13.2.7: Chapter 7
- Page ID
- 117282
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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