Are trigonometric ratios of non-constructible angles transcendental numbers? Can it be said that if an angle is constructible, then all its trigonometric ratios are constructible, and vice versa?
And if this is indeed true, what can be said about the trigonometric ratios of all other angles? 
Since these ratios are not constructible, they clearly cannot be rational. But can they be algebraic? Or must they be transcendental?
 A: A partial answer based on looking up things after I posted the question:


*

*From Wikipedia I find that algebraic numbers include "values of trigonometric functions of rational multiples of $\pi$ (except when undefined): that is, the trigonometric numbers".

*Then there is the Lindemann-Weierstrass theorem that implies that trigonometric ratios of non-zero algebraic numbers are transcendental.


One way to approach the first point is through the expansion of $\cos{(n\theta)}$, which goes like this:
$$
\cos{( n \theta )} = \sum_{\textrm{even}\ k} (-1)^{k\over 2}\cdot{n \choose k}\cdot \cos^{n-k}{\theta}\ \cdot \sin^{k}{\theta}
$$
... which implies:
$$
\cos{( n \theta )} = \sum_{m=0}^{n \over 2} (-1)^m\cdot{n \choose 2m}\cdot \cos^{n-2m}{\theta}\ \cdot (1-\cos^2{\theta})^{m}
$$
... which means that $\cos{(n\theta)}$ is a degree $n$ polynomial of $\cos{\theta}$, which would further mean that the cosine of ${1\over n}$ of any angle is the root of a degree $n$ polynomial. The coefficients of all non-zero powers of $\cos{\theta}$ in the polynomial are clearly integers, so if $\cos{(n\theta)}$ is itself algebraic then the root $\cos{\theta}$ is also algebraic.
So we can say that if the trigonometric ratios of an angle are algebraic, the trigonometric ratios of all rational multiples of that angle are also algebraic. 
