Question

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

Answer 1

Given: ${\cos }^{-1}\left(\cos \frac{13\pi }{6}\right)$
We know that ${\cos }^{-1}\left(\cos x\right)=x,x\in [0,\pi ]$
Since $\frac{13\pi }{6}\notin [0,\pi ]$, which is the range of the principal value branch of ${\cos }^{-1}$
However, $\cos \left(\frac{13\pi }{6}\right)=\cos (2\pi +\frac{\pi }{6})=\cos \frac{\pi }{6}$
and $\frac{\pi }{6}\in [0,\pi ]$
Therefore ${\cos }^{-1}\left(\cos \frac{13\pi }{6}\right)={\cos }^{-1}\left(\cos \frac{\pi }{6}\right)=\frac{\pi }{6}$