Solve this

We know

$\sin \left(\sin ^{-1} \theta\right)=\theta$ if $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$

(i) We have

$\sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6}$

(ii) We have

$\sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)=\sin ^{-1}\left\{\sin \left(\pi+\frac{\pi}{6}\right)\right\}$

$=\sin ^{-1}\left(\sin -\frac{\pi}{6}\right)$

$=-\frac{\pi}{6}$

(iii) We have

$\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{6}\right)\right\}$

$=\sin ^{-1}\left(\sin \frac{\pi}{6}\right)$

$=\frac{\pi}{6}$

(iv) We have

$\sin ^{-1}\left(\sin \frac{13 \pi}{7}\right)=\sin ^{-1}\left\{\sin \left(2 \pi-\frac{\pi}{7}\right)\right\}$

$=\sin ^{-1}\left(\sin -\frac{\pi}{7}\right)$

$=-\frac{\pi}{7}$

(v) We have

$\sin ^{-1}\left(\sin \frac{17 \pi}{8}\right)=\sin ^{-1}\left\{\sin \left(2 \pi+\frac{\pi}{8}\right)\right\}$

$=\sin ^{-1}\left(\sin \frac{\pi}{8}\right)$

$=\frac{\pi}{8}$

(vi) We have

$\sin ^{-1}\left(\sin -\frac{17 \pi}{8}\right)=\sin ^{-1}\left(-\sin \frac{17 \pi}{8}\right)$

$=\sin ^{-1}\left\{-\sin \left(2 \pi+\frac{\pi}{8}\right)\right\}$

$=\sin ^{-1}\left(-\sin \frac{\pi}{8}\right)$

$=\sin ^{-1}\left(\sin -\frac{\pi}{8}\right)$

$=-\frac{\pi}{8}$

(vii) We have

$\sin ^{-1}(\sin 3)=\sin ^{-1}\{\sin (\pi-3)\}$

$=\pi-3$

(viii)We have

$\sin ^{-1}(\sin 4)=\sin ^{-1}\{\sin (\pi-4)\}$

$=\pi-4$

(ix) We have

$\sin ^{-1}(\sin 12)=\sin ^{-1}\{\sin (-\pi+12)\}$

$=12-\pi$

(x) We have

$\sin ^{-1}(\sin 2)=\sin ^{-1}\{\sin (\pi-2)\}$

$=\pi-2$