# Given the point P(-12/13, -5/13), how do you find sintheta and costheta?

Mar 6, 2018

$\sin t = - \frac{5}{13}$
$\cos t = - \frac{12}{13}$

#### Explanation:

$\tan t = \frac{y}{x} = \frac{- \frac{5}{13}}{- \frac{12}{13}} = \frac{5}{12}$.
P and t are located in Quadrant 3.
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{25}{144}} = \frac{144}{169}$
$\cos t = \pm \frac{12}{13}$
Since t is in Quadrant 3, therefor, cos t is negative
$\cos t = - \frac{12}{13}$.
$\sin t = \tan t . \cos t = \left(\frac{5}{12}\right) \left(- \frac{12}{13}\right) = - \frac{5}{13}$