I would like to show that $$\delta(\sin{x})=\sum_{n=-\infty}^{\infty}\delta(x-n\pi),$$ where $\delta(x)$ satisfies the relation $$\int_{-\infty}^{\infty} \delta(x-a)\phi(x) \, dx =\phi(a),$$ for some test function $\phi$.
The $n\pi$ term makes it clear that at some point I need to be considering the roots of $\sin{x}$, but aside from that, I'm not sure what to do.
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$\begingroup$HINT
You should read this, but the idea is basically
$$ \int _{-\infty }^{\infty }f(x)\,\delta (g(x))\,dx=\sum _{i}{\frac {f(x_{i})}{|g'(x_{i})|}}. $$
where $x_i$ are roots of $f$, that is $f(x_i) = 0$. Call $g(x) = \sin(x)$, so that $x_i = i \pi$
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