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My friend and I were discussing this and we couldn't figure out how to prove it one way or another.

The only rational values I can figure out for $\sin(x)$ (or $\cos(x)$, etc...) come about when $x$ is some product of a fraction of $\pi$.

Is $\sin(x) $ (or other trigonometric function) necessarily irrational if $ x $ is rational?

Edit: Excluding the trivial solution of 0.

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And what about $x = 0?$ – Diego Silvera Feb 10 '13 at 2:14
See this. And Niven's book (page 21) here. – David Mitra Feb 10 '13 at 2:19
Well I guess that answers it, David, if you want to submit that as an answer. – Queequeg Feb 10 '13 at 2:23
up vote 15 down vote accepted

In fact each of $\cos x$, $\sin x$, and $\tan x$ are irrational at non-zero rational values of the arguments. This result is Theorem 2.5 and Corollary 2.7 in Ivan Niven's Irrational Numbers.

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If $\sin x$ is rational (or even just algebraic), then $\cos x=\pm \sqrt{1-\sin^2 x}$ is algebraic. Therefore $e^{ix}=\cos x+i\sin x$ is algebraic, so by the Lindemann-Weierstrass theorem, $x$ cannot have been nonzero algebraic -- in particular not nonzero rational.

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