Answer
$\dfrac{-\pi}{3}$
Work Step by Step
Let us substitute:
$f^{-1}(g (\dfrac{ 5 \pi}{6}))=\sin^{-1} (\cos (\dfrac{5 \pi}{6}))$
From the unit circle, we know that:
$\cos \dfrac{5\pi}{6}= (\dfrac{-\sqrt 3}{2})$
We know that the trigonometric function $\sin^{-1} (1)$ has the domain $[-1,1]$ and range $[-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$.
Therefore, $\sin^{-1} (\cos (\dfrac{5 \pi}{6}))=\sin^{-1} (\dfrac{-\sqrt 3}{2}) =\dfrac{-\pi}{3}$ because $\sin \dfrac{-\pi}{3}=\dfrac{-\sqrt 3}{2}$ and $\dfrac{-\pi}{3}$ lies in the range of $\sin^{-1} x$.
