# How do you evaluate sin^-1(sin((13pi)/10))?

Jul 30, 2016

$- \frac{3 \pi}{10}$

#### Explanation:

The inverse sine function has domain $\left[- 1 , 1\right]$ which means it will have range $- \frac{\pi}{2} \le y \le \frac{\pi}{2}$

This means that any solutions we obtain must lie in this interval.

As a consequence of double angle formulae, $\sin \left(x\right) = \sin \left(\pi - x\right)$ so

$\sin \left(\frac{13 \pi}{10}\right) = \sin \left(- \frac{3 \pi}{10}\right)$

Sine is $2 \pi$ periodic so we can say that

${\sin}^{- 1} \left(\sin \left(x\right)\right) = x + 2 n \pi , n \in \mathbb{Z}$

However any solutions must lie in the interval $- \frac{\pi}{2} \le y \le \frac{\pi}{2}$.

There is no integer multiple of $2 \pi$ we can add to $\frac{13 \pi}{10}$ to get it within this interval so the only solution is $- \frac{3 \pi}{10}$.