Question

Show that ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=2{\sin }^{-1}x$

Answer 1

Consider, LHS= ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

Put $x=\sin \theta$

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

L.H.S$={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)$

$={\sin }^{-1}\left(2\sin \theta \sqrt{{\cos }^2\theta }\right)$

$={\sin }^{-1}\left(2\sin \theta \cos \theta \right)$

$={\sin }^{-1}\sin 2\theta$

$=2\theta$

$=2{\sin }^{-1}x=$R.H.S

Hence L.H.S$=$R.H.S

Hence proved.