13.2.7: Chapter 7

Try It

7.1 Solving Trigonometric Equations with Identities

1.

cscθcosθtanθ=(1sinθ)cosθ(sinθcosθ) =cosθsinθ(sinθcosθ) =sinθcosθsinθcosθ =1cscθcosθtanθ=(1sinθ)cosθ(sinθcosθ) =cosθsinθ(sinθcosθ) =sinθcosθsinθcosθ =1

2.

cotθcscθ=cosθsinθ1sinθ =cosθsinθ⋅sinθ1 =cosθcotθcscθ=cosθsinθ1sinθ =cosθsinθ⋅sinθ1 =cosθ

3.

sin2θ−1tanθsinθ−tanθ=(sinθ+1)(sinθ−1)tanθ(sinθ−1)=sinθ+1tanθsin2θ−1tanθsinθ−tanθ=(sinθ+1)(sinθ−1)tanθ(sinθ−1)=sinθ+1tanθ

4.

This is a difference of squares formula: 25−9sin2θ=(5−3sinθ)(5+3sinθ).25−9sin2θ=(5−3sinθ)(5+3sinθ).

5.

cosθ1+sinθ(1−sinθ1−sinθ)=cosθ(1−sinθ)1−sin2θ =cosθ(1−sinθ)cos2θ =1−sinθcosθcosθ1+sinθ(1−sinθ1−sinθ)=cosθ(1−sinθ)1−sin2θ =cosθ(1−sinθ)cos2θ =1−sinθcosθ

7.2 Sum and Difference Identities

1.

2√+6√42+64

2.

2√−6√42−64

3.

1−3√1+3√1−31+3

4.

cos(5π14)cos(5π14)

5.

tan(π−θ)=tan(π)−tanθ1+tan(π)tanθ =0−tanθ1+0⋅tanθ =−tanθtan(π−θ)=tan(π)−tanθ1+tan(π)tanθ =0−tanθ1+0⋅tanθ =−tanθ

7.3 Double-Angle, Half-Angle, and Reduction Formulas

1.

cos(2α)=732cos(2α)=732

2.

cos4θ−sin4θ=(cos2θ+sin2θ)(cos2θ−sin2θ)=cos(2θ)cos4θ−sin4θ=(cos2θ+sin2θ)(cos2θ−sin2θ)=cos(2θ)

3.

cos(2θ)cosθ=(cos2θ−sin2θ)cosθ=cos3θ−cosθsin2θcos(2θ)cosθ=(cos2θ−sin2θ)cosθ=cos3θ−cosθsin2θ

4.

10cos4x=10cos4x=10(cos2x)2 =10[1+cos(2x)2]2 =104[1+2cos(2x)+cos2(2x)] =104+102cos(2x)+104(1+cos2(2x)2) =104+102cos(2x)+108+108cos(4x) =308+5cos(2x)+108cos(4x) =154+5cos(2x)+54cos(4x)Substitute reduction formula for cos2x.Substitute reduction formula for cos2x.10cos4x=10cos4x=10(cos2x)2 =10[ 1+cos(2x)2 ]2Substitute reduction formula for cos2x. =104[1+2cos(2x)+cos2(2x)] =104+102cos(2x)+104(1+cos2(2x)2)Substitute reduction formula for cos2x. =104+102cos(2x)+108+108cos(4x) =308+5cos(2x)+108cos(4x) =154+5cos(2x)+54cos(4x)

5.

−25√−25

7.4 Sum-to-Product and Product-to-Sum Formulas

1.

12(cos6θ+cos2θ)12(cos6θ+cos2θ)

2.

12(sin2x+sin2y)12(sin2x+sin2y)

3.

−2−3√4−2−34

4.

2sin(2θ)cos(θ)2sin(2θ)cos(θ)

5.

tanθcotθ−cos2θ=(sinθcosθ)(cosθsinθ)−cos2θ=1−cos2θ=sin2θtanθcotθ−cos2θ=(sinθcosθ)(cosθsinθ)−cos2θ=1−cos2θ=sin2θ

7.5 Solving Trigonometric Equations

1.

x=7π6,11π6x=7π6,11π6

2.

π3±πkπ3±πk

3.

θ≈1.7722±2πkθ≈1.7722±2πk and θ≈4.5110±2πkθ≈4.5110±2πk

4.

cosθ=−1,θ=πcosθ=−1,θ=π

5.

