Mark the tick against the correct answer in the following:
The value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ is
A. $\frac{13 \pi}{6}$
B.
C. $\frac{5 \pi}{6}$
D. $\frac{\pi}{6} \frac{7 \pi}{6}$
To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$
Now, let $x=\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$
$\Rightarrow \cos x=\cos \left(\frac{13 \pi}{6}\right)$
Here, range of principle value of $\cos$ is $[0, \pi]$
$\Rightarrow x=\frac{13 \pi}{6} \notin[0, \pi]$
Hence for all values of $x$ in range $[0, \pi]$, the value of
$\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$ is
$\Rightarrow \cos x=\cos \left(2 \pi-\frac{\pi}{6}\right)\left(\because \cos \left(\frac{13 \pi}{6}\right)=\cos \left(2 \pi-\frac{\pi}{6}\right)\right)$
$\Rightarrow \cos x=\cos \left(\frac{\pi}{6}\right)(\because \cos (2 \pi-\theta)=\cos \theta)$
$\Rightarrow x=\frac{\pi}{6}$
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