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URLhttps://math.stackexchange.com/questions/2319458/how-to-find-the-degree-of-sin2-pi-n-over-mathbbq
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1 $\begingroup$ 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. asked Jun 12, 2017 at 8:46 ηŽ‹ζŽθΏœ 892 1 gold badge 6 silver badges 19 bronze badges $\endgroup$ 7 You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions See similar questions with these tags.
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[Skip to main content](https://math.stackexchange.com/questions/2319458/how-to-find-the-degree-of-sin2-pi-n-over-mathbbq#content) #### Stack Exchange Network Stack Exchange network consists of 183 Q\&A communities including [Stack Overflow](https://stackoverflow.com/), the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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[Explore Stack Internal](https://stackoverflow.co/internal/?utm_medium=referral&utm_source=math-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover) # [How to find the degree of \$\\sin(2\\pi /n)\$ over \$\\mathbb{Q}\$](https://math.stackexchange.com/questions/2319458/how-to-find-the-degree-of-sin2-pi-n-over-mathbbq) [Ask Question](https://math.stackexchange.com/questions/ask) Asked 8 years, 6 months ago Modified [8 years, 6 months ago](https://math.stackexchange.com/questions/2319458/how-to-find-the-degree-of-sin2-pi-n-over-mathbbq?lastactivity "2017-06-12 08:46:45Z") Viewed 1k times 1 \$\\begingroup\$ 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](https://math.stackexchange.com/questions/1352274/a-question-about-the-degree-of-sin2%CF%80-n-over-the-rationals), but that seems doesn't solve this problem, I need some detail about this quesion. - [field-theory](https://math.stackexchange.com/questions/tagged/field-theory "show questions tagged 'field-theory'") - [extension-field](https://math.stackexchange.com/questions/tagged/extension-field "show questions tagged 'extension-field'") - [cyclotomic-fields](https://math.stackexchange.com/questions/tagged/cyclotomic-fields "show questions tagged 'cyclotomic-fields'") [Share](https://math.stackexchange.com/q/2319458 "Short permalink to this question") Cite Follow [edited Jun 12, 2020 at 10:38](https://math.stackexchange.com/posts/2319458/revisions "show all edits to this post") [![Community's user avatar](https://www.gravatar.com/avatar/a007be5a61f6aa8f3e85ae2fc18dd66e?s=64&d=identicon&r=PG)](https://math.stackexchange.com/users/-1/community) [Community](https://math.stackexchange.com/users/-1/community)Bot 1 asked Jun 12, 2017 at 8:46 [![ηŽ‹ζŽθΏœ's user avatar](https://i.sstatic.net/sDXOr.jpg?s=64)](https://math.stackexchange.com/users/334260/%E7%8E%8B%E6%9D%8E%E8%BF%9C) [ηŽ‹ζŽθΏœ](https://math.stackexchange.com/users/334260/%E7%8E%8B%E6%9D%8E%E8%BF%9C) 89211 gold badge66 silver badges1919 bronze badges \$\\endgroup\$ 7 - \$\\begingroup\$ What do you mean by "it doesn't solve this problem"? It does. \$\\endgroup\$ user23365 – user23365 2017-06-12 08:50:17 +00:00 Commented Jun 12, 2017 at 8:50 - \$\\begingroup\$ Not "the" but "**a**" primitive root. \$\\endgroup\$ DonAntonio – [DonAntonio](https://math.stackexchange.com/users/31254/donantonio "215,485 reputation") 2017-06-12 08:50:22 +00:00 Commented Jun 12, 2017 at 8:50 - \$\\begingroup\$ 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}})\$ \$\\endgroup\$ ηŽ‹ζŽθΏœ – [ηŽ‹ζŽθΏœ](https://math.stackexchange.com/users/334260/%E7%8E%8B%E6%9D%8E%E8%BF%9C "892 reputation") 2017-06-12 08:56:47 +00:00 Commented Jun 12, 2017 at 8:56 - \$\\begingroup\$ Would you please show me some details and post as an answer?@DonAntonio \$\\endgroup\$ ηŽ‹ζŽθΏœ – [ηŽ‹ζŽθΏœ](https://math.stackexchange.com/users/334260/%E7%8E%8B%E6%9D%8E%E8%BF%9C "892 reputation") 2017-06-12 08:58:50 +00:00 Commented Jun 12, 2017 at 8:58 - \$\\begingroup\$ 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\$. 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