I want to show that \begin{equation}\label{eqn}\sum_{n=1}^{\infty}\sin{nx}\sin{nx'} = \pi \delta(x-x') \tag{*} \end{equation} but I have reached the following result:
\begin{aligned} \sum_{n=1}^{\infty} \sin (n x) \sin \left(n x^{\prime}\right) &=\frac{1}{2} \sum_{n=-\infty}^{\infty} \sin n x \sin n x^{\prime} \\ &=\frac{1}{4} \sum_{n=-\infty}^{\infty}\left(\cos n\left(x-x^{\prime}\right)-\cos n\left(x+x^{\prime}\right)\right) \\ &=\frac{1}{8} \sum_{n=-\infty}^{\infty}\left(c^{i n\left(x-x^{\prime}\right)}+e^{-i n\left(x-x^{\prime}\right)}-e^{\operatorname{in}\left(x+x^{\prime}\right)}-e^{-i n\left(x+x^{\prime}\right)}\right) \\ &=\frac{\pi}{4}\left(2 \delta\left(x-x^{\prime}\right)-2 \delta\left(x+x^{\prime}\right)\right) \\ &=\frac{\pi}{2}\left(\delta\left(x-x^{\prime}\right)-\delta\left(x+x^{\prime}\right)\right) \end{aligned}
Where I have used the fact that $$\delta(x-x')=\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}c^{i n\left(x-x^{\prime}\right)}$$
Is the result I have obtained equivalent in any way to \eqref{eqn}? Intuitively, the result I obtained seems to make sense, since $$\sum_{n=1}^{\infty}\sin{nx}\sin{n(-x')} =-\sum_{n=1}^{\infty}\sin{nx}\sin{nx'}$$ i.e. the series is even with respect to $x'$, as is $$\frac{\pi}{2}\left(\delta\left(x-x^{\prime}\right)-\delta\left(x+x^{\prime}\right)\right).$$ I'd like to make sense of the relation between these two seemingly different expressions. Any help would be greatly appreciated.