Question

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

Answer 1

$$\begin{gathered} {\cos }^{-1}\left[\sin \left({\cos }^{-1}\frac{1}{2}\right)\right] \\ ={\cos }^{-1}\left[\sin \left(\frac{\mathrm{\pi }}{3}\right)\right]\left(\cos \frac{\mathrm{\pi }}{3}=\frac{1}{2}\Rightarrow {\cos }^{-1}\frac{1}{2}=\frac{\mathrm{\pi }}{3}\right) \\ ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ =\frac{\mathrm{\pi }}{6}\left(\cos \frac{\mathrm{\pi }}{6}=\frac{\sqrt{3}}{2}\Rightarrow {\cos }^{-1}\frac{\sqrt{3}}{2}=\frac{\mathrm{\pi }}{6}\right) \end{gathered}$$

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

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