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Draw the given angle as a rotation about the origin in the Cartesian plane:
$60^{\circ}$
$135^{\circ}$
$1140^{\circ}$
$-240^{\circ}$
$\frac{\pi}{3}$
$\frac{7\pi}{6}$
$-\frac{\pi}{6}$
$\frac{5\pi}{2}$
Convert from degrees to radians:
$45^{\circ}$
$270^{\circ}$
$1^{\circ}$
$-230^{\circ}$
$540^{\circ}$
$\frac{\pi}{4}$
$\frac{3\pi}{2}$
$\frac{\pi}{180}$
$-\frac{23\pi}{18}$
$3\pi$
Convert from radians to degrees:
$\frac{2\pi}{3}$
$7\pi$
$\frac{\pi}{6}$
$\frac{19\pi}{2}$
$120^{\circ}$
$1260^{\circ}$
$30^{\circ}$
$1710^{\circ}$
Find the values of the following:
$\cos 5\pi$
$\sin(-\frac{7\pi}{6})$
$\cos \frac{23\pi}{4}$
$\sin 9\pi$
$\cos (-\frac{10\pi}{3})$
$\sin (-\frac{4\pi}{3})$
$\cot \frac{13\pi}{6}$
$\tan \frac{9\pi}{2}$
$\csc (-\frac{\pi}{6})$
$\tan \frac{23\pi}{4}$
$\sec \frac{10\pi}{3}$
$\csc 5\pi$
$\cot (-\frac{5\pi}{4})$
$\sin 150^{\circ}$
$\sec (-120^{\circ})$
$\csc 495^{\circ}$
$-1$
$\frac{1}{2}$
$\frac{\sqrt{2}}{2}$
$0$
$-\frac{1}{2}$
$\frac{\sqrt{3}}{2}$
$\sqrt{3}$
does not exist (not in the domain)
$-2$
$-1$
$-2$
does not exist (not in the domain)
$-1$
$\frac{1}{2}$
$-2$
$\sqrt{2}$
Make a table giving the values of all six trigonometric functions for $t = 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{3\pi}{4},\frac{5\pi}{4},\pi,\frac{7\pi}{6},\frac{5\pi}{4},\frac{4\pi}{3},\frac{3\pi}{2},\frac{5\pi}{3},\frac{7\pi}{4},\frac{11\pi}{6},\textrm{ and } 2\pi$.
Find the value of
$\sin \theta$, if $\cot \theta = \frac{3}{4}$ and $\pi \lt \theta \lt \frac{3\pi}{2}$
$\sec \theta$, if $\csc \theta = -3$ and $\frac{3\pi}{2} \lt \theta \lt 2\pi$
$\cos t$, if $\tan t = -\frac{2}{3}$ and $\frac{3\pi}{2} \lt t \lt 2\pi$
$\sin t$, if $\sec t = \frac{13}{5}$ and $0 \lt t \lt \frac{\pi}{2}$
$\tan t$, if $\csc t = \frac{5}{3}$ and $\frac{\pi}{2} \lt t \lt \pi$
$\cot t$, if $\csc t = \frac{5}{4}$ and $0 \lt t \lt \frac{\pi}{2}$
$\sin \theta$, if $\cot \theta = -\frac{4}{9}$ and $\frac{\pi}{2} \lt \theta \lt \pi$
$\cos \theta$, if $\tan \theta = \frac{\sqrt{3}}{2}$ and $\pi \lt \theta \lt \frac{3\pi}{2}$
$\sec \theta$, if $\sin \theta = -\frac{1}{6}$ and $\frac{3\pi}{2} \lt \theta \lt 2\pi$
$\csc \theta$, if $\cot \theta = -\frac{\sqrt{13}}{12}$ and $\frac{\pi}{2} \lt \theta \lt \pi$
As $\cot \theta = \frac{3}{4}$, we know $\frac{\cos \theta}{\sin \theta} = \frac{3}{4}$, so if $x = \sin \theta$, then $\cos \theta = \frac{3x}{4}$. By the Pythagorean identity, $x^2 + \frac{9x^2}{16} = 1$. Solving for $x$ yields $\pm \frac{4}{5}$, but the restriction that $\pi \lt \theta \lt \frac{3\pi}{2}$ tells us $\theta$ is in quadrant III, where the sine is negative. Thus, $\sin \theta = -\frac{4}{5}$
As $\csc \theta = -3$, we know $\sin \theta = -\frac{1}{3}$. Then by the Pythagorean identity, $\cos^2 \theta + \frac{1}{9} = 1$. Solving for $\cos \theta$, we find $\cos \theta = \pm \sqrt{1 - \frac{1}{9}} = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3}$. Thus, $\sec \theta = \pm \frac{3}{2\sqrt{2}} = \pm \frac{3\sqrt{2}}{4}$. Noting the restriction that $\frac{3\pi}{2} \lt \theta \lt 2\pi$, we realize $\theta$ is in quadrant IV, where the cosine (and hence, secant) are positive. Therefore, $\sec \theta = \frac{3\sqrt{2}}{4}$.
