(Edit) Note: $a\in \mathbb{R},0<a\leq\pi$. Also, the sum skips $n=0$ (that's where the other term comes from).
Working through a Fourier analysis exercise I've got stuck in a clearly trigonometical step. I've arrived at:
$$1=\frac{1}{2\pi}+\sum_{n=-\infty}^{\infty} \frac{\sin\frac{na}{2}}{\pi ne^{ina/2}}$$
And I need to reduce the infinite sum to one of the form $\sum_{n=-\infty}^{\infty} \frac{\sin(na)}{\pi n}$.
Euler's formula and half-angle formulae doesn't seem to help. Any idea?
$$\sum_{n=-\infty}^{\infty} \frac{\sin(\frac{na}{2})}{\pi ne^{ina/2}} \\=\sum_{n=-\infty}^{\infty}\frac{\sin(na/2)e^{-ina/2}}{\pi n}=\sum_{n=-\infty}^{\infty}\frac{\sin(na/2)[\cos(na/2)-i\sin(na/2)]}{\pi n} \\=\sum_{n=-\infty}^{\infty}\frac{\frac12\sin(na)}{\pi n}-i\sum_{n=-\infty}^{\infty}\frac{\frac12[1-\cos(na)]}{n\pi} \\=\sum_{n=-\infty}^{\infty}\frac{\frac12\sin(na)}{\pi n}+i\sum_{n=-\infty}^{\infty}\frac{\frac12\cos(na)}{n\pi}-\underbrace{\sum_{n=-\infty}^{\infty}\frac{i}{2n\pi}}_{x+(-x)=0} \\=\sum_{n=-\infty}^{\infty}\frac{e^{ina}}{2\pi n}=S_0+\sum_{n=1}^{\infty}\frac{e^{ian}+e^{-ian}}{2}.\frac{1}{n\pi}=S_0+\sum_{n=1}^{\infty}\frac{\sin(na)}{n\pi}$$ I don't know what to do about $S_0$(i.e., when $n=0$).
Hope rest is clear and it helped you.