Browse Questions

# Find the value of the following:$cos^{-1} \frac {1} {2} +2 sin^{-1}\frac {1} {2}$

Ans : $\frac{\pi}{3}+2.\frac{\pi}{6}=\frac{2\pi}{3}$

Toolbox:
• The range of the principal value of $\sin^{-1}x$ is $\left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
• The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
• In such problems, we find the principal values of each of the inverse trignometric identities and arrive at the final answer.
Let $\sin^{-1}(\large\frac{1}{2}) = x$ $\Rightarrow \sin x = \large\frac{1}{2}$
$\Rightarrow \sin x = \large\frac{1}{x} = \sin \large\frac{\pi}{6}$

$\therefore$ $x =\large\frac{\pi}{6} \; \epsilon \; \left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
Let $\cos^{-1}(\large\frac{1}{2}) = y\:$ $\Rightarrow\: \cos y = \large\frac{1}{2}$
$\Rightarrow \cos y = \large\frac{1}{2} = \cos \large\frac{\pi}{3}$

$\therefore$ $y = \large\frac{\pi}{3} \; \epsilon \; \left [ 0,\pi \right ]$
$\cos^{-1}(\large\frac{1}{2}) + 2 \sin^{-1}(\large\frac{1}{2}) = y + 2x$
$\Rightarrow \large\frac{\pi}{3} + 2\large\frac{\pi}{6} = \large\frac{\pi}{3} + \large\frac{\pi}{3} = 2 \large\frac{\pi}{3}$
$\therefore$ $\cos^{-1}(\large\frac{1}{2}) + 2 \sin^{-1}(\large\frac{1}{2}) = 2 \large\frac{\pi}{3}$

edited Mar 15, 2013