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

$\begin{array}{1 1} \frac{3\pi}{4} \\ -\frac{3\pi}{4} \\ \frac{\pi}{4} \\ - \frac{\pi}{4} \end{array}$

Ans : $$cos^{-1} \bigg( cos \bigg( \pi-\frac{\pi}{4} \bigg) \bigg) = \frac{3\pi}{4}$$
answered Feb 22, 2013

Toolbox:
• The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
• $\cos (\pi - x) = -cos\; x$
Let $\cos^{-1}-(\frac{1}{\sqrt 2}) = x$ $\Rightarrow \cos x = \frac{-1}{\sqrt 2}$
The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$
Therefore, $\cos x = \frac{-1}{\sqrt 2} = - \cos \frac{\pi}{4}$
Because $\cos (\pi - x) = -cos\; x$, $\cos x = cos (\pi - \frac{\pi}{4}) = \cos \frac{3\pi}{4}$
$\Rightarrow x = \frac{3 \pi}{4}$, where $x \;\epsilon \; \left [ 0,\pi \right ]$
Therefore, the principal value of $\cos^{-1} (\frac{-1}{\sqrt 2})$ is $\frac{3\pi}{4}$
answered Mar 2, 2013