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

${\sin}^{-} 1 \left(\sin \left(- \frac{\pi}{10}\right)\right) = - \frac{\pi}{10}$
${\sin}^{-} 1 \left(\sin \left(- \frac{\pi}{10}\right)\right)$
The restriction for the range of ${\sin}^{-} 1 x$ is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ and since $- \frac{\pi}{10}$ is in the range we can apply the property f^-1(f(x)=x. That is,
${\sin}^{-} 1 \left(\sin \left(- \frac{\pi}{10}\right)\right) = - \frac{\pi}{10}$