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 $\frac {\pi}{6}$) = $\frac {\pi}{6}$