# 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?

• If you have an angle, then can't you use your compass to mark the length of the diagonal? Then you have the length of the base (adjacent side) and the hypotenuse, and you can use similar triangles to get this ratio in terms of the base (if you assume the base has unit length). – Toby Mak Jan 27 at 13:45
• Non-constructible numbers intersect both the transcendental and the algebraics. – CogitoErgoCogitoSum Apr 23 at 16:34

1. 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".
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.