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Can $$\sin r\pi $$ be rational if $r$ is irrational? Either a direct or existence proof is fine.

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    I was about to invoke [Niven's theorem](http://mathworld.wolfram.com/NivensTheorem.html), but that assumes both $r$ and $\sin\;r\pi$ are rational.2010-12-29
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    possible duplicate of [Irrationality of Trigonometric Functions](http://math.stackexchange.com/questions/2476/irrationality-of-trigonometric-functions)2010-12-29
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    @Moron: this one divides out the $\pi$, which eliminates some of the easy answers.2010-12-29
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    @Ross: Yes, but the answers there answer this too. Anyway, I guess they are a bit different. I might have been too hasty to cast the dupe close vote.2010-12-29
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    I've created a new Wikipedia article titled [Niven's theorem](http://en.wikipedia.org/wiki/Niven%27s_theorem). Two-and-a-half hours after I created it, I entered "Niven's theorem" into Google and that Wikipedia article was on the first page of results. So contribute to it if you can. Besides contributions _within_ the article, there's the matter of which other articles ought to link to it. I've created three such links; if you think of others that should be there, you can add those too.2011-12-02

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As J. M. said, Niven's theorem does it. There is some $r$ such that $\sin (r\pi) = \frac{1}{3}$ As $\sin (r\pi)$ is rational and not $0, \pm1, \pm \frac{1}{2}$, $r$ is not rational