Have "algebraic angles" been studied before?

I'm writing a geometric software library and I came up with a useful concept.

Let's call a real number $\alpha$ an algebraic angle if $\alpha\in[0,2\pi)$ and $\cos \alpha$ is an algebraic number. The set of algebraic angles has some pretty neat properties:

• The sine, cosine, and tangent of an algebraic angle are algebraic.
• Define negation and addition of angles the usual way, wrapping around at $2\pi$. Then algebraic angles are closed under negation and addition (since $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$).
• Define multiplication of an algebraic angle by a rational number to also wrap around at $2\pi$. Multiplying an algebraic angle by a rational yields another algebraic angle! This can be proved from the identity $\cos(n\alpha)=T_n(\cos\alpha)$, which I learned about here. $T_n$ is a Chebyshev polynomial of the first kind. In particular, it is a polynomial with integer coefficients.
• Since $\pi$ is an algebraic angle, so is any rational multiple of $\pi$ in $[0,2\pi)$.
• Algebraic angles are a vector space over the rationals (I haven't proved anything interesting using this fact though). They're not a vector space because $x(y\alpha)$ is not necessarily equal to $(xy)\alpha$. For example, $(1/2)(2\cdot\pi)$ is $0$ but $(1/2\cdot 2)\pi$ is $\pi$.

Has the set of algebraic angles appeared in academic literature? Individual algebraic angles come up everywhere in geometry: for example, the interior angle between two faces of a regular dodecahedron is $\arccos(-\frac{1}{5}\sqrt{5})$.

Let me elaborate on why an algebraic angle divided by an integer yields an algebraic angle: Suppose that $\cos\alpha$ is algebraic, so that there is some polynomial $P(x)$ with integer coefficients such that $P(\cos\alpha)=0$. Let $n$ be a positive integer. By the Chebyshev identity above, $P(T_n(\cos(\alpha/n)))=0$, so $\cos(\alpha/n)$ is algebraic.

Edited 2017-01-10:

Ok, after reading some of the answers and comments I realized that the interesting properties of algebraic angles become obvious if you consider how they transform under $\alpha\mapsto e^{i\alpha}$. The algebraic angles become algebraic points on the unit circle, angle negation becomes complex conjugation, angle addition becomes complex multiplication, and multiplying by a rational essentially becomes raising to a rational power. It's clear that algebraicity is preserved by each of these operations.

Just to point out, consider the usual Weierstrass substitution. That is to say, start with the point $(-1,0)$ on the unit circle and form the line connecting it to the point $(0,h)$ for $0≤h≤1$. Then it is easy to see that this line meets the unit circle again at the point $P(h)=\left( \frac {1-h^2}{1+h^2},\frac {2h}{1+h^2}\right)$. We see from this that, if $h$ is algebraic, then so are both coordinates of $P(h)$.

Conversely, given a point $Q(\theta)=(\cos \theta,\sin \theta)$ in the first quadrant on the unit circle, we can recover $h$ as $h=\frac {1-\cos \theta}{\sin \theta}$. Thus if both coordinates are algebraic, so is $h$. (N.B. note that $h=\tan \frac {\theta}2$ ).

Thus, up to a few reflections, The coordinates of the points on the circle which correspond to your algebraic angles (that is to say, the $\sin$ and $\cos$ of those "algebraic angles") are just rational transformations of the algebraic numbers on $[0,1]$.

I want to make precise your statement that "Algebraic angles are a vector space over the rationals." since as it is, it is sadly incorrect. First of all, since you can add algebraic angless and multiply them by integers (which is what you proved with the Tchebyshev polynomial argument) and since algebraic angles live within the set $\mathbb R/2 \pi \mathbb Z$ (which is the set in which "real numbers wrapped around $2\pi$" live in), the set of algebraic angles is a $\mathbb Z$-submodule (also known as a subgroup) of the $\mathbb Z$-module (also known as a group) $\mathbb R/2 \pi \mathbb Z$. However, the latter is not a $\mathbb Q$-vector space since when $V$ is a vector space over the field $F$ and $(v,\lambda) \in V \times F$ with $v,\lambda \neq 0$, then $\lambda v \neq 0$. This is not the case with $\mathbb R/2\pi \mathbb Z$ over $\mathbb Q$ since $2 \cdot \pi = 2\pi = 0 \pmod{2\pi}$.

The correct statement concerning multiplication would be as follows : if $\alpha \in \mathbb R/2 \pi \mathbb Z$ is a real number modulo $2\pi$ and $n \in \mathbb Z$ an integer such that $n \alpha$ is an algebraic number which is not a multiple of $2\pi$, then $\alpha$ is an algebraic number.

Note that proving that $n \alpha$ is an algebraic integer when $\alpha$ is can also be proved by induction on $n$ using your formula for addition since $n\alpha = (n-1)\alpha + \alpha$.

Hope that helps,

What you are describing is just the special orthogonal group $SO_2(\Bbb R_{\rm alg})$ of the field of algebraic real numbers, which one could also describe as $U_1(\overline{\Bbb Q})$ for a choice of (complex) conjugation in the algebraic closure $\overline{\Bbb Q}$ of$~\Bbb Q$, and the fact that this is a divisible group.