# How do you evaluate Cos(Arc sin (5/13))?

Jun 23, 2016

$\pm \frac{12}{13}$

#### Explanation:

Let $a = a r c \sin \left(\frac{5}{13}\right)$. Then, $\sin a = \frac{5}{13} > 0$. a is in either 1st

quadrant or in the 2nd. Accordingly, the given expression

$\cos a = \pm \sqrt{1 - {\sin}^{2} a} = \pm \sqrt{1 - {5}^{2} / {13}^{2}} = \pm \frac{12}{13}$,

The negative value is from $\sin \left(\pi - a\right) = \sin a = \frac{5}{13} \mathmr{and} \cos \left(\pi - a\right) = - \cos a$.

This is off, if arc sin (5/13) is restricted to be the principal value in

$\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.