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We know that, for any rational number $p$, we have that $\cos(p\pi)$ is an algebraic number.

Since this property comes from the fact that $e^{ip\pi}$ is algebraic (as a root of unity), I suspect that $\pi$ is the unique transcendental number with such property, in the sense that there does not exists another transcendental number $\alpha\ne q \pi$, for rational $q$, such that $\cos(p\alpha)$ is an algebraic number. But I don't find a proof. It is true?

Emilio Novati
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    Like many "common sense" expectations about transcendental numbers, this can likely be linked to Schaunel's conjecture, which is to say that the answer is "probably". – Ben Grossmann Oct 01 '16 at 18:58
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    "such that $\cos(p\alpha)$ is an algebraic number" for some $p\in\mathbb{Q}$ or for every $p\in\mathbb{Q}$? – barak manos Oct 01 '16 at 19:00
  • Possibly related: https://math.stackexchange.com/questions/1573244 – Watson Oct 01 '16 at 19:01
  • @Watson: Thank you for the link. Reading the answer ''If $x$ is an algebraic irrational multiple of $\pi$, then $\cos x$ is transcendental (this follows from the Gelfond-Schneider theorem)'' it seems that this i the answer to my question, but I 'don't see how we can use the the Gelfond-Schneider theorem to prove the statement. Do you have some other reference? – Emilio Novati Oct 01 '16 at 19:09
  • @Omnomnomnom. This was also my suspect. But there is some proof that, assuming Schaunel's conjecture, it is true? – Emilio Novati Oct 01 '16 at 19:11
  • @EmilioNovati my guess is yes. I don't actually know, though. – Ben Grossmann Oct 01 '16 at 19:12
  • @barakmanos. I'm asking for every rational $p$. But i suspect that if we can find an $\alpha$ that work for some $p$ than we have also the general result. – Emilio Novati Oct 01 '16 at 19:16

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This answer shows that $\alpha=\cos^{-1}(3/5)$ is not a rational multiple of $\pi$. Additionally, the Lindemann-Weierstrass theorem shows that $\alpha$ is transcendental.

Nevertheless, $\cos(p\alpha)$ is algebraic for all rational $p$. If $p=\frac{n}{m}$, then $\cos(p \alpha)$ satisfies the polynomial equation $T_m(\cos(p \alpha))=T_n(3/5)$, where $T_k$ is the $k$th Chebyshev polynomial.

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