10.5E: Exercises

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Exercises

In Exercises 1 - 24, graph one cycle of the given function. State the period, amplitude, phase shift and vertical shift of the function.

$y = 3\sin(x)$
$y = \sin(3x)$
$y = -2\cos(x)$
$y = \cos \left( x - \dfrac{\pi}{2} \right)$
$y = -\sin \left( x + \dfrac{\pi}{3} \right)$
$y = \sin(2x - \pi)$
$y = -\dfrac{1}{3}\cos \left( \dfrac{1}{2}x + \dfrac{\pi}{3} \right)$
$y = \cos (3x - 2\pi) + 4$
$y = \sin \left( -x - \dfrac{\pi}{4} \right) - 2$
$y = \dfrac{2}{3} \cos \left( \dfrac{\pi}{2} - 4x \right) + 1$
$y = -\dfrac{3}{2} \cos \left( 2x + \dfrac{\pi}{3} \right) - \dfrac{1}{2}$
$y = 4\sin (-2\pi x + \pi)$

In Exercises 13 - 24, graph one cycle of the given function. State the period of the function.

$y = \tan \left(x - \dfrac{\pi}{3} \right)$
$y = 2\tan \left( \dfrac{1}{4}x \right) - 3$
$y = \dfrac{1}{3}\tan(-2x - \pi) + 1$
$y = \sec \left( x - \dfrac{\pi}{2} \right)$
$y = -\csc \left( x + \dfrac{\pi}{3} \right)$
$y = -\dfrac{1}{3} \sec \left( \dfrac{1}{2}x + \dfrac{\pi}{3} \right)$
$y = \csc (2x - \pi)$
$y = \sec(3x - 2\pi) + 4$
$y = \csc \left( -x - \dfrac{\pi}{4} \right) - 2$
$y = \cot \left( x + \dfrac{\pi}{6} \right)$
$y = -11\cot \left( \dfrac{1}{5} x \right)$
$y = \dfrac{1}{3} \cot \left( 2x + \dfrac{3\pi}{2} \right) + 1$

In Exercises 25 - 34, show that the function is a sinusoid by rewriting it in the forms $C(x) = A \cos(\omega x + \phi) + B$ and $S(x) = A \sin(\omega x + \phi) + B$ for $\omega > 0$ and $0 \leq \phi < 2\pi$.

$f(x) = \sqrt{2}\sin(x) + \sqrt{2}\cos(x) + 1$
$f(x) = 3\sqrt{3}\sin(3x) - 3\cos(3x)$
$f(x) = -\sin(x) + \cos(x) - 2$
$f(x) = -\dfrac{1}{2}\sin(2x) - \dfrac{\sqrt{3}}{2}\cos(2x)$
$f(x) = 2\sqrt{3} \cos(x) - 2\sin(x)$
$f(x) = \dfrac{3}{2} \cos(2x) - \dfrac{3\sqrt{3}}{2} \sin(2x) + 6$
$f(x) = -\dfrac{1}{2} \cos(5x) -\dfrac{\sqrt{3}}{2} \sin(5x)$
$f(x) = -6\sqrt{3} \cos(3x) - 6\sin(3x) - 3$
$f(x) = \dfrac{5\sqrt{2}}{2} \sin(x) -\dfrac{5\sqrt{2}}{2} \cos(x)$
$f(x) =3 \sin \left(\dfrac{x}{6}\right) -3\sqrt{3} \cos \left(\dfrac{x}{6}\right)$
In Exercises 25 - 34, you should have noticed a relationship between the phases $\phi$ for the $S(x)$ and $C(x)$. Show that if $f(x) = A \sin(\omega x + \alpha) + B$, then $f(x) = A \cos(\omega x + \beta) + B$ where $\beta = \alpha - \dfrac{\pi}{2}$.
Let $\phi$ be an angle measured in radians and let $P(a,b)$ be a point on the terminal side of $\phi$ when it is drawn in standard position. Use Theorem 10.2.4 and the sum identity for sine to show that $f(x) = a \, \sin(\omega x) + b\, \cos(\omega x) + B$ (with $\omega > 0$) can be rewritten as $f(x) = \sqrt{a^{2} + b^{2}}\sin(\omega x + \phi) + B$.
With the help of your classmates, express the domains of the functions in Examples 10.5.4 and 10.5.5 using extended interval notation.

