Search

Text Color

Margin Size

Font Type

Enable Dyslexic Font

17.3: Exercises

Last updated

May 2, 2022
Save as PDF
- 17.2: sin, cos, and tan as functions
- 18: Addition of Angles and Multiple Angles

Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

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

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

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

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

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

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

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

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

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

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

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

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

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

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

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

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

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

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

Exercise $\PageIndex{1}$

Find $\sin(x)$ , $\cos(x)$ , and $\tan(x)$ for the following angles.

$x=120^\circ$
$x=390^\circ$
$x=-150^\circ$
$x=-45^\circ$
$x=1050^\circ$
$x=-810^\circ$
$x=\dfrac{5\pi}{4}$
$x=\dfrac{5\pi}{6}$
$x=\dfrac{10\pi}{3}$
$x=\dfrac{15\pi}{2}$
$x=\dfrac{-\pi}{6}$
$x=\dfrac{-54\pi}{8}$

Answer

$\sin \left(120^{\circ}\right)=\dfrac{\sqrt{3}}{2}, \cos \left(120^{\circ}\right)=-\dfrac{1}{2}, \tan \left(120^{\circ}\right)=-\sqrt{3}$
$\sin \left(390^{\circ}\right)=\dfrac{1}{2}, \cos \left(390^{\circ}\right)=\dfrac{\sqrt{3}}{2}, \tan \left(390^{\circ}\right)=\dfrac{\sqrt{3}}{3}$
$\sin \left(-150^{\circ}\right)=-\dfrac{1}{2}, \cos \left(-150^{\circ}\right)=-\dfrac{\sqrt{3}}{2}, \tan \left(-150^{\circ}\right)=\dfrac{\sqrt{3}}{3}$
$\sin \left(-45^{\circ}\right)=-\dfrac{\sqrt{2}}{2}, \cos \left(-45^{\circ}\right)=\dfrac{\sqrt{2}}{2}, \tan \left(-45^{\circ}\right)=-1$
$\sin \left(1050^{\circ}\right)=-\dfrac{1}{2}, \cos \left(1050^{\circ}\right)=\dfrac{\sqrt{3}}{2}, \tan \left(1050^{\circ}\right)=-\dfrac{\sqrt{3}}{3}$
$\sin \left(-810^{\circ}\right)=-1, \cos \left(-810^{\circ}\right)=0, \tan \left(-810^{\circ}\right)$ is undefined
$\sin \left(\dfrac{5 \pi}{4}\right)=-\dfrac{\sqrt{2}}{2}, \cos \left(\dfrac{5 \pi}{4}\right)=-\dfrac{\sqrt{2}}{2}, \tan \left(\dfrac{5 \pi}{4}\right)=1$
$\sin \left(\dfrac{5 \pi}{6}\right)=\dfrac{1}{2}, \cos \left(\dfrac{5 \pi}{6}\right)=-\dfrac{\sqrt{3}}{2}, \tan \left(\dfrac{5 \pi}{6}\right)=-\dfrac{\sqrt{3}}{3}$
$\sin \left(\dfrac{10 \pi}{3}\right)=-\dfrac{\sqrt{3}}{2}, \cos \left(\dfrac{10 \pi}{3}\right)=-\dfrac{1}{2}, \tan \left(\dfrac{10 \pi}{3}\right)=\sqrt{3}$
$\sin \left(\dfrac{15 \pi}{2}\right)=-1, \cos \left(\dfrac{15 \pi}{2}\right)=0, \tan \left(\dfrac{15 \pi}{2}\right)$ is undefined
$\sin \left(\dfrac{-\pi}{6}\right)=-\dfrac{1}{2}, \cos \left(\dfrac{-\pi}{6}\right)=\dfrac{\sqrt{3}}{2}, \tan \left(\dfrac{-\pi}{6}\right)=-\dfrac{\sqrt{3}}{3}$
$\sin \left(\dfrac{-54 \pi}{8}\right)=-\dfrac{\sqrt{2}}{2}, \cos \left(\dfrac{-54 \pi}{8}\right)=-\dfrac{\sqrt{2}}{2}, \tan \left(\dfrac{-54 \pi}{8}\right)=1$

Exercise $\PageIndex{2}$

Graph the function, and describe how the graph can be obtained from one of the basic graphs $y=\sin(x)$ , $y=\cos(x)$ , or $y=\tan(x)$ .

