I have been asked to determine the degree of the field extension over $\mathbb{Q}$ generated by $\sin(2\pi/n)$ for any natural number $n$. I have been able to show that the degree of extension by $\cos(2\pi/n)$ is $\varphi(n)/2$. However I am going nowhere with $\sin(2\pi/n)$. Can I get some help as to how to approach?
2026-03-26 08:00:37.1774512037
Degree of the field extension generated by $\sin(2\pi/n)$ for any natural number $n$.
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Okay so first observe that $[\mathbb{Q}(\cos(\frac{2k\pi}{n})):\mathbb{Q}]=\frac{\varphi(n)}{2}$, whenever $\gcd(k,n)=1$. I hope you can prove this quite easily. Let me know if you find proving this difficult.
Now let us recollect two basic results:
Next, write $\sin(\frac{2k\pi}{n})=\cos(\frac{\pi}{2}-\frac{2k\pi}{n})=\cos\Big(\frac{(n-4k)\pi}{2n}\Big)$. Choose $k$ to be odd and coprime to $n$ (this will help you later in the proof). The rest of the proof is routine. Roughly speaking, you will have to consider different cases and reduce the above expression suitably to the form $\cos(\frac{2k'\pi}{n'})$ such that in each case $k'$ and $n'$ are coprime (this can be checked using the two results given above). Then you can apply the statement given at the very beginning of this answer to obtain the required degree of extension. I will give an outline. Work out the details yourself. The following are the four cases:
Finally putting $k=1$ (note that this choice of $k$ is odd and is coprime to $n$) in each case we get the required degrees of extension.