Question

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

Answer 1

We know that cos⁻¹(cos x) = x if x ∈ [0, \(\pi\)], which is the principal value branch of cos⁻¹x.
Here, \(\frac {13\pi}{6}\) ∉ [0, \(\pi\)]. 
Now,cos^-1(cos\(\frac {13\pi}{6}\)) can be written as: 
cos^-1(cos\(\frac {13\pi}{6}\)) = cos^-1(cos(2\(\pi\)+\(\frac {\pi}{6}\))) = cos^-1(cos \(\frac {\pi}{6}\)), where \(\frac {\pi}{6}\) ∈ [0, \(\pi\)]. 

Therefore cos^-1(cos\(\frac {13\pi}{6}\)) = cos^-1(cos \(\frac {\pi}{6}\)) = \(\frac {\pi}{6}\)