I have shown that $\mathbb{Q}(\cos\frac{2\pi}{n})$ is Galois over $\mathbb{Q}$ since it's contained in the $n$-th cyclotomic field. But I don't think it's also true for $\mathbb{Q}(\sin\frac{2\pi}{n})$. I tried the case $n=5$ as follows:
Since $\sin\frac{2\pi}{5}=\sqrt\frac{5+\sqrt 5}{8}$, its minimal polynomial over $\mathbb{Q}$ is $f(x)=x^4-\frac{5}{4} x^2+\frac{5}{16}$. So $\pm$$\sqrt\frac{5+\sqrt 5}{8}$ and $\pm$$\sqrt\frac{5-\sqrt 5}{8}$ are its roots. But I don't know whether $\sqrt\frac{5-\sqrt 5}{8}$ is in $\mathbb{Q}(\sin\frac{2\pi}{n})=\mathbb{Q}(\sqrt\frac{5+\sqrt 5}{8})$.
Maybe I should assume the contrast and lead to a contradiction? If $\mathbb{Q}(\sin\frac{2\pi}{n})/\mathbb{Q}$ is Galois, then $\mathbb{Q}(\sin\frac{2\pi}{n},\cos\frac{2\pi}{n},i)/\mathbb{Q}$ is also Galois... But again I get lost. Can someone give me some hints? Many thanks.
Hint: it is a subfield of the abelian extension $\mathbb{Q}(e^{2\pi i / 4n})$ over $\mathbb{Q}$, so...