# Find the value of

Question:

Find the value of

Solution:

We know that $\cos ^{-1}(\cos x)=x$ if $x \in[0, \pi]$, which is the principal value branch of $\cos ^{-1} x$.

Here, $\frac{13 \pi}{6} \notin[0, \pi]$.

Now, $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ can be written as:

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

$\therefore \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}$