Is $ \sin: \mathbb{N} \to \mathbb{R}$ injective? I was trying to show that $\sin(x)$ is non-zero for integers $x$ other than zero and I thought that this result might emerge as a corollary if I managed to show that the result in question is true. 
I think it's possible to demonstrate this by looking at the power series expansion of $\sin(x)$ and assuming that we don't know anything about the existence of $\pi$. 
All of the answers below insist that the proposition '$\exists p,q \in \mathbb{Z}, \sin(p)=\sin(q)$' -where $\sin(x)$ is the power series representation-is undecidable without using the properties of $\pi$. If so this is a truly wonderful conjecture and I would like to be provided with a proof. Until then, I insist that methods for analyzing infinite series from analysis should suffice to show that the proposition is false. 

Note: In a previous version of this post the question said "bijective" instead of "injective". Some of the answers below have answered the first version of this post.

Recall the following facts.
Fact 1. $\sin t=0$ if and only if $t=k\pi$ for some $k\in\Bbb Z$
Fact 2. $\cos t=0$ if and only if $t=k\pi+\frac{\pi}{2}$ for some $k\in\Bbb Z$
Also, recall the identity
$$
\sin u-\sin v=2\,\cos\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right)
$$
Thus $\sin u=\sin v$ if and only if $u+v=k\pi+\frac{\pi}{2}$ or $u-v=k\pi$ for some $k\in\Bbb Z$. But the values
\begin{align*}
k\pi+\frac{\pi}{2} && k\pi
\end{align*}
are necessarily irrational for nonzero $k\in\Bbb Z$. Hence $\sin u\neq\sin v$ for $u,v\in\Bbb Z$ with $u\neq v$.
