I have to find the Fourier cosine series of $ | \sin x |$ on $(-\pi, \pi)$. I understand the trick to use is that the integral of an absolute value function is even. The formula for the coefficients is given by:
$A_m = \frac{2}{L} \int_{0}^{L} \phi (x) \cos\left(\frac{m\pi x}{L}\right) dx $.
I am confused because the question states that I must find the cosine series, but if this takes place over the whole interval $(-\pi, \pi)$ then aren't we looking for the Full Fourier Series?
Also I am quite confused as to how the formula for the coefficients changes when you double the size of the interval. I appreciate the help.
In your formula L is the period of your function. In your case f is $2\pi$ periodic, so you have $L=2\pi$. But Since that your function is $2\pi$ periodic, you could also integrate between $-L/2$ and $L/2$. And if you integrated betwen $0$ and $2L$, the factor in front of the integral would be $\frac{1}{L}$, since you double the size of the interval you divide by 2 the integral.
And you are only calculating the cosine terms because it is a consequence of the fact that f is an even-function.
In the end you have : $$f(t)=\frac {2}{\pi} - \sum\limits_{n = 0}^\infty {\frac{4}{\pi((2n)^2-1)}cos(2nt)}$$