$\frac{3\sqrt{13}}{13}$
$\frac{12}{13}$
$-\frac{3}{4}$
$\frac{3}{4}$
$\frac{9\sqrt{97}}{97}$
$-\frac{2\sqrt{7}}{7}$
$\frac{6\sqrt{35}}{35}$
$\frac{\sqrt{157}}{12}$
Show whether each of the following equations is or is not an identity:
$\displaystyle{\frac{\sin \theta}{\cos \theta} = 1 - \frac{\cos \theta}{\sin \theta}}$
$\displaystyle{1 - \cos^4 \theta = (2 - \sin^2 \theta) \sin^2 \theta}$
$\displaystyle{1 - 2\sin^2 \theta = 2\cos^2 \theta - 1}$
$\displaystyle{\frac{\sec \theta - \csc \theta}{\sec \theta + \csc \theta} = \frac{\tan \theta + 1}{\tan \theta - 1}}$
$\displaystyle{\frac{\sec^4 t - \tan^4 t}{1 - 2\tan^2 t} = 1}$
$\displaystyle{\sin^2 \theta \cot^2 \theta + \cos^2 \theta \tan^2 \theta = 1}$
$\displaystyle{\sec \theta - \frac{\cos \theta}{1 + \sin \theta} = \cot \theta}$
$\displaystyle{\frac{\tan^2 x}{1 + \cos x} = \frac{\sec x - 1}{\cos x}}$
$\displaystyle{(\csc t - \cot t)^2 = \frac{1 - \cos t}{1 + \cos t}}$
$\displaystyle{1 + \frac{1}{\cos \theta} = \frac{\tan^2 \theta}{\sec \theta - 1}}$
not an identity
identity
identity
not an identity
not an identity
identity
not an identity
identity
identity
identity
Graph the following for $-2\pi \le x \le 2\pi$.
$y = 4\cos x$
$y = \sin \frac{2}{3} x$
$y = 4\cos(2x - \frac{3\pi}{2})$
$y = \sin(x - \frac{\pi}{6})$
$y= -\frac{1}{2} \sin x$
Graph the following.
$y = -\frac{8}{5} \cos (\frac{x}{5} + \frac{\pi}{3})$ over $[-5\pi,10\pi]$
$y = 4\sin(2x - \frac{\pi}{6})$ over $[-\pi,2\pi]$
$y = \frac{5}{2} \cos (2x + \frac{\pi}{4})$ from $-\pi$ to $\pi$
$y = \cos(x + \frac{\pi}{4})$ from $-2\pi$ to $2\pi$
Graph the following.
$y = 1 + \cos x$ for $-2\pi \le x \le 2\pi$
$y = 2 - \sin x$ from $-\pi$ to $\frac{3\pi}{2}$
$y = 2 + 2\sin(\frac{x}{3} - \frac{\pi}{6})$ from $-\pi$ to $2\pi$
$y = 2 - 3\cos 2x$ over $[-2\pi,\pi]$ (omit finding the $x$-intercepts)
Graph the following. Label interecepts and other important features (e.g., asymptotes)
$y = -\tan x$ over $[-2\pi,2\pi]$
$y = -\sec x$ from $-\pi$ to $\pi$
$y = \frac{1}{2} \tan 2x$ from $-\frac{5\pi}{4}$ to $\frac{3\pi}{8}$
$y = \csc 3x$ for $-\frac{\pi}{2} \le x \le \frac{5\pi}{6}$
$y = 2\tan \frac{x}{2}$ from $-3\pi$ to $\frac{5\pi}{2}$
$y = -\csc(4x+\pi)$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$
Find all solutions of the following equations:
$\tan x = 0$
$2 \cos x + \sqrt{2} = 0$
$\cos^2 x - 1 = 0$
$2\cos^2 x - 3\cos x - 2 = 0$
$\tan^2 x + (\sqrt{3} - 1)\tan x - \sqrt{3} = 0$
$3\sec^2 x = \sec x$
$2\sin^2 x - \sin x - 1 = 0$
$\cos 2x = \sin x$
$\displaystyle{\frac{1+\cos x}{\cos x} = 2}$
$\displaystyle{\sqrt{\frac{1+2 \sin x}{2}} = 1}$
$0 \pm \pi n, \quad (n = 0, 1, 2, \ldots)$