In Exercises 38 - 43, verify the identity by graphing the right and left hand sides on a calculator.

$\sin^{2}(x) + \cos^{2}(x) = 1$
$\sec^{2}(x) - \tan^{2}(x) = 1$
$\cos(x) = \sin\left(\dfrac{\pi}{2} - x\right)$
$\tan(x+\pi) = \tan(x)$
$\sin(2x) = 2\sin(x)\cos(x)$
$\tan\left(\dfrac{x}{2}\right) = \dfrac{\sin(x)}{1+\cos(x)}$

In Exercises 44 - 50, graph the function with the help of your calculator and discuss the given questions with your classmates.

$f(x) = \cos(3x) + \sin(x)$. Is this function periodic? If so, what is the period?
$f(x) = \frac{\sin(x)}{x}$. What appears to be the horizontal asymptote of the graph?
$f(x) = x \sin(x)$. Graph $y = \pm x$ on the same set of axes and describe the behavior of $f$.
$f(x) = \sin\left(\frac{1}{x}\right)$. What’s happening as $x \rightarrow 0$?
$f(x) = x - \tan(x)$. Graph $y = x$ on the same set of axes and describe the behavior of $f$.
$f(x) = e^{-0.1x} \left( \cos(2x) + \sin(2x)\right)$. Graph $y = \pm e^{-0.1x}$ on the same set of axes and describe the behavior of $f$.
$f(x) = e^{-0.1x} \left( \cos(2x) + 2\sin(x)\right)$. Graph $y = \pm e^{-0.1x}$ on the same set of axes and describe the behavior of $f$.
Show that a constant function $f$ is periodic by showing that $f(x + 117) = f(x)$ for all real numbers $x$. Then show that $f$ has no period by showing that you cannot find a smallest number $p$ such that $f(x + p) = f(x)$ for all real numbers $x$. Said another way, show that $f(x + p) = f(x)$ for all real numbers $x$ for ALL values of $p > 0$, so no smallest value exists to satisfy the definition of ‘period’.

Answers

$y = 3\sin(x)$
Period: $2\pi$
Amplitude: $3$
Phase Shift: $0$
Vertical Shift: $0$

$Screen Shot 2022-05-13 at 4.08.51 PM.png$
$y = \sin(3x)$
Period: $\dfrac{2\pi}{3}$
Amplitude: $1$
Phase Shift: $0$
Vertical Shift: $0$

$Screen Shot 2022-05-13 at 4.10.05 PM.png$
$y = -2\cos(x)$
Period: $2\pi$
Amplitude: $2$
Phase Shift: $0$
Vertical Shift: $0$

$Screen Shot 2022-05-13 at 4.10.54 PM.png$
$y = \cos \left( x - \dfrac{\pi}{2} \right)$
Period: $2\pi$
Amplitude: $1$
Phase Shift: $\dfrac{\pi}{2}$
Vertical Shift: $0$

$Screen Shot 2022-05-13 at 4.33.01 PM.png$
$y = -\sin \left( x + \dfrac{\pi}{3} \right)$
Period: $2\pi$
Amplitude: $1$
Phase Shift: $-\dfrac{\pi}{3}$
Vertical Shift: $0$

$Screen Shot 2022-05-13 at 4.33.49 PM.png$
$y = \sin(2x - \pi)$
Period: $\pi$
Amplitude: $1$
Phase Shift: $\dfrac{\pi}{2}$
Vertical Shift: $0$

