Evaluate cos-1(cos 4π/3)

Given \( cos^{-1}(cos \dfrac{4\pi}{3})\)

But we know that \( cos \dfrac{4\pi}{3}=\dfrac{-1}{2}\)

By substituting these values in \( cos^{-1}(cos \dfrac{4\pi}{3})\)we get,

\(cos^{-1}( \dfrac{-1}{2})\)

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

Therefore cos y = \(- \dfrac{1}{2}\)

\( – Cos (\dfrac{\pi}{3}) = \dfrac{1}{2}\)

\( Cos (\pi-\dfrac{\pi}{3}) =- \dfrac{1}{2}\)

\( Cos (\dfrac{2\pi}{3}) = -\dfrac{1}{2}\)

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

Therefore \( cos^{-1}(cos \dfrac{4\pi}{3})=\dfrac{2\pi}{3}\)