Answer
$-\frac{\pi}{3}$
Work Step by Step
$f^{-1}(g(\frac{5\pi}{6}))=\sin^{-1}(\cos{\frac{5\pi}{6}})$.
$\cos{\frac{5\pi}{6}}=-\frac{\sqrt3}{2}$.
Thus $\sin^{-1}\left(\cos{\frac{5\pi}{6}}\right)=\sin^{-1}{-\frac{\sqrt3}{2}}=-\frac{\pi}{3}$, because $\sin{-\frac{\pi}{3}}=-\frac{\sqrt3}{2}$ and $-\frac{\pi}{3}$ is in the range of $\sin^{-1}{x}$, which is $[-\frac{\pi}{2},\frac{\pi}{2}].$
