Question

Solve:- ${\sin }^{-1}\left(\cos x\right)$

Answer 1

${\sin }^{-1}\left(\cos x\right)=x+\frac{\Pi }{2}$

Explanation:

We know that the cosine function, is nothing more than the $sine\frac{\Pi }{2}$ radians out of phase, as proved below:

$cos(\theta -\frac{\Pi }{2})=cos\left(\theta \right)cos(-\frac{\Pi }{2})-sin\left(\theta \right)sin(-\frac{\Pi }{2})$

$cos(\theta -\frac{\Pi }{2})=cos\left(\theta \right).0-(-sin(\theta \left)sin\right(\frac{\Pi }{2}\left)\right)$

$cos(\theta -\frac{\Pi }{2})=\sin \left(\theta \right)$

So we can say that the sine function, 90 degrees ahead, is the cosine function.

Using the property of inverse functions that \displaystyle $f^{-1}\left(f\left(x\right)\right)=x$

${\sin }^{-1}\left(\cos x\right)=x+\frac{\Pi }{2}$