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Let $\zeta_n$ be the $n$-th primitive root of unity and $4 \mid n$. Consider the field extensions $\mathbb Q \subset \mathbb Q(\sin(2\pi k/n) \subset \mathbb Q(\zeta_n)$.

  1. What is the degree of the extension $[\mathbb Q(\zeta_n):\mathbb Q(\sin(2\pi k/n)]$?
  2. What is $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q(\sin(2\pi k/n))$?
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What is $k{}{}$? – Chris Eagle Apr 25 '12 at 9:17
I don't think that $\sin(2\pi k/n)$ is always an element of the cyclotomic field? For example, if $n=5$, $k=1$, then $\sin 72^\circ=\cos 18^\circ=\sqrt{2(5+\sqrt5)}/4$, which generates a quartic extension (see e.g. this table). But the real subfield of $\mathbb{Q}(\zeta_5)$ is $\mathbb{Q}(\sqrt5)$?? OTOH, if $4\mid n$, then $i\in \mathbb{Q}(\zeta_n)$, and we have the obvious way of writing $\sin 2\pi k/n$ in terms of $\zeta_n^k$. – Jyrki Lahtonen Apr 25 '12 at 9:22
@JyrkiLahtonen: Perhaps $\cos \frac{2 \pi k}{n}$ is meant, since Galois number fields are always closed under real parts. – Zhen Lin Apr 25 '12 at 9:53
@ZhenLin, that would be my guess as well. But in that case the answer to the question is very easy (and probably covered in most books on Galois theory). Let's just wait for the OP to clarify, whether the typo is about $\sin\leftrightarrow \cos$, or about forgetting the assumption $4\mid n$. – Jyrki Lahtonen Apr 25 '12 at 9:59
Duh. The best example showing that the problem is off is the case $n=3$: $\mathbb{Q}(\sin2\pi/3)=\mathbb{Q}(\sqrt3)$ and $\mathbb{Q}(\zeta_3)=\mathbb{Q}(\sqrt{-3})$. – Jyrki Lahtonen Apr 25 '12 at 10:02
up vote 4 down vote accepted

You should first prove that $[\mathbb Q(\cos(2\pi /n)):\mathbb Q] =\phi(n)/2$, which is easy.
Then you should use that $\sin(2\pi k/n)=\cos ((n-4k)\pi/2n)$.
A few calculations necessitating case distinctions will then yield the result: $$[\mathbb Q(\sin(2\pi /n)):\mathbb Q]=\phi(n) \quad \text {if} \; n \not \equiv 0 \;( \text {mod} \; 4) $$ $$[\mathbb Q(\sin(2\pi /n)):\mathbb Q]=\phi(n)/2 \quad \text {if} \; n \equiv 0 \;( \text {mod} \; 8)$$ $$[\mathbb Q(\sin(2\pi /n)):\mathbb Q]=\phi(n)/4 \quad \text {if} \; n \equiv 4 \;( \text {mod} \; 8)$$ [I have answered your question in the above form because (as Jyrki noted) the original question doesn't make sense, since we don't always have an inclusion $\mathbb Q(\sin(2\pi /n))\subset \mathbb Q(\zeta_n).]$

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How can we show $[\mathbb Q(\sin(2\pi /n)):\mathbb Q]=\phi(n)/2 \quad \text {if} \; n \equiv 0 \;( \text {mod} \; 8)$ from $\sin(2\pi k/n)=\cos ((n-4k)\pi/2n)$ – Mohan Apr 25 '12 at 12:48
@Mohan, if $\gcd(k,n)=1$, then $K=\mathbb{Q}(\cos(2\pi k/n))$ is the real subfield of $L=\mathbb{Q}(\zeta_n)$. Therefore $[L:K]=2$. – Jyrki Lahtonen Apr 25 '12 at 12:53

Since $e^{ix} = \cos(x) + i \sin(x)$ we have $$ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}. $$ Thus $\sin(2 \pi k/n)$ is, unless $n$ is a multiple of $4$, an element of ${\mathbb Q}(\zeta_{4n})$. The degrees of your extensions are basically measured by the greatest common divisors of $k$ and $n$. There will be clean formulas for prime values of $n$, or e.g. if $k = 1$.

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