The value of

Question:

The value of $\cos ^{-1}\left\{\sin \left(\cos ^{-1} \frac{1}{2}\right)\right\}$ is ________________________.

Solution:

$\cos ^{-1}\left[\sin \left(\cos ^{-1} \frac{1}{2}\right)\right]$

$=\cos ^{-1}\left[\sin \left(\frac{\pi}{3}\right)\right]$                              $\left(\cos \frac{\pi}{3}=\frac{1}{2} \Rightarrow \cos ^{-1} \frac{1}{2}=\frac{\pi}{3}\right)$

$=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$=\frac{\pi}{6}$                                                     $\left(\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \Rightarrow \cos ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{6}\right)$

Thus, the value of $\cos ^{-1}\left[\sin \left(\cos ^{-1} \frac{1}{2}\right)\right]$ is $\frac{\pi}{6}$.

The value of $\cos ^{-1}\left\{\sin \left(\cos ^{-1} \frac{1}{2}\right)\right\}$ is $\frac{\pi}{6}$

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