(2x)

59. sin(x+2x)=sinxcos(2x)+sin(2x)cosx=sinx(cos2xsin2x)+2sinxcosxcosx=sinxcos2xsin3x+2sinxcos2x=3sinxcos2xsin3xsin(x+2x)=sinxcos(2x)+sin(2x)cosx=sinx(cos2xsin2x)+2sinxcosxcosx=sinxcos2xsin3x+2sinxcos2x=3sinxcos2xsin3x

61. 1+cos(2t)sin(2t)cost=1+2cos2t12sintcostcost=2cos2tcost(2sint1)=2cost2sint11+cos(2t)sin(2t)cost=1+2cos2t12sintcostcost=2cos2tcost(2sint1)=2cost2sint1

63. (cos2(4x)sin2(4x)sin(8x))(cos2(4x)sin2(4x)+sin(8x))= =(cos(8x)sin(8x))(cos(8x)+sin(8x)) =cos2(8x)sin2(8x) =cos(16x)(cos2(4x)sin2(4x)sin(8x))(cos2(4x)sin2(4x)+sin(8x))= =(cos(8x)sin(8x))(cos(8x)+sin(8x)) =cos2(8x)sin2(8x) =cos(16x)

Section 5.4 Solutions

1. Substitute αα into cosine and ββ into sine and evaluate.

3. Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: sin(3x)+sinxcosx=1sin(3x)+sinxcosx=1. When converting the numerator to a product the equation becomes: 2sin(2x)cosxcosx=1

5. 8(cos(5x)cos(27x))

7. sin(2x)+sin(8x)

9. 12(cos(6x)cos(4x))

11. 2cos(5t)cost

13. 2cos(7x)

15. 2cos(6x)cos(3x)

17. 14(1+3)

19. 14(32)

21. 14(31)

23. cos(80)cos(120)

25. 12(sin(221)+sin(205))

27. 2cos(31)

29. 2cos(66.5)sin(34.5)

31. 2sin(1.5)cos(0.5)

33. 2sin(7x)2sinx=2sin(4x+3x)2sin(4x3x)=2(sin(4x)cos(3x)+sin(3x)cos(4x))2(sin(4x)cos(3x)sin(3x)cos(4x))=2sin(4x)cos(3x)+2sin(3x)cos(4x)2sin(4x)cos(3x)+2sin(3x)cos(4x)=4sin(3x)cos(4x)

 

35. sinx+sin(3x)=2sin(4x2)cos(2x2)=
2sin(2x)cosx=2(2sinxcosx)cosx=
4sinxcos2x

37. 2tanxcos(3x)=2sinxcos(3x)cosx=2(.5(sin(4x)sin(2x)))cosx
=1cosx(sin(4x)sin(2x))=secx(sin(4x)sin(2x))

39. 2cos(35)cos(23), 1.5081

41. 2sin(33)sin(11), 0.2078

43. 12(cos(99)cos(71)), 0.2410

45. It is an identity.

47. It is not an identity, but 2cos3x is.

49. tan(3t)

51. 2cos(2x)

53. sin(14x)

55. Start with cosx+cosy. Make a substitution and let x=α+β and let y=αβ, so cosx+cosy becomes

cos(α+β)+cos(αβ)=cosαcosβsinαsinβ+cosαcosβ+sinαsinβ=2cosαcosβ

Since x=α+β and y=αβ, we can solve for α and β in terms of x and y and substitute in for 2cosαcosβ and get 2cos(x+y2)cos(xy2).

57. cos(3x)+cosxcos(3x)cosx=2cos(2x)cosx2sin(2x)sinx=cot(2x)cotx

59. cos(2y)cos(4y)sin(2y)+sin(4y)=2sin(3y)sin(y)2sin(3y)cosy=2sin(3y)sin(y)2sin(3y)cosy=tany

61. cosxcos(3x)=2sin(2x)sin(x)=2(2sinxcosx)sinx=4sin2xcosx

63. tan(π4t)=tan(π4)tant1+tan(π4)tan(t)=1tant1+tant

Section 5.5 Solutions

1. There will not always be solutions to trigonometric function equations. For a basic example, cos(x)=5.

3. If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.

5. π3,2π3

7. 3π4,5π4

9. π4,5π4

11. π4,3π4,5π4,7π4

13. π4,7π4

15. 7π6,11π6

17. π18,5π18,13π18,17π18,25π18,29π18

19. 3π12,5π12,11π12,13π12,19π12,21π12

21. 16,56,136,176,256,296,376

23. 0,π3,π,5π3

25. π3,π,5π3

27. π3,3π2,5π3

29. 0,π

31. πsin1(14),7π6,11π6,2π+sin1(14)

33. 13(sin1(910)),π313(sin1(910)),2π3+13(sin1(910)),π13(sin1(910)),4π3+13(sin1(910)),5π313(sin1(910))

35. 0

37. π6,5π6,7π6,11π6

39. 3π2,π6,5π6

41. 0,π3,π,4π3

43. There are no solutions.

45. cos1(13(17)),2πcos1(13(17))

47. tan1(12(295)),π+tan1(12(295)),π+tan1(12(295)),2π+tan1(12(295))

49. There are no solutions.

51. There are no solutions.

53. 0,2π3,4π3

55. π4,3π4,5π4,7π4

57. sin1(35),π2,πsin1(35),3π2

59. cos1(14),2πcos1(14)

61. π3,cos1(34),2πcos1(34),5π3

63. cos1(34),cos1(23),2πcos1(23),2πcos1(34)

65. 0,π2,π,3π2

67. π3,cos1(14),2πcos1(14),5π3

69. There are no solutions.

71. π+tan1(2),π+tan1(32),2π+tan1(2),2π+tan1(32)

73. 2πk+0.2734,2πk+2.8682

75. πk0.3277

77. 0.6694,1.8287,3.8110,4.9703

79. 1.0472,3.1416,5.2360

81. 0.5326,1.7648,3.6742,4.9064

83. sin1(14),πsin1(14),3π2

85. π2,3π2

87. There are no solutions.

89. 0,π2,π,3π2

91. There are no solutions.

93. 7.2

95. 5.7

97. 82.4

99. 31.0

101. 88.7

103. 59.0

105. 36.9

Section 5.6 Solutions

1. Physical behavior should be periodic, or cyclical.

3. Since cumulative rainfall is always increasing, a sinusoidal function would not be ideal here.

5. y=3cos(π6x)1

7. 5sin(2x)+2

9. 4cos(xπ2)3

11. 58sin(xπ2)

13. tan(xπ12)

15. Answers will vary. Sample answer: This function could model temperature changes over the course of one very hot day in Phoenix, Arizona.

Graph of f(x) = -18cos(x*pi/12) - 5sin(x*pi/12) + 100 on the interval [0,24]. There is a single peak around 12.

17. 9 years from now

19. 56F

21. 1.8024 hours

23. 4:30

25. From July 8 to October 23

27. From day 19 through day 40

29. Floods: July 24 through October 7. Droughts: February 4 through March 27

31. Amplitude: 11, period: 16, frequency: 6 Hz

33. Amplitude: 5, period: 130, frequency: 30 Hz

35. P(t)=15cos(π6t)+650+556t

37. P(t)=40cos(π6t)+800(1.04)t

39. D(t)=7(0.89)tcos(40πt)

41. D(t)=19(0.9265)tcos(26πt)

43. 20.1 years

45. 17.8 seconds

47. Spring 2 comes to rest first after 8.0 seconds.

49. 500 miles, at 90

51. y=6(5)x+4sin(π2x)

53. y=8(12)xcos(π2x)+3