Problem: find Fourier series of $$f(x) = \frac{1}{2}\,\arctan\left(\frac{2a \sin(x)}{1-a^2}\right)$$ where $|a|<1$ and $ -\pi<x<\pi$.
Solution: $$a\sin(x) + \frac{a^3}{3}\,\sin(3x) + \frac{a^5}{5}\,\sin(5x)+\dots $$ My question is: how to find that solution?
My approach:
Write $\arctan$ as $\arctan(y)=\frac{1}{2i}\ln\left(\frac{i-y}{i+y}\right)=\frac{1}{2i}\ln\left(\frac{1+iy}{1-iy}\right)$ and then: $$\arctan(y)=\frac{1}{2i}(\ln(1+iy)-\ln(1-iy)). $$ In my case $y\in\mathbb{R}$, $y=\frac{2a \sin(x)}{1-a^2}$. Since $\ln(z)=\ln|z|+i\theta$ and $y\in\mathbb{R}$, we have: $$\arctan(y)=\frac{1}{2i}[(\ln|1+iy|+i\theta)-(\ln|1-iy|+i(-\theta))], $$ $$\arctan(y)=\frac{1}{2i}[2i\theta]=\theta. $$ So $f(x)$ is: $$f(x) = \frac{1}{2} \theta.$$ For $\theta$ we have that it is imaginary part of $(1+iy)$: $$ \theta = \operatorname{Im}\{\ln(1+iy)\}. $$ Using the Taylor series for $$\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k+1} \frac{x^{2k-1}}{2k-1}$$ we can rewrite $\theta$, but that does not produce a solution (it becomes very messy and does not give anything useful, as far as I can see).
Another strategy was to do something with $\frac{2a}{1-a^2}=\frac{1}{1-a} - \frac{1}{1+a}$ and geometric series, but it does not give anything useful (as far as I can see).
Thank you for any help!
If $|a|<1$ then $$ \arctan\left(\frac{2a\cos(x)}{1-a^2}\right) = \text{Im}\log\left(1+2ia \cos(x)-a^2\right)=\text{Im}\log\left[(1+ia e^{ix})(1+iae^{-ix})\right] $$ and from $$ \log\left(1+ia e^{\pm ix}\right) = \sum_{n\geq 1}\frac{(-1)^{n+1} i^n a^n}{n}e^{\pm nix} $$ we get $$ \text{Im}\log\left(1+ia e^{\pm ix}\right) = \sum_{n\geq 1}\frac{(-1)^n a^{2n+1}}{2n+1}\cos((2n+1)x)-\sum_{n\geq 1}\frac{(-1)^n a^{2n}}{2n}\sin(\pm 2nx)$$ and $$ \arctan\left(\frac{2a\cos(x)}{1-a^2}\right) = 2\sum_{n\geq 1}\frac{(-1)^n a^{2n+1}}{2n+1}\cos((2n+1)x). $$ To get the Fourier series of the wanted function, it is enough to map $x$ into $\frac{\pi}{2}-x$.