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. −1−1
13. sec2xsec2x
15. sin2x+1sin2x+1
17. 1sinx1sinx
19. 1cotx1cotx
21. tanxtanx
23. −4secxtanx−4secxtanx
25. ±√1cot2x+1±√1cot2x+1
27. ±√1−sin2xsinx±√1−sin2xsinx
29. Answers will vary. Sample proof:
cosx−cos3x=cosx(1−cos2x)cosx−cos3x=cosx(1−cos2x)
=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:
cos2x−tan2x=1−sin2x−(sec2x−1)=1−sin2x−sec2x+1=2−sin2x−sec2xcos2x−tan2x=1−sin2x−(sec2x−1)=1−sin2x−sec2x+1=2−sin2x−sec2x
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 π2−xπ2−x. Then sinx=cos(π2−x)sinx=cos(π2−x). 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(0−x)=cosxcos(−x)=cos(0−x)=cosx, so cosxcosx is even.
5. √2+√64√2+√64
7. √6−√24√6−√24
9. −2−√3−2−√3
11. −√22sinx−√22cosx−√22sinx−√22cosx
13. −12cosx−√32sinx−12cosx−√32sinx
15. cscθcscθ
17. cotxcotx
19. tan(x10)tan(x10)
21. sin(a−b)=(45)(13)−(35)(2√23)=4−6√215sin(a−b)=(45)(13)−(35)(2√23)=4−6√215
cos(a+b)=(35)(13)−(45)(2√23)=3−8√215cos(a+b)=(35)(13)−(45)(2√23)=3−8√215
23. √2−√64√2−√64
25. sinxsinx
27. cot(π6−x)cot(π6−x)
29. cot(π4+x)cot(π4+x)
31. sinx√2+cosx√2sinx√2+cosx√2
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)=2tanθ1−tan2θg(θ)=2tanθ1−tan2θ.
41. They are different, try g(x)=tanx−tan(2x)1+tanxtan(2x)g(x)=tanx−tan(2x)1+tanxtan(2x).
43. −√3−12√2, or −0.2588−√3−12√2, or −0.2588
45. 1+√32√21+√32√2, or 0.9659
47. tan(x+π4)=tanx+tan(π4)1−tanxtan(π4)=tanx+11−tanx(1)=tanx+11−tanxtan(x+π4)=tanx+tan(π4)1−tanxtan(π4)=tanx+11−tanx(1)=tanx+11−tanx
49. cos(a+b)cosacosb=cosacosbcosacosb−sinasinbcosacosb=1−tanatanbcos(a+b)cosacosb=cosacosbcosacosb−sinasinbcosacosb=1−tanatanb
51. cos(x+h)−cosxh=cosxcosh−sinxsinh−cosxh=cosx(cosh−1)−sinxsinhh=cosxcosh−1h−sinxsinhhcos(x+h)−cosxh=cosxcosh−sinxsinh−cosxh=cosx(cosh−1)−sinxsinhh=cosxcosh−1h−sinxsinhh
53. True
55. True. Note that sin(α+β)=sin(π−γ)sin(α+β)=sin(π−γ) and expand the right hand side.
Section 5.3 Solutions
1. Use the Pythagorean identities and isolate the squared term.
3. 1−cosxsinx,sinx1+cosx1−cosxsinx,sinx1+cosx, multiplying the top and bottom by √1−cosx√1−cosx and √1+cosx√1+cosx, respectively.