$f(x)=\sin(x)+2$
$f(x)=\cos(x-\pi)$
$f(x)=\tan(x)-4$
$f(x)=5\cdot \sin(x)$
$f(x)=\cos(2\cdot x)$
$f(x)=\sin(x-2)-5$

Answer

shift $y = \sin(x)$ up by $2$ $clipboard_e87a9fec74db13512dc88d00c5a408b50.png$
$y = \cos(x)$ shifted to the right by $\pi$ $clipboard_ea49e58b7b0e659038bf9635b9ba96f8e.png$
$y = \tan(x)$ shifted down by $4$ $clipboard_e4568fbe75b058d26e053e360fb41df95.png$
$y = \sin(x)$ stretched away from the $x$ -axis by a factor $5$ $clipboard_ea70bf9f705034f50ed58be4f251a6281.png$
$y = \cos(x)$ compressed towards the $y$ -axis by a factor $2$ $clipboard_e080744b55078ea167a3afba862c78862.png$
$y = \sin(x)$ shifted to the right by $2$ and down by $5$ $clipboard_e7b1355ee1b59ad066326d55ba2459d2a.png$

Exercise $\PageIndex{3}$

Identify the formulas with the graphs.

$\begin{array}{lll} f(x)=\sin (x)+2, & g(x)=\tan (x-1), & h(x)=3 \sin (x), \\ i(x)=3 \cos (x), & j(x)=\cos (x-\pi), & k(x)=\tan (x)-1 \end{array} \nonumber$

$clipboard_ed9f7f6deab20c9a085dbd10a02f684dd.png$
$clipboard_e020683b23064247032aac69fc1c7a107.png$
$clipboard_e5469166b94ce909f729ca288a1aed298.png$
$clipboard_e06decdf4a038417807f651176e3dd831.png$
$clipboard_e51f2385d56e4e68d886ca212744a7316.png$
$clipboard_ecf741760cde398a86b7a8ec8776b2587.png$

Answer

$g(x)$
$h(x)$
$j(x)$
$k(x)$
$i(x)$
$f(x)$

Exercise $\PageIndex{4}$

Find the formula of a function whose graph is the one displayed below.

$clipboard_ea430c8007e6555ae5fafacf64920c0b9.png$
$clipboard_eef7dfc0304d8751afc660afd06d91a26.png$
$clipboard_e3cdb414f98f2d1e143a135131b7718ba.png$
$clipboard_ea748c6feb830be4ec4d0323292225119.png$
$clipboard_ec442cb9f2d0e69954a33c055884fafe1.png$
$clipboard_ecf5d99db8f8e0d021c4d07ffcfef9c6e.png$

Answer

$y = 5 \cos(x)$
$y = −5 \cos(x)$
$y = −5 \sin(x)$
$y = \cos(x) + 5$
$y = \sin(x) + 5$
$y = 2 \sin(x) + 3$

Exercise $\PageIndex{5}$

Find the amplitude, period, and phase-shift of the function.

$f(x)=5\sin(2x+3)$
$f(x)=\sin(\pi x-5)$
$f(x)=6\sin(4x)$
$f(x)=-2\cos\left(x+\dfrac{\pi}{4}\right)$
$f(x)=8\cos(2x-6)$
$f(x)=3\sin\left(\dfrac{x}{4}\right)$
$f(x)=-\cos(x+2)$
$f(x)=7\sin \left(\dfrac{2\pi}{5}x-\dfrac{6\pi}{5}\right)$
$f(x)=\cos(-2x)$

Answer

amplitude $5$ , period $\pi$ , phase-shift $\dfrac{−3}{2}$
amplitude $1$ , period $2$ , phase-shift $\dfrac 5 \pi$
amplitude $6$ , period $\dfrac \pi 2$ , phase-shift $0$
amplitude $2$ , period $\dfrac 2 \pi$ , phase-shift $\dfrac{−\pi}{4}$
amplitude $8$ , period $\pi$ , phase-shift $3$
amplitude $3$ , period $\dfrac 8 \pi$ , phase-shift $0$
amplitude $1$ , period $\dfrac 2 \pi$ , phase-shift $−2$
amplitude $7$ , period $5$ , phase-shift $3$
amplitude $1$ , period $\pi$ , phase-shift $0$

Exercise $\PageIndex{6}$

Find the amplitude, period, and phase-shift of the function. Use this information to graph the function over a full period. Label all maxima, minima, and zeros of the function.