$\displaystyle{\left. \begin{array}{c} \frac{3\pi}{4} \pm 2\pi n\\ \frac{5\pi}{4} \pm 2\pi n \end{array} \right\} \quad (n = 0, 1, 2, \ldots)}$
$\displaystyle{\left. \begin{array}{c} 0 \pm 2\pi n\\ \pi \pm 2\pi n \end{array} \right\} \quad (n = 0, 1, 2, \ldots)}$ or more compactly, $\pm \pi n, \quad (n = 0, 1, 2, \ldots)$
$\displaystyle{\left. \begin{array}{c} \frac{2\pi}{3} \pm 2\pi n\\ \frac{4\pi}{3} \pm 2\pi n \end{array} \right\} \quad (n = 0, 1, 2, \ldots)}$
$\displaystyle{\left. \begin{array}{c} \frac{\pi}{4} \pm \pi n\\ \frac{2\pi}{3} \pm \pi n \end{array} \right\} \quad (n = 0, 1, 2, \ldots)}$
no solutions
$\displaystyle{\left. \begin{array}{c} \frac{7\pi}{6} \pm 2\pi n\\ \frac{11\pi}{6} \pm 2\pi n\\ \frac{\pi}{2} \pm 2\pi n \end{array} \right\} \quad (n = 0, 1, 2, \ldots)}$
$\displaystyle{\left. \begin{array}{c} \frac{\pi}{6} \pm 2\pi n\\ \frac{5\pi}{6} \pm 2\pi n\\ \frac{3\pi}{2} \pm 2\pi n \end{array} \right\} \quad (n = 0, 1, 2, \ldots)}$
$0 \pm 2\pi n, \quad (n = 0, 1, 2, \ldots)$
$\displaystyle{\left. \begin{array}{c} \frac{\pi}{6} \pm \pi n\\ \frac{5\pi}{6} \pm \pi n \end{array} \right\} \quad (n = 0, 1, 2, \ldots)}$
Find the values of the following:
$\arcsin 0$
$\textrm{arccot}\, (-\frac{\sqrt{3}}{3})$
$\cot(\textrm{arccot}\,(-3))$
$\cos(\arccos \frac{4}{5})$
$\textrm{arccsc}\, 2$
$\arccos (-1)$
$\csc(\arcsin \frac{3}{5})$
$\textrm{arcsec}\, (\sin \frac{\pi}{2})$
$\sin (\textrm{arcsec}\, 2)$
$\arccos (-\frac{1}{2})$
$\arctan^3 (-\sqrt{3})$
$3\arcsin^2 (\frac{\sqrt{3}}{2})$
$\textrm{arcsec}\, 0$
$\sin(\arctan 2)$
$\arccos(\sin(-\frac{\pi}{6}))$
$\tan(\arccos(-\frac{2}{3}))$
$\arccos 2$
$\cos(\arcsin(-\frac{4}{5}))$
$4 \arctan 1$
$\csc(\textrm{arcsec}\, 12)$
$\textrm{arccsc}\, \sqrt{2}$
$\textrm{arcsec}\, 2$
$\arctan(\sin \frac{\pi}{2})$
$\arctan(\cos \pi)$
$\arcsin (\tan \frac{\pi}{4})$
$0$
$\frac{2\pi}{3}$
$-3$
$\frac{4}{5}$
$\frac{\pi}{6}$
$\pi$
$\frac{5}{3}$
$0$
$\frac{\sqrt{3}}{2}$
$\frac{2\pi}{3}$
$-\frac{\pi^3}{27}$
$\frac{\pi^2}{3}$
no value
$\frac{2}{\sqrt{5}}$
$\frac{2\pi}{3}$
$-\frac{\sqrt{5}}{2}$
no value
$\frac{3}{5}$
$\pi$
$\frac{12}{\sqrt{143}}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$-\frac{\pi}{4}$
$\frac{\pi}{2}$
Write the given expression in terms of $x$ without any trigonometric functions:
$\sin(\arctan x)$
$\tan(\arcsin x)$
$\cot(\arcsin x)$
$\cos(\arcsin x)$
$\cos(\textrm{arcsec}\, x)$
$\csc(\textrm{arccot}\, \frac{x}{4})$
$\displaystyle{\frac{x}{\sqrt{x^2 + 1}}}$
$\displaystyle{\frac{x}{\sqrt{1-x^2}}}$
$\displaystyle{\frac{\sqrt{1-x^2}}{x}}$
$\displaystyle{\sqrt{1-x^2}}$
$\displaystyle{\frac{1}{x}}$
$\displaystyle{\frac{\sqrt{x^2 + 16}}{4}}$