# How do you simplify sin^-1 (cos ((7pi)/6))?

${\sin}^{- 1} \left(\cos \left(\frac{7 \pi}{6}\right)\right) = - \frac{\pi}{3}$
As $\cos \left(\frac{7 \pi}{6}\right) = \cos \left(\pi + \frac{\pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right)$
= $- \sin \left(\frac{\pi}{2} - \frac{\pi}{6}\right) = - \sin \left(\frac{\pi}{3}\right) = \sin \left(- \frac{\pi}{3}\right)$
Hence, ${\sin}^{- 1} \left(\cos \left(\frac{7 \pi}{6}\right)\right) = - \frac{\pi}{3}$