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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.

Queequeg
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    See [this](http://someclassicalmaths.wordpress.com/2009/07/17/nivens-proof-that-the-trigonometric-and-inverse-trigonometric-functions-are-irrational-for-rational-non-zero-arguments/). And Niven's book (page 21) [here](http://books.google.com/books?id=ov-IlIEo47cC&pg=PA15&source=gbs_toc_r&cad=4#v=onepage&q&f=false). – David Mitra Feb 10 '13 at 02:19

2 Answers2

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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.

hmakholm left over Monica
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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.

David Mitra
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    is it true for $sin(x)/x$ as well? – m_power Sep 27 '18 at 14:10
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    If $x\in\mathbb{Q}\setminus\{0\}$, then $\sin(x)\not\in\mathbb{Q}$. If $\frac{\sin(x)}x\in\mathbb{Q}$, then $\sin(x)=\frac{\sin(x)}x\cdot x\in\mathbb{Q}$ (contradiction). Thus, $\frac{\sin(x)}x\not\in\mathbb{Q}$. – robjohn May 11 '19 at 00:07