Fourier series representation for $\sin(x/2)$ Is there a faster approach for finding the Fourier series of $$\sin(x/2)~,\cos(x/2)~~~\text{etc}$$
other than the usual approach? 
 A: I'll give a non-standard derivation for $\cos (x/2)$ on $[-\pi,\pi]$. Faster or not, you tell.  
The function being even,  we work with cosines. Observe   that if $$f(x)\sim \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx\tag1$$ then
$$f''(x)\sim -\sum_{n=1}^\infty n^2 a_n \cos nx\tag2$$ 
I am writing $\sim $ instead of $=$ here  to avoid getting arrested by the 
convergence police. Naturally, we want to plug (1) and (2) into $f''= (-1/4)f$ and equate coefficients. This does not work.
The problem is, $f''$ is not exactly $(-1/4)f$. The values of $f'$ at the endpoints don't match, and this
discontinuity contributes a point mass to $f''$. The correct formula is $$f''=(-1/4)f+\delta_{\pi}+\delta_{-\pi}\tag3$$ 
It's easy to find the coefficients for the delta function:
$$\delta_\pi +\delta_{-\pi}= \frac{c_0}{2}+ \sum_{n=1}^\infty c_n\cos nx, \quad c_n= (-1)^n\frac{2}{\pi}\tag4$$
because "integrating"  against a delta function amounts to evaluation.  Now plug into (3).
$$-n^2 a_n = (-1/4)a_n +  c_n\tag5$$
which yields
$$ a_n = \frac{-c_n}{n^2-1/4}\tag6$$
with $c_n$ as in (4).  
