Question

Proof of arcsin(2xsqrt(1-x^2))=2arcsinx, x^2<= 1/2 .

Answer 1

\[\theta = \sin^{-1} x \rightarrow \sin \theta = x\]
\[2 x \sqrt{1-x^2} = 2 \sin \theta \sqrt{1-\sin^2 \theta} = 2 \sin \theta \cos \theta\]
double angle identity
\[\sin (2 \theta) = 2 \sin \theta \cos \theta\]
therefore
\[\sin^{-1} (\sin 2\theta) = 2 \sin^{-1} (\sin \theta)\]
\[2 \theta = 2 \theta\]