Evaluate cos-1(cos 13π/6)

Given \( cos^{-1}(cos \dfrac{13\pi}{6})\)

But we know that  \( cos \dfrac{13\pi}{6}=\dfrac{ \sqrt{3}}{2}\)

By substituting these values in \( cos^{-1}(cos \dfrac{13\pi}{6}) \)we get,

Cos^-1 \( \dfrac{ \sqrt{3}}{2}\)

Now let y = cos^-1 \( \dfrac{ \sqrt{3}}{2}\)

Therefore cos y = \( \dfrac{ \sqrt{3}}{2}\)

 \(Cos (\dfrac{\pi}{6}) = \dfrac{ \sqrt{3}}{2}\)

Hence range of principal value of cos^-1 is [0, π] and \( Cos (\dfrac{\pi}{6}) = \dfrac{ \sqrt{3}}{2}\)

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