π2,2π3,4π3,3π2π2,2π3,4π3,3π2

7.6 Modeling with Trigonometric Functions

1.

The amplitude is 3,3, and the period is 23.23.

2.

x	3sin(3x)3sin(3x)
0	0
π6π6	3
π3π3	0
π2π2	−3−3
2π32π3	0

3.

y=8sin(π12t)+32y=8sin(π12t)+32
The temperature reaches freezing at noon and at midnight.

4.

initial displacement =6, damping constant = -6, frequency = 2π2π

5.

y=10e−0.5tcos(πt)y=10e−0.5tcos(πt)

6.

y=5cos(6πt)y=5cos(6πt)

7.1 Section Exercises

1.

All three functions, FF, GG, and H,H, 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=0,cost=0, then sect=10,sect=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−sin2x√sinx±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

7.2 Section Exercises

1.

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x,x, the second angle measures π2−x.π2−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)=−sinx,sin(−x)=−sinx, so sinxsinx is odd. cos(−x)=cos(0−x)=cosx,cos(−x)=cos(0−x)=cosx, so cosxcosx is even.

5.

2√+6√42+64

7.

6√−2√46−24

9.

−2−3–√−2−3

11.

−2√2sinx−2√2cosx−22sinx−22cosx

13.

−12cosx−3√2sinx−12cosx−32sinx

15.

cscθcscθ

17.

cotxcotx

19.

tan(x10)tan(x10)

21.

sin(a−b)=(45)(13)−(35)(22√3)=4−62√15sin(a−b)=(45)(13)−(35)(223)=4−6215
cos(a+b)=(35)(13)−(45)(22√3)=3−82√15cos(a+b)=(35)(13)−(45)(223)=3−8215

23.

2√−6√42−64

25.

sinxsinx

27.

cot(π6−x)cot(π6−x)

29.

cot(π4+x)cot(π4+x)

31.

sinx2√+cosx2√sinx2+cosx2

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−tan2θ.g(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√−122√, or −0.2588−3−122, or −0.2588

45.

1+3√22√,1+322, 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.

7.3 Section Exercises

1.

Use the Pythagorean identities and isolate the squared term.

3.

1−cosxsinx,sinx1+cosx,1−cosxsinx,sinx1+cosx, multiplying the top and bottom by 1−cosx−−−−−−−−√1−cosx and 1+cosx−−−−−−−−√,1+cosx, respectively.

5.

a) 37√323732 b) 31323132 c) 37√313731

7.

a) 3√232 b) −12−12 c) −3–√−3

9.

cosθ=−25√5,sinθ=5√5,tanθ=−12,cscθ=5–√,secθ=−5√2,cotθ=−2cosθ=−255,sinθ=55,tanθ=−12,cscθ=5,secθ=−52,cotθ=−2

11.

2sin(π2)2sin(π2)

13.

2−2√√22−22

15.

2−3√√22−32

17.

2+3–√2+3

19.

−1−2–√−1−2

21.

a) 313√1331313 b) −213√13−21313 c) −32−32

23.

a) 10√4104 b) 6√464 c) 15√3153

25.

120169,–119169,–120119120169,–119169,–120119

27.

213√13,313√13,2321313,31313,23

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+cos2θ−sin2θtan2θ=2sin(θ)cos(θ)2cos2θtan2θ=sin(θ)cosθtan2θ=tanθtan2θ=tan3θsin(2θ)1+cos(2θ)tan2θ=2sin(θ)cos(θ)1+cos2θ−sin2θtan2θ=2sin(θ)cos(θ)2cos2θtan2θ=sin(θ)cosθtan2θ=tanθtan2θ=tan3θ

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)

7.4 Section Exercises

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=1.sin(3x)+sinxcosx=1. When converting the numerator to a product the equation becomes: 2sin(2x)cosxcosx=12sin(2x)cosxcosx=1

5.

8(cos(5x)−cos(27x))8(cos(5x)−cos(27x))

7.

sin(2x)+sin(8x)sin(2x)+sin(8x)

9.

12(cos(6x)−cos(4x))12(cos(6x)−cos(4x))

11.

2cos(5t)cost2cos(5t)cost

13.

2cos(7x)2cos(7x)

15.

2cos(6x)cos(3x)2cos(6x)cos(3x)

17.

14(1+3–√)14(1+3)

19.

14(3–√−2)14(3−2)

21.

