Let $$f(x)= \begin{cases}\frac{x}{a} & \text{ if } 0\leq x\leq a \\ \frac{2\pi - x}{2\pi - a} &\text{ if } a \leq x \leq 2\pi\end{cases}$$ and prolonged over $\mathbb R$ by $2\pi-$periodicity.
1) Compute it's Fourier coeficients.
2) Deduce $\sum_{k=1}^\infty \frac{\cos(ak)}{k^2}$
Attempt
1) $$c_n=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx}\mathrm d x=\frac{1}{2\pi}\int_0^a\frac{x}{a}e^{-inx}\mathrm d x+\frac{1}{2\pi}\int_a^{2\pi}\frac{2\pi-x}{2\pi -a}e^{-inx}\mathrm d x.$$ Integrating by part, we get $$\frac{1}{2\pi}\int_0^a\frac{x}{a}e^{-inx}\mathrm d x=\frac{e^{-ian}(1+ian)-1}{2\pi an^2}$$ and $$\frac{1}{2\pi}\int_a^{2\pi}\frac{2\pi-x}{2\pi-a}e^{-inx}\mathrm d x=\frac{e^{-ian}(ian-2i\pi n+1)-1}{2\pi(2\pi-a)n^2}.$$
2) I don't see how to do it. I know that $$f(x)=\sum_{k\in\mathbb Z}c_ke^{-ikx}$$ for all $x\in [0,2\pi]$, but I don't know how to conclude.