# Euler formula and $\sin^3$

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. – Mark Bennet Jun 29 '11 at 15:06
@Mark: you should make that an answer! – The Chaz 2.0 Jun 29 '11 at 15:08
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. – Mark Bennet Jun 29 '11 at 15:33

$$\sin^3 x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^3 = \frac{e^{i3x}-3e^{ix}+3e^{-ix}-e^{-i3x}}{-8i}$$

$$= -\frac{1}{4}\left(\frac{e^{i3x}-e^{-i3x}}{2i}\right) +\frac{3}{4}\left(\frac{e^{ix}-e^{-ix}}{2i}\right) = -\frac{\sin 3x - 3\sin x}{4}$$

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+1 for showing that no ingenunity is required; the equation $e^{ix} = \cos x + i\sin x$ is really a "proof machine" for most trigonometric identities that can be carried out mechanically. – ShreevatsaR Jun 29 '11 at 16:07
@ShreevatsaR : Agreed, I think this way is the most straightforward. – Joel Cohen Jun 29 '11 at 17:07
another answer gets bumped, and yours gets upvoted :-) – robjohn Nov 11 '11 at 17:38
But it seems CuStud never came back to look at the answers... – GEdgar Nov 11 '11 at 17:56

You don't need to know de Moivre's Theorem, in fact it is implicit in the first formula.

Try $(e^{i\omega t})^3 = (\cos {\omega t} + i\sin{\omega t})^3$.

But you also have

$$(e^{i\omega t})^3 = e^{i3\omega t} = \cos {3\omega t} + i\sin{3\omega t}$$

using the first formula with $3\omega t$ instead of $\omega t$.

Equate the two, take real and imaginary parts, and use $1 - \cos^2 x = \sin^2 x$ to get everything in terms of the $\sin$ function.

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Here's how you do it:

By de Moivre's theorem, you should know that

$(\cos 3x + i \sin 3x) = (\cos x + i \sin x)^3$.

Expand the term on the right hand side using the binomial theorem, viz $(\cos 3x + i \sin 3x) = \cos^3 x +3i\cos^2 x \sin x - 3\sin^2 x \cos x - i \sin^3 x.$

So equating imaginary parts and you should get

$\sin^3 x = 3\cos^2 x \sin x - \sin 3x$.

Then you can use the fact that $1 - \cos^2 x = \sin^2 x$ which gives:

$\sin^3 x = (3 - 3\sin^2 x)\sin x - \sin3x$, hence

$4\sin^3 x = 3\sin x - \sin 3x$

which is what you want.

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But you ignored the "Using" clause in the question! See my answer... – GEdgar Jun 29 '11 at 15:28