Question

${\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)$ is equal to

• A. $\frac{\mathrm{\pi }}{4}$
• B. $\frac{\mathrm{\pi }}{6}$
• C. $\frac{\mathrm{\pi }}{3}$
• D. $\frac{2\mathrm{\pi }}{3}$

Answer 1

The correct option is D

$\frac{2\mathrm{\pi }}{3}$

Explanation for the correct option:

Given, ${\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)$

$={\cos }^{-1}\left(\cos \frac{\pi }{3}\right)+2{\sin }^{-1}\left(\sin \frac{\pi }{6}\right)$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{\pi }}{\mathbf{6}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\right]$

$=\frac{\pi }{3}+2\left(\frac{\pi }{6}\right)$

$=\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{3}$

$=\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{3}}$

Hence, Option ‘D’ is Correct.