Is this a property of the sine function? Because of the equidistribution property of $\{n\mod{2\pi}\}_{n\in\mathbb{Z}}$ in $[0,2\pi]$, my intuition tells me that the following statement about the sine function must be true.  
We can find an absolute constant $C>0$ such that for all $\epsilon>0$ and all $n\in\mathbb{Z}$ there exists a $n' \in \mathbb{Z}$  with $|n'-n|<C\epsilon^{-1}$ and $|\sin{n'}|<\epsilon$.
First, is this true? And if it is/isn't, how can it be shown? And is this statement equivalent to (or follows from) a well-known theorem?
 A: Your question is equivalent to $\{n \mod 2\pi\}$ equidistributing quickly.
Writing $n' = n+k$, we want some $|k| \le C\epsilon^{-1}$ with $\epsilon > \sin(n') = \sin(n+k) = \sin(2\pi(\frac{n+k}{2\pi}))$ $= \sin\left(2\pi(\lfloor \frac{n}{2\pi}\rfloor+\lfloor\frac{k}{2\pi}\rfloor+\{\frac{n}{2\pi}\}+\{\frac{k}{2\pi}\})\right)$ $= \sin(2\pi \{\frac{n}{2\pi}\}+2\pi\{\frac{k}{2\pi}\})$. Since $n$ is arbitrary, $\{\frac{n}{2\pi}\}$ can basically be anything, so you need/want $\{\frac{k}{2\pi}\}$ to create an $\asymp\epsilon$-net in $[0,1]$ (note $\sin(x) \asymp x$ for small $x$) for $|k| \lesssim \epsilon^{-1}$.
If you didn't understand that and want full rigor, here ya go.

Potential Claim: There exists an absolute $C > 0$ such that for any $\epsilon > 0$, the set $\{\{\frac{k}{2\pi}\} : |k| \le C\epsilon^{-1}\}$ intersects each interval of size $\frac{\epsilon}{10}$ in $[0,1]$.

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If Potential Claim is true, then answer to your question is "yes".
Proof: Take $C$ as in potential claim and $\epsilon > 0$. Take any $n \in \mathbb{Z}$. Take $|k| \le C\epsilon^{-1}$ such that $\{\frac{k}{2\pi}\} \in [1-\{\frac{n}{2\pi}\}-\frac{\epsilon}{20},1-\{\frac{n}{2\pi}\}+\frac{\epsilon}{20}]$ (if $1-\{\frac{n}{2\pi}\} < \frac{\epsilon}{20}$, just use $[1-\{\frac{n}{2\pi}\},1-\{\frac{n}{2\pi}\}+\frac{\epsilon}{10}]$ and analogously if $1-\{\frac{n}{2\pi}\} > 1-\frac{\epsilon}{20}$). Let $n' = n+k$. Then $|\sin(n')| = |\sin\left(2\pi(\{\frac{n}{2\pi}\}+\{\frac{k}{2\pi}\})\right)| < \epsilon$ since $|2\pi(\{\frac{n}{2\pi}\}+\{\frac{k}{2\pi}\})| \le 2\pi\frac{\epsilon}{20} \le \frac{\epsilon}{2}$.

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If Potential Claim is false, then the answer to your question is "no".
Proof: I'll leave this to you. I basically already explained it at the top of my answer.

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So, is the Potential Claim true? Well, this is related to the discrepancy of an irrational rotation on the torus (note that we are adding $\frac{1}{2\pi}$ mod $1$ each time we increase $k$ by $1$). And this is related to the continued fraction expansion of the rotation, in this case $\frac{1}{2\pi}$. I just googled and found this for one such reference (containing many other references).
