Proving a trigonometric identity: $\frac{2\sin^{3}x}{1-\cos x} = 2\sin x + \sin 2x$ I really need some help with this question. I need to prove this identity:
$$\frac{2\sin^{3}x}{1-\cos x} = 2\sin x + \sin 2x.$$

$\displaystyle\frac{2\sin^3x}{1-\cos x}=2\sin^2x\cdot\frac{\sin x}{1-\cos x}=2\sin^2x\cdot\frac{1+\cos x}{\sin x}$(using this)
$\displaystyle=2\sin x(1+\cos x)=2\sin x+2\sin x\cos x=\cdots$
