# How do you sketch the angle whose terminal side in standard position passes through (-9,-40) and how do you find sin and cos?

Jun 26, 2018

$\sin t = - \frac{40}{41}$
$\cos t = - \frac{9}{41}$

#### Explanation:

The point (x = -9, y = -40) is on the terminal side of the angle t, that lies in Quadrant 3.
tan t = y/x = -40/-9 = 40/9
To find cos t, use trig identity:
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{1600}{81}} = \frac{81}{1681}$
$\cos t = - \frac{9}{41}$ (because t lies in Quadrant 3).
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{81}{1681} = \frac{1600}{1681}$
$\sin t = - \frac{40}{41}$ (because t lies in Quadrant 3)
Another way to find sin t:
$\sin t = \tan t . \cos t = \left(- \frac{40}{9}\right) \left(- \frac{9}{41}\right) = \frac{40}{41}$.