Answer
a) $\sin^{-1}(-\frac{1}{\sqrt2})=-\frac{\pi}{4}$
b) $\cos^{-1}\frac{\sqrt3}{2}=\frac{\pi}{6}$
Work Step by Step
a) let $\sin^{-1}(-\frac{1}{\sqrt2})=\theta, -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
$\sin\theta=-\frac{1}{\sqrt2}$
$\theta=-\frac{\pi}{4}+2k\pi, -\frac{3\pi}{4}+2k\pi$, where k is an integer
$\because -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2},$ let $k=0, \theta=-\frac{\pi}{4}$
$\therefore \sin^{-1}(-\frac{1}{\sqrt2})=-\frac{\pi}{4}$
b) let $\cos^{-1}\frac{\sqrt3}{2}=\theta, 0\leq\theta\leq\pi$
$\cos\theta=\frac{\sqrt3}{2}$
$\theta=\frac{\pi}{6}+2k\pi, -\frac{\pi}{6}+2k\pi$, where k is an integer
$\because 0\leq\theta\leq\pi, $ let $k=0, \theta=\frac{\pi}{6}$
$\therefore \cos^{-1}\frac{\sqrt3}{2}=\frac{\pi}{6}$
