# How do you find the exact values of costheta and sintheta when tantheta=-1/2?

Dec 21, 2016

$\sin t = \pm \frac{\sqrt{5}}{5}$
$\cos t = \pm \frac{2 \sqrt{5}}{5}$

#### Explanation:

Use trig identity:
${\cos}^{2} x = \frac{1}{1 + {\tan}^{2} x}$
${\cos}^{2} t = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \left(\frac{4}{5}\right)$
$\cos t = \pm \frac{2}{\sqrt{5}} = \pm \frac{2 \sqrt{5}}{5}$
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{4}{5} = \frac{5 - 4}{5} = \frac{1}{5}$
$\sin t = \pm \frac{1}{\sqrt{5}} = \pm \frac{\sqrt{5}}{5}$
Because tan $t = - \frac{1}{2}$, then t is in Quadrant II, or in Quadrant IV.
sin t and cos t in these 2 Quadrants have opposite signs.
a. If t is in Quadrant II --> $\sin t = \frac{\sqrt{5}}{5}$, and $\cos t = - \frac{2 \sqrt{5}}{5}$, and $\tan t = - \frac{1}{2}$
b. If t is in Quadrant IV --> $\sin t = - \frac{\sqrt{5}}{5}$, and $\cos t = \frac{2 \sqrt{5}}{5}$, and $\tan t = - \frac{1}{2}$.