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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