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

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Thank you Ian Mateus for fixing my equation. I will try to figure out how to do that. I really appreciate it! :) –  Paula Nov 16 '13 at 18:06
    
For future reference, there is a quick tutorial about how I did it here. –  Ian Mateus Nov 23 '13 at 17:36

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

$\displaystyle=2\sin x(1+\cos x)=2\sin x+2\sin x\cos x=\cdots$

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and sin2x=2sinxcosx I believe that double angle identity –  Paula Nov 16 '13 at 18:04
    
I understand it now. THANK YOU SO MUCH!! :) –  Paula Nov 16 '13 at 18:04
    
@Paula, mathworld.wolfram.com/Double-AngleFormulas.html. Nice to hear that –  lab bhattacharjee Nov 16 '13 at 18:05

Write $$\sin^3(x) = \sin(x) \sin^2(x) = \sin(x)(1-\cos^2(x)) = \sin(x)(1+\cos(x))(1-\cos(x))$$ and finally make use of $\sin(2x) = 2\sin(x) \cos(x)$ to get what you want.

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