Summing a Trigonometric Series
\[e^{i \theta}= cos \theta +i sin \theta\]
(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.\[\begin{equation} \begin{aligned} S &=Im(\frac{a}{1-r}) \\ &= Im(\frac{e^{i(2x)}}{1-e^{i(2x)}}) \\ &= Im( \frac{e^{i(2x)}}{1-e^{i(2x)}} \times \frac{1+e^{i(2x)}}{1+e^{i(2x)}}) \\ &= Im( \frac{e^{i(2x)}(1+e^{i(2x)})}{1-e^{i(4x)}}) \\ &= Im(\frac{e^{i(3x)}(e^{-i(x)}+e^{i(x)})}{e^{i(2x)}(e^{i(-2x)}-e^{i(2x)})}) \\ &= Im(\frac{e^{i(x)}(e^{-i(x)}+e^{i(x)})}{e^{i(2x)}(e^{i(-2x)}-e^{i(2x)})}) \\ &= Im( \frac{cosx+isinx)2cosx}{-2isin2x})= \\ &= \frac{2cos^2x}{-4sinxcosx} \\ &= - \frac{1}{2}cotx \end{aligned} \end{equation}\]
