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, $\sec \theta = \frac{3\sqrt{2}}{4}$.
$\frac{3\sqrt{13}}{13}$
$\frac{12}{13}$
$-\frac{3}{4}$
$\frac{3}{4}$
$\frac{9\sqrt{97}}{97}$
$-\frac{2\sqrt{7}}{7}$
$\frac{6\sqrt{35}}{35}$
$\frac{\sqrt{157}}{12}$