$Screen Shot 2022-05-13 at 4.36.54 PM.png$
$y = -\dfrac{1}{3}\cos \left( \dfrac{1}{2}x + \dfrac{\pi}{3} \right)$
Period: $4\pi$
Amplitude: $\dfrac{1}{3}$
Phase Shift: $-\dfrac{2\pi}{3}$
Vertical Shift: $0$

$Screen Shot 2022-05-13 at 4.37.31 PM.png$
$y = \cos (3x - 2\pi) + 4$
Period: $\dfrac{2\pi}{3}$
Amplitude: $1$
Phase Shift: $\dfrac{2\pi}{3}$
Vertical Shift: 4

$Screen Shot 2022-05-13 at 4.39.27 PM.png$
$y = \sin \left( -x - \dfrac{\pi}{4} \right) - 2$
Period: $2\pi$
Amplitude: $1$
Phase Shift: $-\dfrac{\pi}{4}$ (You need to use
$y = -\sin \left( x + \dfrac{\pi}{4} \right) - 2$ to find this.)¹⁵
Vertical Shift: $-2$

$Screen Shot 2022-05-13 at 4.40.28 PM.png$
$y = \dfrac{2}{3} \cos \left( \dfrac{\pi}{2} - 4x \right) + 1$
Period: $\dfrac{\pi}{2}$
Amplitude: $\dfrac{2}{3}$
Phase Shift: $\dfrac{\pi}{8}$ (You need to use
$y = \dfrac{2}{3} \cos \left( 4x - \dfrac{\pi}{2} \right) + 1$ to find this.)¹⁶
Vertical Shift: $1$

$Screen Shot 2022-05-13 at 4.41.47 PM.png$
$y = -\dfrac{3}{2} \cos \left( 2x + \dfrac{\pi}{3} \right) - \dfrac{1}{2}$
Period: $\pi$
Amplitude: $\dfrac{3}{2}$
Phase Shift: $-\dfrac{\pi}{6}$
Vertical Shift: $-\dfrac{1}{2}$

$Screen Shot 2022-05-13 at 4.42.32 PM.png$
$y = 4\sin (-2\pi x + \pi)$
Period: $1$
Amplitude: $4$
Phase Shift: $\dfrac{1}{2}$ (You need to use
$y = -4\sin (2\pi x - \pi)$ to find this.)¹⁷
Vertical Shift: $0$

$Screen Shot 2022-05-13 at 4.56.14 PM.png$
$y = \tan \left(x - \dfrac{\pi}{3} \right)$
Period: $\pi$

$Screen Shot 2022-05-13 at 4.59.18 PM.png$
$y = 2\tan \left( \dfrac{1}{4}x \right) - 3$
Period: $4\pi$

$Screen Shot 2022-05-14 at 5.13.37 PM.png$
$y = \dfrac{1}{3}\tan(-2x - \pi) + 1$
is equivalent to
$y = -\dfrac{1}{3}\tan(2x + \pi) + 1$
via the Even / Odd identity for tangent.
Period: $\dfrac{\pi}{2}$

$Screen Shot 2022-05-14 at 5.16.30 PM.png$
$y = \sec \left( x - \frac{\pi}{2} \right)$
Start with $y = \cos \left( x - \frac{\pi}{2} \right)$
Period: $2\pi$

$Screen Shot 2022-05-14 at 5.17.09 PM.png$
$y = -\csc \left( x + \dfrac{\pi}{3} \right)$
Start with $y = -\sin \left( x + \dfrac{\pi}{3} \right)$
Period: $2\pi$

$Screen Shot 2022-05-14 at 5.17.47 PM.png$
$y = -\dfrac{1}{3} \sec \left( \dfrac{1}{2}x + \dfrac{\pi}{3} \right)$
Start with $y = -\dfrac{1}{3}\cos \left( \dfrac{1}{2}x + \dfrac{\pi}{3} \right)$
Period: $4\pi$

$Screen Shot 2022-05-14 at 5.18.29 PM.png$
$y = \csc (2x - \pi)$
Start with $y = \sin(2x - \pi)$
Period: $\pi$

