# How do you solve and find the value of sin^-1(1/sqrt2)?

Mar 2, 2017

The angle = $\frac{\pi}{2} = {45}^{\circ}$
${\sin}^{-} 1 \left(\frac{1}{\sqrt{2}}\right)$ says, give me the angle that has a sin of $\left(\frac{1}{\sqrt{2}}\right) = \frac{\sqrt{2}}{2}$
Either from a trig circle or from a ${45}^{\circ} - {45}^{\circ} - {90}^{\circ}$ triangle, that angle would be either $\frac{\pi}{4} = {45}^{\circ}$ or $\frac{3 \pi}{4} = {135}^{\circ}$. However, the $\arcsin$ $\left({\sin}^{-} 1\right)$ function has a limited domain $\left[- 1 , 1\right]$and range $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right] = \left[- {90}^{\circ} , {90}^{\circ}\right]$. This means $\frac{3 \pi}{4} = {135}^{\circ}$ is not a valid answer.
You can see this from the graph $f \left(x\right) = \arcsin \left(x\right) = {\sin}^{-} 1 \left(x\right)$: