Question

Simplify: ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)=\dots \dots (|x|<\frac{1}{\sqrt{2}})$

Answer 1

$\begin{array}{c}{\sin }^{-1}\left(2x\sqrt{1-x^2}\right)\\ letx=\cos \theta ,and\theta ={\cos }^{-1}x\\ \Rightarrow {\sin }^{-1}\left[2\cos \theta \sqrt{1-{\cos }^2\theta }\right]\\ \Rightarrow {\sin }^{-1}\left[2\cos \theta \sin \theta \right]\\ \Rightarrow {\sin }^{-1}\left[\sin 2\theta \right]\\ \Rightarrow 2\theta =2{\cos }^{-1}x\\ \end{array}$