Question

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

Answer 1

`x = sqrt(2)/2`

Explanation:

By definition:

`sin^(-1) x` is in the range `[-pi/2, pi/2]`

`cos^(-1) x` is in the range `[0, pi]`

So if:

`sin^(-1) x = cos^(-1) x = theta`

then:

`theta in [0, pi/2]`

and:

`sin theta = cos theta = x`

Note that `sin theta` is strictly monotonically increasing in `[0, pi/2]` and `cos theta` is strictly monotonically decreasing in `[0, pi/2]`.

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

`theta = pi/4`

`sin (pi/4) = cos (pi/4) = sqrt(2)/2`

So: `" "x = sqrt(2)/2`