# What is DeMoivre's theorem?

Mar 5, 2018

DeMoivre's Theorem expand's on Euler's formula:
${e}^{i x} = \cos x + i \sin x$

DeMoivre's Theorem says that:

• ${\left({e}^{i x}\right)}^{n} = {\left(\cos x + i \sin x\right)}^{n}$
• ${\left({e}^{i x}\right)}^{n} = {e}^{i n x}$
• ${e}^{i n x} = \cos \left(n x\right) + i \sin \left(n x\right)$
• $\cos \left(n x\right) + i \sin \left(n x\right) \equiv {\left(\cos x + i \sin x\right)}^{n}$

Example:
$\cos \left(2 x\right) + i \sin \left(2 x\right) \equiv {\left(\cos x + i \sin x\right)}^{2}$

${\left(\cos x + i \sin x\right)}^{2} = {\cos}^{2} x + 2 i \cos x \sin x + {i}^{2} {\sin}^{2} x$

However, ${i}^{2} = - 1$

${\left(\cos x + i \sin x\right)}^{2} = {\cos}^{2} x + 2 i \cos x \sin x - {\sin}^{2} x$

Resolving for real and imaginary parts of $x$:
${\cos}^{2} x - {\sin}^{2} x + i \left(2 \cos x \sin x\right)$

Comparing to $\cos \left(2 x\right) + i \sin \left(2 x\right)$

$\cos \left(2 x\right) = {\cos}^{2} x - {\sin}^{2} x$
$\sin \left(2 x\right) = 2 \sin x \cos x$
These are the double angle formulae for $\cos$ and $\sin$

This allows us to expand $\cos \left(n x\right)$ or $\sin \left(n x\right)$ in terms of powers of $\sin x$ and $\cos x$

DeMoivre's theorem can be taken further:
Given $z = \cos x + i \sin x$

${z}^{n} = \cos \left(n x\right) + i \sin \left(n x\right)$

${z}^{- n} = {\left(\cos x + i \sin x\right)}^{- n} = \frac{1}{\cos \left(n x\right) + i \sin \left(n x\right)}$

${z}^{- n} = \frac{1}{\cos \left(n x\right) + i \sin \left(n x\right)} \times \frac{\cos \left(n x\right) - i \sin \left(n x\right)}{\cos \left(n x\right) - i \sin \left(n x\right)} = \frac{\cos \left(n x\right) - i \sin \left(n x\right)}{{\cos}^{2} \left(n x\right) + {\sin}^{2} \left(n x\right)} = \cos \left(n x\right) - i \sin \left(n x\right)$

${z}^{n} + {z}^{- n} = 2 \cos \left(n x\right)$
${z}^{n} - {z}^{- n} = 2 i \sin \left(n x\right)$

So, if you wanted to express ${\sin}^{n} x$ in terms of multiple angles of $\sin x$ and $\cos x$:
${\left(2 i \sin x\right)}^{n} = {\left(z - \frac{1}{z}\right)}^{n}$

Expand and simply, then input values for ${z}^{n} + {z}^{- n}$ and ${z}^{n} - {z}^{- n}$ where necessary.

However, if it involved ${\cos}^{n} x$, then you would do ${\left(2 \cos x\right)}^{n} = {\left(z + \frac{1}{z}\right)}^{n}$ and follow the similar steps.