$y=5\cos(2x)$
$y=4\sin(\pi x)$
$y=2\sin\left(\dfrac{2\pi}{3}x\right)$
$y=\cos(2x-\pi)$
$y=\cos(\pi x-\pi)$
$y=-6\cos(-\dfrac{x}{4})$
$y=-\cos(4x+\pi)$
$y=7\sin\left(x+\dfrac{\pi}{4}\right)$
$y=5\cos\left(x+\dfrac{3\pi}{2}\right)$
$y=4\sin(5x-\pi)$
$y=-3\cos(2\pi x-4)$
$y=7\sin\left(\dfrac 1 4 x+\dfrac{\pi}{4}\right)$
$y=\cos(3x-4\pi)$
$y=2\sin\big(\dfrac 1 5 x-\dfrac{\pi}{10}\big)$
$y=\dfrac 1 3 \cos\left(\dfrac{14}{5}x-\dfrac{6\pi}{5}\right)$

Answer

amplitude $5$ , period $\pi$ , phase-shift $0$ $clipboard_e113627ff02d07cc9bea67038b4d8a974.png$
amplitude $4$ , period $2$ , phase-shift $0$ $clipboard_edef39b269c155473be986ec93e2577c2.png$
amplitude $2$ , period $3$ , phase-shift $0$ $clipboard_e6ab47a8cab9e849a888344707a04ecda.png$
amplitude $1$ , period $\pi$ , phase-shift $\dfrac \pi 2$ $clipboard_e64231c2bd27d118a37761a9bddfafb9d.png$
amplitude $1$ , period $2$ , phase-shift $1$ $clipboard_e7ec6ef7ac4ef9917dd623cbfe2a5ef3d.png$
amplitude $6$ , period $\dfrac 8 \pi$ , phase-shift $0$ $clipboard_ed1774e04d9474194870114a705c9b522.png$
amplitude $1$ , period $\dfrac \pi 2$ , phase-shift $\dfrac {−\pi}{4}$ $clipboard_e3eb28effa7317eab90f2b0de9a48367d.png$
amplitude $7$ , period $\dfrac 2 \pi$ , phase-shift $\dfrac {−\pi}{4}$ $clipboard_ec0268bfd6521121da67ce54e9eba246b.png$
amplitude $5$ , period $\dfrac 2 \pi$ , phase-shift $\dfrac {−3\pi}{2}$ $clipboard_e41db0af7dfe18c464fc5480985a46120.png$
amplitude $4$ , period $\dfrac {2\pi}{5}$ , phase-shift $\dfrac \pi 5$ $clipboard_e1fcc37d96d30151b0cf20a260141868c.png$
amplitude $3$ , period $1$ , phase-shift $\dfrac 2 \pi$ $clipboard_e6675a6425361cf8d163fd4982cde4056.png$
amplitude $7$ , period $\dfrac 8 \pi$ , phase-shift $-\pi$ $clipboard_ea018243cb0d56c545263af605094776f.png$
amplitude $1$ , period $\dfrac {2\pi}{3}$ , phase-shift $\dfrac {4\pi}{3}$ $clipboard_e47af696f68a6368c441fb4b5a6f6c833.png$
amplitude $2$ , period $\dfrac 10 \pi$ , phase-shift $\dfrac \pi 2$ $clipboard_e32bbf0c4182ae4dc97ce0163f8be391b.png$
amplitude $\dfrac 1 3$ , period $\dfrac {5\pi}{7}$ , phase-shift $\dfrac {3\pi}{7}$ $clipboard_ef80d92a6baf65f7530888a753667da1c.png$