I was looking at the graph for $\dbinom{n}{x}$ and I found using Wolfram Alpha that for $n\in\mathbb{Z}$ and $x\neq0,1,\dots,n$, $$\dbinom{n}{x}=\dfrac{\left(-1\right)^nn!\sin(\pi x)}{\pi\prod_{m=0}^{n}\left(x-m\right)}.$$ I'm curious why the sine function would appear in something like a binomial coefficient.
Edit: By $\dbinom{n}{x}$, I mean $\dfrac{\Gamma(n+1)}{\Gamma(x+1)\Gamma(n-x+1)}$.
Presumably, we are defining $$\binom nx=\frac{n!}{\Gamma(x+1)\Gamma(n+1-x)}$$
But $$\begin{align}\Gamma(n+1-x)&=(n-x)(n-1-x)\cdots (0-x)\Gamma(-x)\\ &=\Gamma(-x)(-1)^{n+1}\prod_{m=0}^n (x-m) \end{align}$$
The key identity is $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}$$ when $z=1+x$ gives your formula:
$$\begin{align}\binom nx&=\frac{n!}{\Gamma(x+1)\Gamma(-x)(-1)^{n+1}\prod_m(x-m)}\\&=\frac{(-1)^{n+1}n!\sin(1+x)\pi}{\pi\prod_m (x-m)}\\ &=\frac{(-1)^nn!\sin \pi x}{\pi\prod_m (x-m)} \end{align}$$
As $x\to k,$ for $k=0,1,\dots,n$ we have that $$\prod_{m\neq k} (k-m)=(-1)^{n-k} k! (n-k)!$$ and $$\lim_{x\to k} \frac{\pi x}{\pi(x-k)}=\lim_{y\to 0}\frac{\sin((y+k)\pi)}{\pi y}=(-1)^k$$
So this all agrees with the usual definition of $\binom nk.$