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Using the formula:

$$e^{i\omega t} = \cos {\omega t} + i\sin{\omega t}$$

I would like to prove that:

$$\sin^3\;x = -\frac{\sin{3x} - 3\sin{x}}{4} $$

However I haven't found any approach to this question. Just converting the first formula to $\sin^3$ doesn't seem to help as I'm still getting $\cos^3$ on the other side. Can anyone help me to guide me on the right way?

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    Try cubing the first formula as it is and taking the imaginary part on each side. You may also need to use $sin^2(x)+cos^2(x)=1$ Taking the real part gives the corresponding formula for cos.2011-06-29
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    @Mark: you should make that an answer!2011-06-29
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    I was trying to give a hint rather than an answer, because I reckon you get more value working through the calculation at least once in your life.2011-06-29

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