Skip to main content
Mathematics LibreTexts

13.2.7: Chapter 7

  • Page ID
    117282
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    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

    Graph of y=3sin(3x) using the five key points: intervals of equal length representing 1/4 of the period. Here, the points are at 0, pi/6, pi/3, pi/2, and 2pi/3. 3.

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

    Graph of the function y=8sin(pi/12 t) + 32 for temperature. The midline is at 32. The times when the temperature is at 32 are midnight and noon.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

    Graph of y=sin(x) from -2pi to 2pi.27.

    cot(π6−x)cot(π6−x)

    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√+cosx2√sinx2+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)=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.

    6a2d68e5d517a87a1389e657248f0b4cf158be55

    15.

    9cc3241a9fa38c26dbe57f003f8d88232d1cdb6b

    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


    13.2.7: Chapter 7 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?