Find the value of the following:    1. $\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )$

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,

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

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