In cyclotomic theory, $\cos (2\pi /n)$) is shown to have degree $\varphi (n)/2$ over the rationals $\mathbb{Q}$, while $\sin (2\pi /n)$has degree $\varphi (n)$ (as long as n is not divisible by 4).$Cos(2\pi /n)=(\zeta +{{\zeta }^{-1}})/2$ where $\zeta$ is a primitive root of unity in the nth cyclotomic field, so $\mathbb{Q}(2\cos (2\pi /n))$= $\mathbb{Q}(\zeta +{{\zeta }^{-1}})$ and this extension field has degree 2 over $\mathbb{Q}(\zeta )$and hence degree $\varphi (n)/2 $ over $\mathbb{Q}$. The extension field $\mathbb{Q}(\zeta +{{\zeta }^{-1}})$ is sometimes called the maximal real extension of $\mathbb{Q}(\zeta )$.
$Sin(2\pi /n)=(\zeta -{{\zeta }^{-1}})/i$ so the matching extension field is $\mathbb{Q}(\zeta -{{\zeta }^{-1}})$which is totally complex and this extension is the maximal complex extension. When n is not divisible by 4 only the identity map ${{\sigma }_{{{1}_{{}}}}}:\zeta \to {{\zeta }^{1}}$ fixes $\zeta - {\zeta ^{ - 1}}$, so this extension is degree 1 over $\mathbb{Q}(\zeta )$ and hence degree $\varphi (n)$over $\mathbb{Q}$ as stated above.
But as a vector space $\mathbb{Q}(\zeta )$is the direct sum of $\mathbb{Q}(\zeta +{{\zeta }^{-1}})$ and $\mathbb{Q}(\zeta -{{\zeta }^{-1}})$ and since $\mathbb{Q}(\zeta +{{\zeta }^{-1}})$ is degree $\varphi (n)/2$over $\mathbb{Q}$, $\mathbb{Q}(\zeta -{{\zeta }^{-1}})$ should also be degree $\varphi (n)/2$ over $\mathbb{Q}$, yet the degree of sine over $\mathbb{Q}$is $\varphi (n)$ (I suspect the issue is that $\sin (2\pi /n)$is really in $\mathbb{Q}(\zeta ,i)$.)