5. a) 3√7323√732 b) 31323132 c) 3√7313√731
7. a) √32√32 b) −12−12 c) −√3−√3
9. cosθ=−2√55,sinθ=√55,tanθ=−12,cscθ=√5,secθ=−√52,cotθ=−2cosθ=−2√55,sinθ=√55,tanθ=−12,cscθ=√5,secθ=−√52,cotθ=−2
11. 2sin(π2)=22sin(π2)=2
13. √2−√22√2−√22
15. √2−√32√2−√32
17. 2+√32+√3
19. −1−√2−1−√2
21. a) 3√13133√1313 b) −2√1313−2√1313 c) −32−32
23. a) √104√104 b) √64√64 c) √153√153
25. 120169,−119169,−120119120169,−119169,−120119
27. 2√1313,3√1313,232√1313,3√1313,23
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+cos2θ−sin2θtan2θ=2sin(θ)cos(θ)2cos2θtan2θ=sin(θ)cosθtan2θ=cot(θ)tan2θ=tanθsin(2θ)1+cos(2θ)tan2θ=2sin(θ)cos(θ)1+cos2θ−sin2θtan2θ=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. 1−cos(4x)81−cos(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(cos2x−sin2x)4sinxcosx(cos2x−sin2x)
55. 2tanx1+tan2x=2sinxcosx1+sin2xcos2x=2sinxcosxcos2x+sin2xcos2x=2tanx1+tan2x=2sinxcosx1+sin2xcos2x=2sinxcosxcos2x+sin2xcos2x=
2sinxcosx.cos2x1=2sinxcosx=sin(2x)2sinxcosx.cos2x1=2sinxcosx=sin(2x)
57. 2sinxcosx2cos2x−1=sin(2x)cos(2x)=tan(2x)2sinxcosx2cos2x−1=sin(2x)cos(2x)=tan(2x)
59. sin(x+2x)=sinxcos(2x)+sin(2x)cosx=sinx(cos2x−sin2x)+2sinxcosxcosx=sinxcos2x−sin3x+2sinxcos2x=3sinxcos2x−sin3xsin(x+2x)=sinxcos(2x)+sin(2x)cosx=sinx(cos2x−sin2x)+2sinxcosxcosx=sinxcos2x−sin3x+2sinxcos2x=3sinxcos2x−sin3x
61. 1+cos(2t)sin(2t)−cost=1+2cos2t−12sintcost−cost=2cos2tcost(2sint−1)=2cost2sint−11+cos(2t)sin(2t)−cost=1+2cos2t−12sintcost−cost=2cos2tcost(2sint−1)=2cost2sint−1
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(√3−2)
21. 14(√3−1)
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(4x−3x)=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=α−β, so cosx+cosy becomes
cos(α+β)+cos(α−β)=cosαcosβ−sinαsinβ+cosαcosβ+sinαsinβ=2cosαcosβ
Since x=α+β and y=α−β, we can solve for α and β in terms of x and y and substitute in for 2cosαcosβ and get 2cos(x+y2)cos(x−y2).
57. cos(3x)+cosxcos(3x)−cosx=2cos(2x)cosx−2sin(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. cosx−cos(3x)=−2sin(2x)sin(−x)=2(2sinxcosx)sinx=4sin2xcosx
63. tan(π4−t)=tan(π4)−tant1+tan(π4)tan(t)=1−tant1+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. π−sin−1(−14),7π6,11π6,2π+sin−1(−14)
33. 13(sin−1(910)),π3−13(sin−1(910)),2π3+13(sin−1(910)),π−13(sin−1(910)),4π3+13(sin−1(910)),5π3−13(sin−1(910))
35. 0
37. π6,5π6,7π6,11π6
39. 3π2,π6,5π6
41. 0,π3,π,4π3
43. There are no solutions.
45. cos−1(13(1−√7)),2π−cos−1(13(1−√7))
47. tan−1(12(√29−5)),π+tan−1(12(−√29−5)),π+tan−1(12(√29−5)),2π+tan−1(12(−√29−5))
49. There are no solutions.
51. There are no solutions.
53. 0,2π3,4π3
55. π4,3π4,5π4,7π4
57. sin−1(35),π2,π−sin−1(35),3π2
59. cos−1(−14),2π−cos−1(−14)
61. π3,cos−1(−34),2π−cos−1(−34),5π3
63. cos−1(34),cos−1(−23),2π−cos−1(−23),2π−cos−1(34)
65. 0,π2,π,3π2
67. π3,cos−1(−14),2π−cos−1(−14),5π3
69. There are no solutions.
71. π+tan−1(−2),π+tan−1(−32),2π+tan−1(−2),2π+tan−1(−32)
73. 2πk+0.2734,2πk+2.8682
75. πk−0.3277
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. sin−1(14),π−sin−1(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. 5−8sin(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.
17. 9 years from now
19. 56∘F
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