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?
$\endgroup$ 33 Answers
$\begingroup$$$ \sin^3 x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^3 = \frac{e^{i3x}-3e^{ix}+3e^{-ix}-e^{-i3x}}{-8i} $$
$$ = -\frac{1}{4}\left(\frac{e^{i3x}-e^{-i3x}}{2i}\right) +\frac{3}{4}\left(\frac{e^{ix}-e^{-ix}}{2i}\right) = -\frac{\sin 3x - 3\sin x}{4} $$
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ $\begingroup$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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