I'm looking for the Fourier series of $|\sin(8\pi x)| $ from the interval $-\frac{1}{8} $ to $ \frac{1}{8}$
I couldnt seem to simplify my working. Would be great if someone could point me in the right direction Here are my steps: $$a_n = \frac{1}{\frac{1}{8}}\int_{1/8}^{1/8}|\sin(8\pi x)|\cos(\frac{n\pi x}{\frac{1}{8}})dx$$ $$= 8(2) \int_{0}^{1/8}\sin(8 \pi x)\cos(8 n \pi x)dx $$
using identity $ 2\sin x \cos(nx)= \sin[(n+1)x]-\sin[(n-1)x] $
$$ a_n = 8[\frac{-\cos[(n+1)8\pi x]}{n+1}]_{0}^{\frac{1}{8}} + 8[\frac{-\cos[(n-1)8\pi x]}{n-1}]_{0}^{\frac{1}{8}}$$ $$ = \frac{1}{\pi}[\frac{\cos[(n-1)\pi]}{n-1} -\frac{\cos[(n+1)\pi]}{n+1}] $$
How should i simplify the $a_n$ term from here onwards?
The answer at the back of the textbook has $$ f(x) = \frac{2}{\pi}-\frac{2}{\pi}\sum_{1}^{\infty}\frac{1}{n^2 -1}[1+(-1)^n]\cos(8n\pi x)$$
and i'm not sure how to arrive at $(-1)^n$
All your textbook did was note that for an integer $n$, $$ \cos[(n-1) \pi] = \cos[(n+1)\pi] = -(-1)^n $$ one could verify this directly, or deduce it via Euler's formula.