14(3–√−1)14(3−1)

23.

cos(80°)−cos(120°)cos(80°)−cos(120°)

25.

12(sin(221°)+sin(205°))12(sin(221°)+sin(205°))

27.

2–√cos(31°)2cos(31°)

29.

2cos(66.5°)sin(34.5°)2cos(66.5°)sin(34.5°)

31.

2sin(−1.5°)cos(0.5°)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)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)=sinx+sin(3x)=2sin(4x2)cos(−2x2)=
2sin(2x)cosx=2(2sinxcosx)cosx=2sin(2x)cosx=2(2sinxcosx)cosx=
4sinxcos2x4sinxcos2x

37.

2tanxcos(3x)=2sinxcos(3x)cosx=2(.5(sin(4x)−sin(2x)))cosx2tanxcos(3x)=2sinxcos(3x)cosx=2(.5(sin(4x)−sin(2x)))cosx
=1cosx(sin(4x)−sin(2x))=secx(sin(4x)−sin(2x))=1cosx(sin(4x)−sin(2x))=secx(sin(4x)−sin(2x))

39.

2cos(35∘)cos(23∘), 1.50812cos(35∘)cos(23∘), 1.5081

41.

−2sin(33∘)sin(11∘),−0.2078−2sin(33∘)sin(11∘),−0.2078

43.

12(cos(99∘)−cos(71∘)),−0.241012(cos(99∘)−cos(71∘)),−0.2410

45.

It is and identity.

47.

It is not an identity, but 2cos3x2cos3x is.

49.

tan(3t)tan(3t)

51.

2cos(2x)2cos(2x)

53.

−sin(14x)−sin(14x)

55.

Start with cosx+cosy.cosx+cosy. Make a substitution and let x=α+βx=α+β and let y=α−β,y=α−β, so cosx+cosycosx+cosy becomes
cos(α+β)+cos(α−β)=cosαcosβ−sinαsinβ+cosαcosβ+sinαsinβ=2cosαcosβcos(α+β)+cos(α−β)=cosαcosβ−sinαsinβ+cosαcosβ+sinαsinβ=2cosαcosβ

Since x=α+βx=α+β and y=α−β,y=α−β, we can solve for αα and ββ in terms of x and y and substitute in for 2cosαcosβ2cosαcosβ and get 2cos(x+y2)cos(x−y2).2cos(x+y2)cos(x−y2).

57.

cos(3x)+cosxcos(3x)−cosx=2cos(2x)cosx−2sin(2x)sinx=−cot(2x)cotxcos(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=tanycos(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=4sin2xcosxcosx−cos(3x)=−2sin(2x)sin(−x)=2(2sinxcosx)sinx=4sin2xcosx

63.

tan(π4−t)=tan(π4)−tant1+tan(π4)tan(t)=1−tant1+tanttan(π4−t)=tan(π4)−tant1+tan(π4)tan(t)=1−tant1+tant

7.5 Section Exercises

1.

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

π3,2π3π3,2π3

7.

3π4,5π43π4,5π4

9.

π4,5π4π4,5π4

11.

π4,3π4,5π4,7π4π4,3π4,5π4,7π4

13.

π4,7π4π4,7π4

15.

7π6,11π67π6,11π6

17.

π18π18, 5π185π18, 13π1813π18, 17π1817π18, 25π1825π18, 29π1829π18

19.

3π12,5π12,11π12,13π12,19π12,21π123π12,5π12,11π12,13π12,19π12,21π12

21.

16,56,136,176,256,296,37616,56,136,176,256,296,376

23.

0,π3,π,5π30,π3,π,5π3

25.

π3,π,5π3π3,π,5π3

27.

π3,3π2,5π3π3,3π2,5π3

29.

0,π0,π

31.

π−sin−1(−14),7π6,11π6,2π+sin−1(−14)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))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.

00

37.

θ=sin−1(23),π−sin−1(23),π+sin−1(23),2π−sin−1(23)θ=sin-123,π-sin-123,π+sin-123,2π-sin-123

39.

3π2,π6,5π63π2,π6,5π6

41.

0,π3,π,4π30,π3,π,4π3

43.

There are no solutions.

45.

cos−1(13(1−7–√)),2π−cos−1(13(1−7–√))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))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.

0,2π3,4π30,2π3,4π3

55.

