Question

${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)$ is equal to

• A. $\frac{7\pi }{6}$
• B. $\frac{5\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (B)

${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)={\cos }^{-1}\left[\cos \left(2\pi -\frac{5\pi }{6}\right)\right]$
where, $\frac{5\pi }{6}\in [0,\pi ]$
$\Rightarrow {\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)={\cos }^{-1}\left[\cos \left(\frac{5\pi }{6}\right)\right]$
$[\because \cos (2\pi -\theta )=\cos \theta ]$