# How do you find the solutions to sin^-1x=cos^-1x?

Sep 30, 2016

$x = \frac{\sqrt{2}}{2}$

#### Explanation:

By definition:

${\sin}^{- 1} x$ is in the range $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

${\cos}^{- 1} x$ is in the range $\left[0 , \pi\right]$

So if:

${\sin}^{- 1} x = {\cos}^{- 1} x = \theta$

then:

$\theta \in \left[0 , \frac{\pi}{2}\right]$

and:

$\sin \theta = \cos \theta = x$

Note that $\sin \theta$ is strictly monotonically increasing in $\left[0 , \frac{\pi}{2}\right]$ and $\cos \theta$ is strictly monotonically decreasing in $\left[0 , \frac{\pi}{2}\right]$.

Hence the only value of $\theta$ for which they are equal is:

$\theta = \frac{\pi}{4}$

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

So: $\text{ } x = \frac{\sqrt{2}}{2}$