π4,3π4,5π4,7π4π4,3π4,5π4,7π4

57.

sin−1(35),π2,π−sin−1(35),3π2sin−1(35),π2,π−sin−1(35),3π2

59.

cos−1(−14)cos−1(−14), 2π−cos−1(−14)2π−cos−1(−14)

61.

π3,cos−1(−34),2π−cos−1(−34),5π3π3,cos−1(−34),2π−cos−1(−34),5π3

63.

cos−1(34),cos−1(−23),2π−cos−1(−23)cos−1(34),cos−1(−23),2π−cos−1(−23), 2π−cos−1(34)2π−cos−1(34)

65.

0,π2,π,3π20,π2,π,3π2

67.

π3,cos−1(−14),2π−cos−1(−14),5π3π3,cos−1(−14),2π−cos−1(−14),5π3

69.

There are no solutions.

71.

π+tan−1(−2)π+tan−1(−2), π+tan−1(−32),2π+tan−1(−2),2π+tan−1(−32)π+tan−1(−32),2π+tan−1(−2),2π+tan−1(−32)

73.

2πk+0.2734,2πk+2.86822πk+0.2734,2πk+2.8682

75.

πk−0.3277πk−0.3277

77.

0.6694,1.8287,3.8110,4.97030.6694,1.8287,3.8110,4.9703

79.

1.0472,3.1416,5.23601.0472,3.1416,5.2360

81.

0.5326,1.7648,3.6742,4.90640.5326,1.7648,3.6742,4.9064

83.

sin−1(14),π−sin−1(14),3π2sin−1(14),π−sin−1(14),3π2

85.

π2,3π2π2,3π2

87.

There are no solutions.

89.

0,π2,π,3π20,π2,π,3π2

91.

There are no solutions.

93.

7.2∘7.2∘

95.

5.7∘5.7∘

97.

82.4∘82.4∘

99.

31.0∘31.0∘

101.

88.7∘88.7∘

103.

59.0∘59.0∘

105.

36.9∘36.9∘

7.6 Section Exercises

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)−1y=−3cos(π6x)−1

7.

5sin(2x)+25sin(2x)+2

8.

y=4−6cos(xπ2)y=4-6cos(xπ2)

10.

y=tan(xπ8)y=tan(xπ8)

12.

tan(xπ12)tan(xπ12)

13.

15.

17.

75 °F

19.

8 a.m.

21.

2:49

23.

From June 15 through November 16

25.

From day 31 through day 58

27.

Floods: April 16 to July 15. Drought: October 16 to January 15.

29.

Amplitude: 8, period: 13,13, frequency: 3 Hz

31.

Amplitude: 4, period: 4,4, frequency: 1414 Hz

33.

P(t)=−19cos(π6t)+800+16012tP(t)=−19cos(π6t)+800+403tP(t)=−19cos(π6t)+800+16012tP(t)=−19cos(π6t)+800+403t

35.

P(t)=−33cos(π6t)+900+(1.07)tP(t)=−33cos(π6t)+900+(1.07)t

37.

D(t)=10(0.85)tcos(36πt)D(t)=10(0.85)tcos(36πt)

39.

D(t)=17(0.9145)tcos(28πt)D(t)=17(0.9145)tcos(28πt)

41.

6 years

43.

15.4 seconds

45.

Spring 2 comes to rest first after 7.3 seconds.

47.

234.3 miles, at 72.2°

49.

y=6(4)x+5sin(π2x)y=6(4)x+5sin(π2x)

51.

y=4(–2)x+8sin(π2x)y=4(–2)x+8sin(π2x)

53.

y=3(2)xcos(π2)+1y=3(2)xcos(π2x)+1

Review Exercises

1.

sin−1(3√3),π−sin−1(3√3),π+sin−1(3√3),2π−sin−1(3√3)sin−1(33),π−sin−1(33),π+sin−1(33),2π−sin−1(33)

3.

7π6,11π67π6,11π6

5.

sin−1(14),π−sin−1(14)sin−1(14),π−sin−1(14)

7.

11

9.

Yes

11.

−2−3–√−2−3

13.

