# Question #66955

Jun 29, 2016

De Moivre's Theorem states that:
${\left(\cos x + i \sin x\right)}^{n} = \cos n x + i \sin n x$
This can be used (preferably with binomial expansion for higher values) to find double, triple (etc) angle rules by substituting in different values of $n$ and then equating the real and imaginary parts of both sides.

Example:
Let $n = 3$
Therefore, ${\left(\cos x + i \sin x\right)}^{3} = \cos 3 x + i \sin 3 x$
${\cos}^{3} x + 3 {\cos}^{2} x \left(i \sin x\right) + 3 \cos x {\left(i \sin x\right)}^{2} + {\left(i \sin x\right)}^{3} = \cos 3 x + i \sin 3 x$
${\cos}^{3} x + 3 i {\cos}^{2} x \sin x - 3 \cos x {\sin}^{2} x - i {\sin}^{3} x = \cos 3 x + i \sin 3 x$

You can now equate the Re and Im parts of both sides, the idea here being that if you have two equal complex numbers:
$a + b i = c + \mathrm{di} \implies a = c \mathmr{and} b = d$ since the two domains can't interfere. It's a similar idea to vectors in more than one dimension, how different components in different dimensions can't interfere.

Therefore, Equating Re:
${\cos}^{3} x - 3 \cos x {\sin}^{2} x = \cos 3 x$
Equating Im:
$3 {\cos}^{2} x \sin x - {\sin}^{3} x = \sin 3 x$

These rules, especially for $n = 2$, make computing integrals like $\int {\sin}^{2} x \mathrm{dx}$ significantly easier.