Question

(i) ${\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{\sqrt{2}}$
(ii) ${\sin }^{-1}\left\{\cos \left({\sin }^{-1}\frac{\sqrt{3}}{2}\right)\right\}$

Answer 1

(i)

$$\begin{gathered} {\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{\sqrt{2}}={\sin }^{-1}\frac{1}{2}-{\sin }^{-1}2\times \frac{1}{\sqrt{2}}\sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^2} \\ ={\sin }^{-1}\frac{1}{2}-{\sin }^{-1}\sqrt{2}\times \frac{1}{\sqrt{2}} \\ ={\sin }^{-1}\frac{1}{2}-{\sin }^{-1}1 \\ ={\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{6}\right)-{\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{2}\right) \\ =\frac{\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{2} \\ =-\frac{\mathrm{\pi }}{3} \end{gathered}$$
(ii)
$$\begin{gathered} {\sin }^{-1}\left\{\cos \left({\sin }^{-1}\frac{\sqrt{3}}{2}\right)\right\}={\sin }^{-1}\left\{\cos \left({\sin }^{-1}\sin \frac{\mathrm{\pi }}{3}\right)\right\} \\ ={\sin }^{-1}\left\{\cos \left(\frac{\mathrm{\pi }}{3}\right)\right\} \\ ={\sin }^{-1}\left\{\frac{1}{2}\right\} \\ ={\sin }^{-1}\left\{\sin \frac{\mathrm{\pi }}{6}\right\} \\ =\frac{\mathrm{\pi }}{6} \end{gathered}$$