# How do you find cos(theta) if sin(theta)= 3/5 and 90 < theta < 180?

Mar 31, 2018

$\cos \theta = - \frac{4}{5}$

#### Explanation:

$90 < \theta < 180$ implies that we are in the second quadrant, where the cosine is negative and the sine is positive. So, keep in mind, $\cos \theta$ will be negative.

Recall the identity

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

This tells us that

${\cos}^{2} \theta = 1 - {\sin}^{2} \theta \to \cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$

In our situation, as we are in the second quadrant, $\cos \theta = - \sqrt{1 - {\sin}^{2} \theta}$

We have $\sin \theta = \frac{3}{5} , {\sin}^{2} \theta = {\left(\frac{3}{5}\right)}^{2} = \frac{9}{25}$

So,

$\cos \theta = - \sqrt{1 - \frac{9}{25}}$
$\cos \theta = - \sqrt{\frac{25}{25} - \frac{9}{25}}$

$\cos \theta = - \sqrt{\frac{16}{25}} = - \frac{4}{5}$