Question

The value of ${\cos }^{-1}\left(\cos \frac{13\pi }{6}\right)$ is

• A. $\frac{13\pi }{6}$
• B. $\frac{7\pi }{6}$
• C. $\frac{5\pi }{6}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is D $\frac{\pi }{6}$

${\cos }^{-1}\left(\cos \frac{13\pi }{6}\right)={\cos }^{-1}\left[\cos (2\pi +\frac{\pi }{6})\right]={\cos }^{-1}\left(\cos \frac{\pi }{6}\right)=\frac{\pi }{6}(\because {\cos }^{-1}(\cos x)=x,x\in [0,\pi \left]\right)$

Alternate solution:
$\because 2\pi \le \frac{13\pi }{6}\le 3\pi$From the above graph
${\cos }^{-1}\left(\cos x\right)=x-2\pi ,x\in [2\pi ,3\pi ]$
$\therefore {\cos }^{-1}\left(\cos \left(\frac{13\pi }{6}\right)\right)=\frac{13\pi }{6}-2\pi =\frac{\pi }{6}$