# How do you find the exact value of cos^-1 (-sqrt2/2)?

Jun 28, 2015

${\cos}^{- 1} \left(- \frac{\sqrt{2}}{2}\right) = {135}^{o} \mathmr{and} {225}^{o}$
$\textcolor{w h i t e}{\text{XXXX}}$if we restrict the range to $\left[0 , {360}^{o}\right)$

#### Explanation:

$\left(- \frac{\sqrt{2}}{2}\right) = \left(- \frac{1}{\sqrt{2}}\right)$

If $\cos \left(\frac{1}{\sqrt{2}}\right) = \theta$ then $\cos \left(\theta\right) = \left(- \frac{1}{\sqrt{2}}\right)$

If $\theta \epsilon \left[0 , 2 \pi\right)$
we have the two possibilities indicated in the diagram below:

with $\theta$ (measuring from the positive x-axis) as either
${180}^{o} - {45}^{o} = {135}^{o}$
or
${180}^{o} + {45}^{o} = {225}^{o}$

For people who prefer their angles in radians that is
$\pi - \frac{\pi}{4} = \frac{3 \pi}{4}$
or
$\pi + \frac{\pi}{4} = \frac{5 \pi}{4}$