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.

$ \mathbb{N} $ is countable and $ \mathbb{R} $ is uncountable, so there can never be a bijection
