Question

The value of ${\sin }^{-1}\left(\cos \frac{19\pi }{6}\right)$ is equal to

• A. $\frac{-\pi }{3}$
• B. $\frac{-\pi }{6}$
• C. $\frac{4\pi }{3}$
• D. $\frac{5\pi }{3}$

Answer 1

Answer: (A)

$\cos \left(\frac{19\pi }{6}\right)=\cos \left(\frac{7\pi }{6}\right)=\sin \left(\frac{-\pi }{3}\right)$
$\cos \left(\frac{7\pi }{6}\right)=\cos \left(\pi +\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{6}\right)=\frac{-\sqrt{3}}{2}$
$\therefore {\sin }^{-1}\left(\cos \frac{19\pi }{6}\right)={\sin }^{-1}\left(\sin \left(\frac{-\pi }{3}\right)\right)=\frac{-\pi }{3}$