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

Mar 14, 2017

See below

#### Explanation:

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

$\cos \theta = \frac{\sqrt{2}}{2}$

So in an imaginary right-angled triangle, the length of the $\text{hyp}$ is $2$ and the length of the $\text{adj}$ is $\sqrt{2}$. This means that the length of the $\text{opp}$ is $\sqrt{{2}^{2} - {\left(\sqrt{2}\right)}^{2}} = \sqrt{2}$.

Since the $\text{adj}$ and $\text{opp}$ are equal lengths, our triangle is isosceles. This means that it also has two angles of equal lengths. Since one angle is $\frac{\pi}{2}$, the other two must be $\frac{\pi}{4}$.

So if we say that $\theta = \frac{\pi}{4}$, then $\cos \left(\frac{\pi}{4}\right) = \text{adj"/"hyp} = \frac{\sqrt{2}}{2}$
$\therefore \frac{\pi}{4} = {\cos}^{-} 1 \left(\frac{\sqrt{2}}{2}\right)$