text: # Find the value of
  
  Question:
  
  Find the value of $\cos ^{-1}\left(\cos \frac{15 \pi}{6}\right)$
  
  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 $\frac{\pi}{6} \in[0, \pi]$
  
  $\therefore \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}$