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For integers $n \neq 0$ is $\sin n$ irrational or transcendental?

This arose from another question. I would hypothesize yes and yes, possibly with proof for irrationality existing and but not for the more difficult property of transcendentality.

Any ideas?

Community
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Simon S
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1 Answers1

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If $n\in\mathbb{N}\setminus\{0\}$, $\sin(n)$ is trascendental (hence irrational) by the Lindemann-Weierstrass theorem, since $2i\sin(n)=e^{in}-e^{-in}$.

Jack D'Aurizio
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    Ah, very nice. Thanks – Simon S Oct 01 '15 at 15:57
  • @Jack : I think this difference of trascendentals being a trascendental deserves an explanation (for example $(e+5)-(e+4)=1$ and nobody knows if $\pi+e$ is trascendental or algebraic). Besides you have assumed that $e$ is trascendental which can be accepted as well known of course. Regards. – Ataulfo Oct 01 '15 at 19:31
  • @Ataulfo: you got it the wrong way. I was not claiming that the difference of two trascendental number is trascendental, but that if we assume that $\sin(n)$ is algebraic, by exploiting De Moivre's identity we contradict the Lindemann-Weierstrass theorem, Baker's formulation. – Jack D'Aurizio Oct 01 '15 at 19:33
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    We may say it also in the following way: if $\sin(n)$ is algebraic, $e^i$ is algebraic. But $e^i$ is not algebraic, hence neither it is $\sin(n)$. – Jack D'Aurizio Oct 01 '15 at 19:35
  • @Jack: I like more this last answer.Thanks. – Ataulfo Oct 01 '15 at 19:37