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I saw the following statement and I'm not sure how to prove it:

Given a constant value $\beta \in \mathbb{R}$, if $\frac{\beta}{\pi}$ is irrational, then for some value $\alpha \in [0, 1]$, $\forall\epsilon \in \mathbb{R}:\epsilon > 0$, $\exists n \in \mathbb{N} : |\sin(n\beta) - \alpha| < \epsilon$

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    Should be very close to [this](http://math.stackexchange.com/questions/341378/can-sin-n-get-arbitrarily-close-to-1-for-n-in-mathbbn). – 2017-01-04
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    Where did you see the statement and what is the context? – 2017-01-04
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    The modern proof of this result is with the unique ergodicity of $x \to x + \beta$ on $S^1 = \mathbb{R}/\mathbb{Z}$ for $\beta$ irrational. – 2017-01-04

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Let $\alpha=\sin \theta$ then given $\epsilon$ there is (by a well known theorem) an $n$ such that $n\beta =k\pi+\gamma$ with $0\leq \gamma<\pi$ and $|\gamma-\theta|<\epsilon$ now use continuity of the $\sin $ function.

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    It sounds like someone doesnt understand my proof. – 2017-01-04
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    Did you mean $n\beta = 2k\pi + \gamma$, and also, what is the name of the theorem, or do you have a link to it? – 2017-01-04
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I found out that this principle is demonstrated in Kronecker's Approximation Theorem.