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

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Answer 1 · 2016-09-30T23:47:35

$x = sqrt(2)/2$

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: $" "x = sqrt(2)/2$

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