Question

(i) ${\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{6}\right)$
(ii) ${\sin }^{-1}\left(\sin \frac{7\mathrm{\pi }}{6}\right)$
(iii) ${\sin }^{-1}\left(\sin \frac{5\mathrm{\pi }}{6}\right)$
(iv) ${\sin }^{-1}\left(\sin \frac{13\mathrm{\pi }}{7}\right)$
(v) ${\sin }^{-1}\left(\sin \frac{17\mathrm{\pi }}{8}\right)$
(vi) ${\sin }^{-1}\left\{\left(\sin -\frac{17\mathrm{\pi }}{8}\right)\right\}$
(vii) ${\sin }^{-1}\left(\sin 3\right)$
(viii) ${\sin }^{-1}\left(\sin 4\right)$
(ix) ${\sin }^{-1}\left(\sin 12\right)$
(x) ${\sin }^{-1}\left(\sin 2\right)$

Answer 1

We know

$\sin \left({\sin }^{-1}\theta \right)=\theta \mathrm{if}-\frac{\mathrm{\pi }}{2}\le \theta \le \frac{\mathrm{\pi }}{2}$
(i) We have

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

(ii) We have

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

(iii) We have

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

(iv) We have

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

(v) We have

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

(vi) We have

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

(vii) We have

$$\begin{gathered} {\sin }^{-1}\left(\sin 3\right)={\sin }^{-1}\left\{\sin \left(\mathrm{\pi }-3\right)\right\} \\ =\mathrm{\pi }-3 \end{gathered}$$

(viii)We have

$$\begin{gathered} {\sin }^{-1}\left(\sin 4\right)={\sin }^{-1}\left\{\sin \left(\mathrm{\pi }-4\right)\right\} \\ =\mathrm{\pi }-4 \end{gathered}$$

(ix) We have

$$\begin{gathered} {\sin }^{-1}\left(\sin 12\right)={\sin }^{-1}\left\{\sin \left(-\mathrm{\pi }+12\right)\right\} \\ =12-\mathrm{\pi } \end{gathered}$$

(x) )We have

$$\begin{gathered} {\sin }^{-1}\left(\sin 2\right)={\sin }^{-1}\left\{\sin \left(\mathrm{\pi }-2\right)\right\} \\ =\mathrm{\pi }-2 \end{gathered}$$