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Let $\zeta$ be the n-th primitive unit root. with not too much effort I have proved that $\cos(2\pi /n)$ has degree $\varphi (n)/2$ over $\mathbb{Q}$, but failed to the degree of $\sin(2\pi /n)$ over $\mathbb{Q}$.

I can myself only proof that $\sin(2\pi /n)=(\zeta -{{\zeta }^{-1}})/2i$, but don't know how to proceed . I am waiting for your help, thank you!

I have searched this website and find a related question, but that seems doesn't solve this problem, I need some detail about this quesion.

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王李远
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  • What do you mean by "it doesn't solve this problem"? It does. –  Jun 12 '17 at 08:50
  • Not "the" but "a" primitive root. – DonAntonio Jun 12 '17 at 08:50
  • yes , I should say "a". I don't know what he means by saying :$Sin(2\pi /n)=(\zeta -{{\zeta }^{-1}})/i$ so the matching extension field is $\mathbb{Q}(\zeta -{{\zeta }^{-1}})$ – 王李远 Jun 12 '17 at 08:56
  • Would you please show me some details and post as an answer?@DonAntonio – 王李远 Jun 12 '17 at 08:58
  • I present two approaches for solving a specific case of the $\cos(2\pi/n)$ problem in the link below. The first approach might be tough to adapt given the generality of "$n$", but the Galois-theoretic approach generalizes nicely if you can determine the elements of $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ in terms of their action on $\zeta$. https://math.stackexchange.com/questions/2230280/find-the-degree-of-extension-bbb-q%CE%B6-9-%CE%B6%E2%88%921-9-over-bbb-q/2230302#2230302 – Kaj Hansen Jun 12 '17 at 09:06
  • @王李远 You can read in section 3, page 5 of the following all about this: http://www2.oberlin.edu/faculty/jcalcut/tanpap.pdf – DonAntonio Jun 12 '17 at 09:07
  • Thank you very very much for providing such useful and rich materials !!! I will take time to study that @DonAntonio – 王李远 Jun 12 '17 at 09:24

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