# How do you evalute sin^-1 (-sqrt(3)/2)?

##### 1 Answer
Oct 23, 2014

Let

$\theta = {\sin}^{- 1} \left(- \frac{\sqrt{3}}{2}\right)$.

By rewriting it in terms of sine,

$R i g h t a r r o w \sin \theta = - \frac{\sqrt{3}}{2}$

So, we need to find the angle $\theta$ between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ that makes sine equal to $- \frac{\sqrt{3}}{2}$.

Since

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

we have

$\theta = {\sin}^{- 1} \left(- \frac{\sqrt{3}}{2}\right) = - \frac{\pi}{3}$.

I hope that this was helpful.