Find the value of cos ^{-1}left(cos frac{13 p | vedclass

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(A) We know that $\cos ^{-1}(\cos x) = x$ if $x \in [0, \pi]$,which is the principal value branch of $\cos ^{-1} x$. Here,$\frac{13 \pi}{6} 
otin [0,

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) We know that $\cos ^{-1}(\cos x) = x$ if $x \in [0, \pi]$,which is the principal value branch of $\cos ^{-1} x$. Here,$\frac{13 \pi}{6} 
otin [0, \pi]$. Now,$\cos ^{-1}\left(\cos \frac{13 \pi}{6}ight)$ can be written as: $\cos ^{-1}\left(\cos \frac{13 \pi}{6}ight) = \cos ^{-1}\left[\cos \left(2 \pi + \frac{\pi}{6}ight)ight]$. Since $\cos(2 \pi + 	heta) = \cos 	heta$,we have: $\cos ^{-1}\left[\cos \left(2 \pi + \frac{\pi}{6}ight)ight] = \cos ^{-1}\left[\cos \left(\frac{\pi}{6}ight)ight]$. Since $\frac{\pi}{6} \in [0, \pi]$,we get: $\cos ^{-1}\left[\cos \left(\frac{\pi}{6}ight)ight] = \frac{\pi}{6}$.

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

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