Question

Write the value of ${\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)$.

Answer 1

We have
$$\begin{gathered} {\cos }^{-1}\frac{1}{2}+2{\sin }^{-1}\frac{1}{2} \\ ={\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{3}\right)+2{\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{6}\right)\left[\because \mathrm{The}\mathrm{range}\mathrm{of}\mathrm{sine}\mathrm{is}\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right];\frac{\mathrm{\pi }}{6}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right] \\ \mathrm{and}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{cosine}\mathrm{is}\left[0,\mathrm{\pi }\right];\frac{\mathrm{\pi }}{3}\in \left[0,\mathrm{\pi }\right]\right] \\ =\frac{\mathrm{\pi }}{3}+2\left(\frac{\mathrm{\pi }}{6}\right) \\ =\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{3} \\ =\frac{2\mathrm{\pi }}{3} \end{gathered}$$
∴ ${\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{2\mathrm{\pi }}{3}$