I'm having difficulty with the following question:
Find the Fourier series of the $2\pi$-periodic function $g(x)$ given by $$g(x)=\begin{cases}\cos(px)&\text{if }|x|\le\frac\pi2,\\0&\text{if }\frac\pi2<|x|\le\pi\end{cases}$$ where $p$ is a positive integer, and hence show that $$\sum_{n=1}^\infty\frac1{4n^2-1}=\frac12$$
What I've done so far:
$$\begin{align} a_0 &= \frac1\pi\int_{i=-\pi}^\pi\cos(px)\ \mathrm dx\\ &= \frac1\pi\left(\frac{\sin\left(\frac{p\pi}2\right)}p-\frac{\sin\left(\frac{-p\pi}2\right)}p\right)\\ &= 2\frac{\sin\left(\frac{p\pi}2\right)}{p\pi} \end{align}$$
$b_0$=0 because the function is even
$$\begin{align} a_n&=\frac{1}{\pi}\int_{i=-\pi}^\pi f(x)\cos(nx)\ \mathrm dx\\ &=\frac{2}{\pi}\int_{i=0}^\pi \cos(px)\cos(nx)\ \mathrm dx\\ &=\frac{2}{2\pi}\int_{i=0}^{\pi/2} \cos((p+n)x)+\cos((p-n)x)\ \mathrm dx\\ &=\frac{1}{\pi}\left[\frac{\sin((p+n)x)}{p+n}+\frac{\sin((p-n)x)}{p-n}\right]_0^{\pi/2}\\ &=\frac{1}{\pi}\left(\frac{\sin((p+n){π/2})}{p+n}+\frac{\sin((p-n){\pi/2})}{p-n}\right)\\ &=\frac{1}{\pi}\left(\frac{(p-n)\sin((p+n){π/2})}{p^2-n^2}+\frac{(p+n)\sin((p-n){\pi/2})}{p^2-n^2}\right) \end{align}$$
I don't know where to go from here. I can't see how to simplify the $\sin$ terms seeing as $\sin\left(\dfrac{x\pi}2\right)$ can take the values $0, -1$, and $1$ for an integer $x$. Have I gone wrong somewhere?
Hint:
Let's put $p=1$, so $$\sin\left(\frac{p+n}{2}\pi\right)=\begin{cases}0&\text{if }n=2m-1\text{ for some integer }m\\ (-1)^{m}& n=2m\; \text{ for some integer }m\end{cases}$$ Then, $$a_n=\begin{cases}0&\text{if }\;n=2m-1\;\text{ for some integer }m\\[3pt]\frac{2(-1)^{m+1}}{\left((2m)^2-1\right)\pi}&n=2m\; \text{ for some integer }m\end{cases}$$