# How do you sketch the angle whose terminal side in standard position passes through (sqrt5,-sqrt15) and how do you find sin and cos?

Aug 25, 2016

$\sin a = - \frac{\sqrt{3}}{2}$
$\cos a = \frac{1}{2}$

#### Explanation:

Call M the point $M \left(\sqrt{5} , - \sqrt{15}\right)$
M is in Quadrant IV. On the trig unit circle, M is the terminal point of the arc a, that $\tan a = - \frac{\sqrt{15}}{\sqrt{5}} = - \sqrt{3}$
Trig table gives --> $a = - \frac{\pi}{3} \mathmr{and} a = \frac{5 \pi}{3}$ (co -terminal arcs).

$\sin a = \sin \left(- \frac{\pi}{3}\right) = - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$
$\cos a = \cos \left(- \frac{\pi}{3}\right) = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$