Answer
$\frac{3\pi}{4}$
Work Step by Step
$g^{-1}(f(\frac{7\pi}{4}))=\cos^{-1}(\sin{\frac{7\pi}{4}})$.
$\sin{\frac{7\pi}{4}}=-\frac{\sqrt2}{2}$.
Thus $\cos^{-1}\left(\sin{\frac{7\pi}{4}}\right)=\cos^{-1}{-\frac{\sqrt2}{2}}=\frac{3\pi}{4}$, because $\cos{\frac{3\pi}{4}}=-\frac{\sqrt2}{2}$ and $\frac{3\pi}{4}$ is in the range of $\cos^{-1}{x}$, which is $[0,\pi].$
