True or False: The principal value of $\sin^{-1}\begin{bmatrix}\cos\bigg(\sin^{-1}\frac{1}{2}\bigg)\end{bmatrix}\;is\;\Large {\frac{\pi}{3}}$

$\begin{array}{1 1} \large {\frac{\pi}{3}} \\ \large {\frac{\pi}{6}} \\ \large {\frac{2\pi}{3}} \\ \large {\frac{\pi}{4}} \end{array}$

Toolbox:
• $$sin\large\frac{\pi}{6}=\large\frac{1}{2}$$
• $$cos\large\frac{\pi}{6}=\large\frac{\sqrt3}{2}$$
• $$sin\large\frac{\pi}{3}=\large\frac{\sqrt3}{2}$$
Ans - True
$$sin^{-1}\large\frac{1}{2}=sin^{-1}sin\large\frac{\pi}{6}=\large\frac{\pi}{6}$$
$$cos(sin^{-1}\large\frac{1}{2})=cos\large\frac{\pi}{6}=\large\frac{\sqrt 3}{2}$$
$$sin^{-1}\bigg(cos(sin^{-1}\large\frac{1}{2})\bigg)=sin^{-1}\large\frac{\sqrt3}{2}$$

But $$\large\frac{\sqrt3}{2}=sin\large\frac{\pi}{3}$$
$$\Rightarrow\:sin^{-1}\large\frac{\sqrt3}{2}=sin^{-1}sin\large\frac{\pi}{3}=\large\frac{\pi}{3}$$
$$\Rightarrow\:sin^{-1}\bigg(cos(sin^{-1}\large\frac{1}{2})\bigg)=sin^{-1}sin\large\frac{\pi}{3}=\large\frac{\pi}{3}$$

answered Feb 18, 2013
edited Mar 19, 2013