Inverse sum representation of sine The other day I was playing with functions of the form
$$
f(x) = \frac{1}{\frac{1}{a_0(x-b_0)} + \frac{1}{a_1(x-b_1)} + \cdots + \frac{1}{a_n(x-b_n)}}
$$
and I found particularly that
$$
\frac{1}{\cdots \frac{1}{x+2\pi} + \frac{-1}{x+\pi} + \frac{1}{x-0} + \frac{-1}{x-\pi} + \frac{1}{x-2\pi} + \cdots}
$$
seems to converge to $\sin x$ rather quickly (at least in comparison to the Maclaurin series representation). In addition, I found quite similar expressions for $\cos$ and $\tan$. How can it be shown that this expression is or is not equal to $\sin x$ (disregarding when $x=n \pi$ for some integer $n$)?

That's right. Consider the functions $f(z)=\dfrac 1 z+\sum\limits_{k\geqslant 1}(-1)^k\dfrac{2z}{z^2-k^2}$ and $g(z)=\dfrac{1}{\sin(\pi z)}$. You should be able to show that $f(z)$ is meromorphic in $\Bbb C$ with simple poles at the integers, that the same happens with $\sin(\pi z)$, and that $f(z)-g(z)$ is bounded and entire, so that indeed $f(z)=g(z)$ identically since they agree at a point. 
