# How do you evaluate arcsin(sqrt 2/2)?

May 24, 2015

$\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ is the length of one side of the isoceles right angled triangle with sides $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{2}}{2}$ and $1$, which has internal angles $\frac{\pi}{4}$, $\frac{\pi}{4}$ and $\frac{\pi}{2}$.

($\frac{\pi}{4}$ radians = ${45}^{o}$ and $\frac{\pi}{2}$ radians = ${90}^{o}$ if you prefer)

To show this is right angled, check with Pythagoras:

${\left(\frac{\sqrt{2}}{2}\right)}^{2} + {\left(\frac{\sqrt{2}}{2}\right)}^{2}$

$= {\sqrt{2}}^{2} / {2}^{2} + {\sqrt{2}}^{2} / {2}^{2}$

$= \frac{2}{4} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1 = {1}^{2}$

So since $\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ and $\frac{\pi}{4}$ is in the

required range for $\arcsin$ viz $- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, we find

$\arcsin \left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$