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Find the value of the following:

    1. \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )

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If x \epsilon [0,\pi]  then  \cos^{-1}(\cos x) = x , which is principal value of \cos^{-1} x.

So, we have \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )

 where \:\frac{13\pi}{6} \notin \left [ 0, \pi \right ].

Hence\: we \:can\: write\: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) \:as

=\cos^{-1}\left (\cos\left(2\pi+\frac{\pi}{6} \right ) \right )

=\cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right )

   \frac{\pi}{6}\ \epsilon \left [ 0, \pi \right ]

Therefore we have,

\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) = \cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right ) = \frac{\pi}{6}.

Posted by

Divya Prakash Singh

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