Show that (i) \(sin^{-1}(2x\sqrt{1-x^2}) =
2sin^{-1}x, -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}\)
(ii) \(sin^{-1}(2x\sqrt{1-x^2}) = 2cos^{-1}x,
\frac{1}{\sqrt{2}} \le x \le 1\)
Solution
Let \(x = sin\theta\). Then
\(sin^{-1}x = \theta\). We have \(sin^{-1}(2x\sqrt{1-x^2}) =
sin^{-1}(2sin\theta\sqrt{1-sin^2\theta})\)\(= sin^{-1}(2sin\theta cos\theta) =
sin^{-1}(sin2\theta) = 2\theta\)\(=
2sin^{-1}x\)
Take \(x = cos\theta\), then
proceeding as above, we get, \(sin^{-1}(2x\sqrt{1-x^2}) =
2cos^{-1}x\)