Kronecker-Weber theorem for $\sin \left(\frac{2 \pi}{n}\right)$

Question

By the Kronecker-Weber theorem, every abelian extension of $\mathbb{Q}$ is contained in a cyclotomic extension, i.e. $\mathbb{Q}(e^{2\pi i/n})$.

Noting that $$e^{2\pi i/n}=\pm\sqrt{1-\sin \left(\frac{2\pi}{n}\right)^2}+i \sin\left(\frac{2\pi}{n}\right),$$

is it true that every abelian extension of $\mathbb{Q}$ is contained in $\mathbb{Q}(\sin \left(\frac{2\pi}{n}\right))$?

Edit: Is every abelian extension of $\mathbb{Q}$ contained $\mathbb{Q}(i,\sin \left(\frac{2\pi}{n}\right))$?

Answer 1

A more structured assertion here is that the subfield $k$ of $\mathbb Q(\zeta_n)$ fixed by $-1$ (identifying the Galois group with $\mathbb Z/n^\times$) is $\mathbb Q(\cos{2\pi\over n})$. It is "totally real", in the sense that every imbedding of it into $\mathbb C$ actually has image in $\mathbb R$.

For odd $n$, it is not quite that adjoining $i$ to $k$ gives the full cyclotomic field, etc., but we're only off by a little... EDIT: namely, $i\in \mathbb Q(\zeta_n)$ if and only if $4\mid n$. And, if we're imagining that $\mathbb Q(\zeta_n)=\mathbb Q(\cos{2\pi\over n})(\sqrt{d})$ for some negative (square-free) integer $d$ (for example, $-1$), then this entails things... Of course, there is a lot to be said about cyclotomic fields... :)

Answer 2

Assume that $4\mid n$. Then $-ie^{2i\pi/n}$ is still a primitive $n$-th root of unity, and $-ie^{2i\pi/n}=\alpha+\sin{\frac{2\pi}{n}}$ with $\alpha^2=\sin^2{\frac{2\pi}{n}}-1$. Thus $K_n=\mathbb{Q}(e^{2i\pi/n})$ has degree at most $2$ over $R_n=\mathbb{Q}\left(\sin{\frac{2\pi}{n}}\right)$. But $i \in K_n \setminus R_n$. So $K_n=R_n(i)$ and $K_n=\mathbb{Q}\left(i,\sin{\frac{2\pi}{n}}\right)$. We’re done by Kronecker-Weber.