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Hint: We are going to solve the given problem using $$\cos^{-1} ({\cos }\theta ) = \theta $$ if $\left( {0 \leqslant \theta \leqslant \pi } \right)$

$\because $5 >$\pi $(radian measure), we have

${\cos ^{ - 1}}\left( {\cos 5} \right) = {\cos ^{ - 1}}\left\{ {\cos \left( {2\pi - 5} \right)} \right\}$

[ $\because \cos (2\pi - \theta ) = \cos \theta $ ]

${\cos ^{ - 1}}\left( {\cos 5} \right) = 2\pi - 5$

$\therefore $ The value of ${\cos ^{ - 1}}\left( {\cos 5} \right) = 2\pi - 5$

Note:

The measure of 5 radians lie in the fourth quadrant.

We have $$\cos^{-1} ({\cos }\theta ) = \theta $$ only for $\left( {0 \leqslant \theta \leqslant \pi } \right)$

So we converted that as $\cos 5 = cos\left( {2\pi - 5} \right)$. The value of $\left( {2\pi - 5} \right)$ is less than $\pi $.

$\because $5 >$\pi $(radian measure), we have

${\cos ^{ - 1}}\left( {\cos 5} \right) = {\cos ^{ - 1}}\left\{ {\cos \left( {2\pi - 5} \right)} \right\}$

[ $\because \cos (2\pi - \theta ) = \cos \theta $ ]

${\cos ^{ - 1}}\left( {\cos 5} \right) = 2\pi - 5$

$\therefore $ The value of ${\cos ^{ - 1}}\left( {\cos 5} \right) = 2\pi - 5$

Note:

The measure of 5 radians lie in the fourth quadrant.

We have $$\cos^{-1} ({\cos }\theta ) = \theta $$ only for $\left( {0 \leqslant \theta \leqslant \pi } \right)$

So we converted that as $\cos 5 = cos\left( {2\pi - 5} \right)$. The value of $\left( {2\pi - 5} \right)$ is less than $\pi $.

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