In optics, the following expression needs to be evaluated in calculating the intensity of light transmitted through a film after multiple reflections at the surfaces of the film: $$\left(\sum_{n=0}^\infty r^{2n}\cos n\theta\right)^2+\left(\sum_{n=0}^\infty r^{2n}\sin n\theta\right)^2.$$ Show that this is equal to $|\sum_{n=0}^\infty r^{2n}e^{in\theta}|^2$ and so evaluate it assuming $|r|<1$ ($r$ is the fraction of light reflected each time).
Beginning from the second part of the problem.
Since $\vert\; \sum_{n=0}^\infty\; r^{2n}\;e^{in\theta}\vert^2$ looks like the geometric progression, and the sum of geometric series is $S=\frac{a}{1-r},$ when $r<1$, I assume that $a=1$ and ratio is $r^2e^{i\theta}.$
Also we need only the imaginary part of the series, namely sine part.
Corrected version: \begin{align*}\quad \vert S \vert^2\;&=\left\vert \frac{1}{1-r^2e^{i\theta}}\right \vert ^2 \\ &=\vert1-r^2e^{i\theta}\vert^{-2} \\ &=\vert1-r^2(\cos\theta+isin\theta)\vert^{-2} \\&= \vert1-r^2\cos\theta-r^2isin\theta\vert^{-2} \\ &= ((1-r^2\cos\theta)^2+(r^2sin\theta)^2)^{-1}\\ &= (1-2r^2\cos\theta+r^4\cos^2\theta+r^4\sin\theta)^{-1}\\ &=(1+r^4-2r^2\cos\theta)^{-1}.\end{align*}
With $$S=\frac {1}{1-r^2e^{i\theta }}$$ we have $$|S^2|=|S|^2 = (|1-r^2e^{i\theta}|^2)^{-1}$$
Note that
$$ |1-r^2e^{i\theta}|^2 = |1-r^2 (\cos \theta + i \sin\theta)|^2 =$$
$$(1-r^2 (\cos \theta)^2 + (r^2 \sin \theta)^2=$$
$$1-2r^2\cos \theta +r^4$$
Thus the result follows.