The value of cos ^{-1}left(cos frac{8 pi}{3}r | vedclass

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Question

(C) We know that the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$.Given expression is $\cos ^{-1}\left(\cos \frac{8 \pi}{3}\right)$.Since $\

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(C) We know that the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$.Given expression is $\cos ^{-1}\left(\cos \frac{8 \pi}{3}\right)$.Since $\frac{8 \pi}{3} = 2 \pi + \frac{2 \pi}{3}$,we have $\cos \frac{8 \pi}{3} = \cos \left(2 \pi + \frac{2 \pi}{3}\right) = \cos \frac{2 \pi}{3}$.Since $\frac{2 \pi}{3} \in [0, \pi]$,we have $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right) = \frac{2 \pi}{3}$.Therefore,the value is $\frac{2 \pi}{3}$.

The value of $\cos ^{-1}\left(\cos \frac{8 \pi}{3}\right)$ is

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