$Screen Shot 2022-05-14 at 5.19.07 PM.png$
$y = \sec(3x - 2\pi) + 4$
Start with $y = \cos (3x - 2\pi) + 4$
Period: $\dfrac{2\pi}{3}$

$Screen Shot 2022-05-14 at 5.20.58 PM.png$
$y = \csc \left( -x - \dfrac{\pi}{4} \right) - 2$
Start with $y = \sin \left( -x - \dfrac{\pi}{4} \right) - 2$
Period: $2\pi$

$Screen Shot 2022-05-14 at 5.21.40 PM.png$
$y = \cot \left( x + \dfrac{\pi}{6} \right)$
Period: $\pi$

$Screen Shot 2022-05-14 at 5.22.22 PM.png$
$y = -11\cot \left( \dfrac{1}{5} x \right)$
Period: $5\pi$

$Screen Shot 2022-05-14 at 5.23.00 PM.png$
$y = \dfrac{1}{3} \cot \left( 2x + \dfrac{3\pi}{2} \right) + 1$
Period: $\dfrac{\pi}{2}$

$Screen Shot 2022-05-14 at 5.23.36 PM.png$
$f(x) = \sqrt{2}\sin(x) + \sqrt{2}\cos(x) + 1 = 2\sin\left(x + \dfrac{\pi}{4}\right) + 1 = 2\cos\left(x + \dfrac{7\pi}{4}\right) + 1$
$f(x) = 3\sqrt{3}\sin(3x) - 3\cos(3x) = 6\sin\left(3x + \dfrac{11\pi}{6}\right) = 6\cos\left(3x + \dfrac{4\pi}{3}\right)$
$f(x) = -\sin(x) + \cos(x) - 2 = \sqrt{2}\sin\left(x + \dfrac{3\pi}{4}\right) - 2 = \sqrt{2}\cos\left(x + \dfrac{\pi}{4}\right) - 2$
$f(x) = -\dfrac{1}{2}\sin(2x) - \dfrac{\sqrt{3}}{2}\cos(2x) = \sin\left(2x + \dfrac{4\pi}{3}\right) = \cos\left(2x + \dfrac{5\pi}{6}\right)$
$f(x) = 2\sqrt{3} \cos(x) - 2\sin(x) = 4\sin\left(x+\dfrac{2\pi}{3} \right) = 4\cos\left(x + \dfrac{\pi}{6}\right)$
$f(x) = \dfrac{3}{2} \cos(2x) - \dfrac{3\sqrt{3}}{2} \sin(2x) + 6 =3\sin\left(2x + \dfrac{5\pi}{6}\right) + 6 = 3\cos\left(2x + \dfrac{\pi}{3}\right) + 6$
$f(x) = -\dfrac{1}{2} \cos(5x) -\dfrac{\sqrt{3}}{2} \sin(5x) = \sin\left(5x + \dfrac{7\pi}{6}\right) = \cos\left(5x + \dfrac{2\pi}{3}\right)$
$f(x) = -6\sqrt{3} \cos(3x) - 6\sin(3x) - 3 = 12\sin\left(3x + \dfrac{4\pi}{3}\right) - 3 = 12\cos\left(3x + \dfrac{5\pi}{6}\right) - 3$
$f(x) = \dfrac{5\sqrt{2}}{2} \sin(x) -\dfrac{5\sqrt{2}}{2} \cos(x) = 5\sin\left(x + \dfrac{7\pi}{4}\right)= 5\cos\left(x + \dfrac{5\pi}{4}\right)$
$f(x) =3\sin\left(\dfrac{x}{6}\right) -3\sqrt{3} \cos\left(\dfrac{x}{6}\right) = 6\sin\left( \dfrac{x}{6}+\dfrac{5\pi}{3}\right)= 6\cos\left( \dfrac{x}{6}+\dfrac{7\pi}{6}\right)$

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