# How do you calculate arcsin (-1/ sqrt 2)?

Jul 20, 2015

Use properties of right angled triangle with sides $\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$ and $1$ to find $\arcsin \left(- \frac{1}{\sqrt{2}}\right) = - \frac{\pi}{4}$

#### Explanation:

$\arcsin \left(- \frac{1}{\sqrt{2}}\right) = \alpha$ where $\alpha \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ such that $\sin \left(\alpha\right) = - \frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}$ is one side of the right angled triangle with sides:

$\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$ and $1$

(Note ${\left(\frac{1}{\sqrt{2}}\right)}^{2} + {\left(\frac{1}{\sqrt{2}}\right)}^{2} = \frac{1}{2} + \frac{1}{2} = 1 = {1}^{2}$)

This triangle has angles $\frac{\pi}{4}$, $\frac{\pi}{4}$ and $\frac{\pi}{2}$.

So $\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Now $\sin \left(- \theta\right) = - \sin \left(\theta\right)$

So $\sin \left(- \frac{\pi}{4}\right) = - \frac{1}{\sqrt{2}}$

So $\arcsin \left(- \frac{1}{\sqrt{2}}\right) = - \frac{\pi}{4}$