2√222

15.

cos(4x)−cos(3x)cosx=cos(2x+2x)−cos(x+2x)cosx =cos(2x)cos(2x)−sin(2x)sin(2x)−cosxcos(2x)cosx+sinxsin(2x)cosx =(cos2x−sin2x)2−4cos2xsin2x−cos2x(cos2x−sin2x)+sinx(2)sinxcosxcosx =(cos2x−sin2x)2−4cos2xsin2x−cos2x(cos2x−sin2x)+2sin2xcos2x =cos4x−2cos2xsin2x+sin4x−4cos2xsin2x−cos4x+cos2xsin2x+2sin2xcos2x =sin4x−4cos2xsin2x+cos2xsin2x =sin2x(sin2x+cos2x)−4cos2xsin2x =sin2x−4cos2xsin2xcos(4x)−cos(3x)cosx=cos(2x+2x)−cos(x+2x)cosx =cos(2x)cos(2x)−sin(2x)sin(2x)−cosxcos(2x)cosx+sinxsin(2x)cosx =(cos2x−sin2x)2−4cos2xsin2x−cos2x(cos2x−sin2x)+sinx(2)sinxcosxcosx =(cos2x−sin2x)2−4cos2xsin2x−cos2x(cos2x−sin2x)+2sin2xcos2x =cos4x−2cos2xsin2x+sin4x−4cos2xsin2x−cos4x+cos2xsin2x+2sin2xcos2x =sin4x−4cos2xsin2x+cos2xsin2x =sin2x(sin2x+cos2x)−4cos2xsin2x =sin2x−4cos2xsin2x

17.

tan(58x)tan(58x)

19.

3√333

21.

−2425,−725,247−2425,−725,247

23.

2(2+2–√)−−−−−−−−−√2(2+2)

25.

2√10,72√10,17,35,45,34210,7210,17,35,45,34

27.

cotxcos(2x)=cotx(1−2sin2x) =cotx−cosxsinx(2)sin2x =−2sinxcosx+cotx =−sin(2x)+cotxcotxcos(2x)=cotx(1−2sin2x) =cotx−cosxsinx(2)sin2x =−2sinxcosx+cotx =−sin(2x)+cotx

29.

10sinx−5sin(3x)+sin(5x)8(cos(2x)+1)10sinx−5sin(3x)+sin(5x)8(cos(2x)+1)

31.

3√232

33.

−2√2−22

35.

12(sin(6x)+sin(12x))12(sin(6x)+sin(12x))

37.

2sin(132x)cos(92x)2sin(132x)cos(92x)

39.

3π4,7π43π4,7π4

41.

0,π6,5π6,π0,π6,5π6,π

43.

3π23π2

45.

No solution

47.

0.2527,2.8889,4.71240.2527,2.8889,4.7124

49.

1.36941.3694, 1.91061.9106, 4.37264.3726, 4.91374.9137

51.

3sin(xπ2)−23sin(xπ2)−2

53.

71.6∘71.6∘

55.

P(t)=950−450sin(π6t)P(t)=950−450sin(π6t)

57.

Amplitude: 3, period: 2, frequency: 1212 Hz

59.

C(t)=20sin(2πt)+100(1.4427)tC(t)=20sin(2πt)+100(1.4427)t

Practice Test

1.

1

3.

2√−6√42−64

5.

−2–√−3–√−2−3

7.

0,π0,π

9.

π2,3π2π2, 3π2

11.

2cos(3x)cos(5x)2cos(3x)cos(5x)

13.

x=cos–1(15)x=cos–1(15)

15.

35,−45,−3435,−45, −34

17.

tan3x–tanxsec2x=tanx(tan2x–sec2x)=tanx(tan2x–(1+tan2x))=tanx(tan2x–1–tan2x)=–tanx=tan(–x)=tan–x)tan3x–tan x sec2x=tanx(tan2x–sec2x) =tanx(tan2x–(1+tan2x)) =tanx(tan2x–1–tan2x) =–tanx=tan(–x)=tan–x)

19.

sin(2x)sinx–cos(2x)cosx=2sinxcosxsinx–2cos2x–1cosx=2cosx–2cosx+1cosx=1cosx=secx=secxsin(2x)sinx–cos(2x)cosx=2sin x cosxsinx–2cos2x–1cosx =2cosx–2cosx+1cosx =1cosx=secx=secx

21.

Amplitude: 1414 , period 160160 , frequency: 60 Hz

23.

Amplitude: 88 , fast period: 15001500 , fast frequency: 500 Hz, slow period: 110110 , slow frequency: 10 Hz

25.

D(t)=20(0.9086)tcos(4πt)D(t)=20(0.9086)tcos(4πt), 31 seconds