Prove: 2sin⁻¹x = sin⁻¹(2x√(1 - x²)).

Prove: $2\sin^{-1}x = \sin^{-1}(2x\sqrt{1 - x^2})$.

Given: $x = \sin\theta$

To show: $2\sin^{-1}x = \sin^{-1}(2x\sqrt{1 - x^2})$

Answer

Proof with $x=\sin\theta$, for $x\in\left[-\frac1{\sqrt2},\frac1{\sqrt2}\right]$.

1. Let $x=\sin\theta$, so $\theta=\sin^{-1}x$ with $\theta\in\left[-\frac\pi4,\frac\pi4\right]$ (valid for $|x|\le\frac1{\sqrt2}$).
2. $\sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\cos\theta$ (as $\cos\theta\ge0$).
3. $2x\sqrt{1-x^2}=2\sin\theta\cos\theta=\sin2\theta$ (double angle identity).
4. $\sin^{-1}(2x\sqrt{1-x^2})=\sin^{-1}(\sin2\theta)=2\theta$, since $2\theta\in\left[-\frac\pi2,\frac\pi2\right]$, the principal range.
5. $2\theta=2\sin^{-1}x$, hence $2\sin^{-1}x=\sin^{-1}(2x\sqrt{1-x^2})$.