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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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    $\begingroup$ I was about to invoke Niven's theorem, but that assumes both $r$ and $\sin\;r\pi$ are rational. $\endgroup$ – J. M. is a poor mathematician Dec 29 '10 at 16:03
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    $\begingroup$ possible duplicate of Irrationality of Trigonometric Functions $\endgroup$ – Aryabhata Dec 29 '10 at 16:28
  • $\begingroup$ @Moron: this one divides out the $\pi$, which eliminates some of the easy answers. $\endgroup$ – Ross Millikan Dec 29 '10 at 16:32
  • $\begingroup$ @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. $\endgroup$ – Aryabhata Dec 29 '10 at 16:34
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    $\begingroup$ I've created a new Wikipedia article titled Niven's 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. $\endgroup$ – Michael Hardy Dec 2 '11 at 23:06
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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

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