# How do you evaluate arcsin((sqrt3)/2) or sin(Arccos(-15/17))?

Apr 23, 2015

In this way.

The range of the function $y = \arcsin x$ is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, so there is only this solution:

$x = \frac{\pi}{3}$.

$\sin \left(\arccos \left(- \frac{15}{17}\right)\right) = \sin \alpha$

where $\alpha = \arccos \left(- \frac{15}{17}\right)$, and the angle is in the second quadrant because the range of the function $y = \arccos x$ is $\left[0 , \pi\right]$ and the value is negative.

So

$\cos \alpha = - \frac{15}{17}$ and than

$\sin \alpha = + \sqrt{1 - {\cos}^{2} \alpha} = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{289 - 225}{289}} =$

$= \sqrt{\frac{64}{289}} = \frac{8}{17}$

(with the $+$ because in the second quadrant the sinus is positive).