The terminal side of theta in standard position contains (-8,-15), how do you find the exact values of the six trigonometric functions of theta?

May 23, 2018

Call t the arc (or angle). t lies in Quadrant 3
$\tan t = \frac{y}{x} = - \frac{15}{- 8} = \frac{15}{8}$
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{225}{64}} = \frac{64}{289}$
$\cos t = \pm \frac{8}{17}$
Since t lies in Quadrant 3, cos t is negative
$\cos t = - \frac{8}{17}$
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{64}{289} = \frac{225}{289}$
$\sin t = \pm \frac{15}{17}$
Since t is in Quadrant 3, sin t is negative
$\sin t = - \frac{15}{17}$
$\tan t = \sin \frac{t}{\cos t} = \left(- \frac{15}{17}\right) \left(- \frac{17}{8}\right) = \frac{15}{8}$
$\cot = \frac{1}{\tan} = \frac{8}{15}$
$\sec = \frac{1}{\cos} = - \frac{17}{8}$
$\csc t = \frac{1}{\sin} = - \frac{17}{15}$