## Summing a Trigonometric Series

We can sum many trigonometric series using complex number, specicifically the exponential form.
$e^{i \theta}= cos \theta +i sin \theta$
(De Moivre's Theorem)
The ral part of
$e^{i \theta}$
is
$cos atheta$
and the imaginary part is
$sin \theta$
.
Example::
$S=sin 2x+sin4x+sin6x+...sin2kx+...$

Use
$sin2kx=Im(e^{i(2kx)})$
.
Then
$S=sin 2x+sin4x+sin6x+...sin2kx+...=Im(e^{i(2x)})+Im(e^{i(4x)})+Im(e^{i(6x)})+...+Im(e^{i(2kx)})+...=Im(e^{i(2x)})+e^{i(4x)}+e^{i(6x)}+...+e^{i(2kx)}+...)$
.
$e^{i(2x)})+e^{i(4x)}+e^{i(6x)}+...+e^{i(2kx)}+...$
is a geometric series with firsat term
$a=e^{i(2x)}$
and common ratio
$r=e^{i(2x)}$
. We can sum this series and the imaginary part will be equal to the desired sum.
\begin{aligned} S &=Im(\frac{a}{1-r}) \\ &= Im(\frac{e^{i(2x)}}{1-e^{i(2x)}}) \\ &= Im( \frac{e^{i(2x)}}{1-e^{i(2x)}} \times \frac{1+e^{i(2x)}}{1+e^{i(2x)}}) \\ &= Im( \frac{e^{i(2x)}(1+e^{i(2x)})}{1-e^{i(4x)}}) \\ &= Im(\frac{e^{i(3x)}(e^{-i(x)}+e^{i(x)})}{e^{i(2x)}(e^{i(-2x)}-e^{i(2x)})}) \\ &= Im(\frac{e^{i(x)}(e^{-i(x)}+e^{i(x)})}{e^{i(2x)}(e^{i(-2x)}-e^{i(2x)})}) \\ &= Im( \frac{cosx+isinx)2cosx}{-2isin2x})= \\ &= \frac{2cos^2x}{-4sinxcosx} \\ &= - \frac{1}{2}cotx \end{aligned}