Browse Questions

Find the principal values of the following: $cos^{-1} \bigg(\large\frac { \sqrt 3} {2}\bigg)$

$\begin{array}{1 1} -\frac{\pi}{6} \\ \frac{\pi}{6} \\ \frac{\pi}{\sqrt 3} \\ \frac{\pi}{2} \end{array}$

Ans : $cos^{-1} \bigg( cos\large\frac{\pi}{6} \bigg) = \large\frac{\pi}{6}$

edited Mar 15, 2013

Toolbox:
• The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
Let $\cos^{-1}(\large\frac{\sqrt 3}{2}) = x$ $\Rightarrow \cos x = \large\frac{\sqrt 3}{2}$
We know that the range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
Therefore, $\cos x = \large\frac{\sqrt 3}{2} = \cos \large\frac{\pi}{6}$
$\Rightarrow x = \large\frac{\pi}{6}$, where $x \;\epsilon \; \left [ 0,\pi \right ]$
Therefore, the principal value of $\cos^{-1}(\large\frac{\sqrt 3}{2})$ is $\large\frac{\pi}{6}$

edited Mar 15, 2013