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How do I prove:
$ \sin^2 {2x} - \sin^2 {x} = \sin {3x}\sin {x} $
?

I'm lost

Neta Rubin
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    In which place are you lost, precisely? Take $\sin^2 x$ to the rhs and decompose the sum of sines. – zhoraster Sep 03 '15 at 10:18
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    HINT: Recall some formula: $\cos 2x = 1 - 2\sin^2 x$ and $\cos x - \cos y = 2\sin\frac{x+y}{2}\sin \frac{x-y}{2}$. – GAVD Sep 03 '15 at 10:20
  • Hint: $\sin(2x)=2\sin x\cos x$ so on both sides $\sin x$ shows up. – drhab Sep 03 '15 at 10:22

2 Answers2

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LHS=$(\sin2x+\sin x)(\sin2x-\sin x )=2\sin\frac{3x}{2}\cos \frac x2.2\cos\frac{3x}{2}\sin \frac x2=\sin 3x\sin x$ =RHS

David Quinn
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By complex numbers, setting $z=e^{ix}$ and omitting the common denominator $(2i)^2$: $$(z^2-z^{-2})^2-(z-z^{-1})^2=z^4+4+z^{-4}-z^2-4-z^{-2},$$ $$(z^3-z^{-3})(z-z^{-1})=z^4-z^2-z^{-2}+z^{-4}.$$