Question

Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]

Answer 1

We have
\[LHS = \sin^{- 1} \left( 2x\sqrt{1 - x^2} \right)\]
\[\text{Putting }x = \sin a, \text{we get}\]
\[ = \sin^{- 1} \left( 2 \sin a\sqrt{1 - \sin^2 a} \right) \]
\[ = \sin^{- 1} \left( 2\sin a \cos a \right)\]
\[ = \sin^{- 1} \left( \sin 2a \right)\]
\[ = 2a\]
\[ = 2 \sin^{- 1} x \left( \because x = \sin a \right)\]
\[\]