Question

Write the following in the simplest form:

${\displaystyle {\sin }^{-1}\left\{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right\},-1<x<1}$

Answer 1

Let x = sin θ

Now,

${\displaystyle {\sin }^{-1}\left\{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right\}={\sin }^{-1}\left\{\frac{\sin \theta +\sqrt{1-{\sin }^2\theta }}{\sqrt{2}}\right\}}$

${\displaystyle ={\sin }^{-1}\left\{\frac{\sin \theta +\cos \theta }{\sqrt{2}}\right\}}$

${\displaystyle ={\sin }^{-1}\{\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta \}}$

${\displaystyle ={\sin }^{-1}\{\cos \frac{\pi }{4}\sin \theta +\sin \frac{\pi }{4}\cos \theta \}}$

${\displaystyle ={\sin }^{-1}\left\{\sin (\theta +\frac{\pi }{4})\right\}}$

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

${\displaystyle =\frac{\pi }{4}+{\sin }^{-1}x}$

${\displaystyle \therefore {\sin }^{-1}\left\{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right\}={\cos }^{-1}x+\frac{\pi }{4}}$