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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.

Mikhail Hogrefe
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    In regards to your last remark: Every angle you can construct with a compass and straight-edge is algebraic, so these are likely to show up implicitly in a lot of places and the rules for algebraic manipulations of trigonometric functions are certainly important. – Milo Brandt Jan 10 '17 at 01:02
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    Algebraic points on the unit circle are studied frequently in number theory, particularly by analytic number theorists. They have many interesting properties. The fact that some of the points are not simply roots of unity is one of the classical results, for example. I'm not sure how much of this is related to geometry, but then the study of algebraic numbers is more the domain of number theory than geometry anyways. – Adam Hughes Jan 10 '17 at 01:12
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    Here is a cute irony: if $\alpha$ is itself algebraic, then $\sin\alpha$, $\cos\alpha$, and $\tan\alpha$ are transcendental. (This follows from the Lindemann-Weierstrass theorem.) In other words, if $\alpha$ is algebraic, then $\alpha$ is not an algebraic angle! – symplectomorphic Jan 10 '17 at 01:26
  • Good point @symplectomorphic! Maybe "algebraic angle" is too confusing a term... – Mikhail Hogrefe Jan 10 '17 at 01:39
  • @AdamHughes: Could you point me to some papers about algebraic points on the unit circle? Google isn't much help. I suppose I added the geometry tag because I'm interested in doing geometrical computation with algebraic numbers and angles. Rational numbers are too limited, arbitrary-precision reals have too many undecidable properties, but algebraic numbers are just right! – Mikhail Hogrefe Jan 10 '17 at 01:49
  • @MikhailHogrefe here is one example http://www.sciencedirect.com/science/article/pii/S0022314X00926350?np=y – Adam Hughes Jan 10 '17 at 01:53
  • @symplectomorphic : ​ 0 and sin(0) and cos(0) and tan(0) are all algebraic. ​ ​ ​ ​ –  Jan 10 '17 at 06:21
  • @RickyDemer: add the word "nonzero." This is obvious. – symplectomorphic Jan 10 '17 at 15:01

3 Answers3

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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 $[0,1]$.

lulu
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    @Rahul No...as I remark, $h=\tan \frac {\theta}2$. Very different from $\theta$. I meant "rational", as the transformations that take $h$ to $\cos \theta, \sin \theta$ are rational functions. – lulu Jan 10 '17 at 03:10
  • @Rahul Sorry, not following you. I only mentioned angles at all because the OP did. $h$ is a point on an interval. $P(h)$ is a point on the unit circle. If we write $(x_0,y_0)$ for some point on the circle (first quadrant) then we can recover $h$ as $\frac {1-x_0}{y_0}$. No angles needed. The argument I gave shows that $x_0,y_0$ are algebraic iff $h$ is. If, as the OP does, you choose to parametrize the circle with an angle $\theta$ in the usual way then $h=\tan \frac {\theta}2$. The OP calls $\theta$ "algebraic" if it's sine and cosine are algebraic...his term, not mine...... – lulu Jan 10 '17 at 11:40
  • @Rahul .... Anyway, that definition of an "algebraic angle" (again, the OP's, not mine) is equivalent to saying "$h$ is algebraic". To stress, the OP does not mean that the angle itself is algebraic (so perhaps it's not the best phrasing). Indeed other that $\theta =0$ there aren't any algebraic numbers such that the trig functions are also algebraic. Personally, I wouldn't choose this terminology...but I wanted to tie what I wrote to what appears in the question. – lulu Jan 10 '17 at 11:44
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    @Rahul Ah, so are you just objecting that I am using "algebraic angle" to refer to the algebraic points on the circle whereas the OP is using the term to refer to the angle which gives rise to those points? If so, fair enough. I will edit accordingly. – lulu Jan 10 '17 at 17:44
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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,

Patrick Da Silva
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  • Thanks Patrick! I added what I think is a proof that algebraic angles are closed under division by integers. I'll need some time to chew over your comment - I'm self-studying Pinter's undergrad algebra text and I have a lot to learn still. – Mikhail Hogrefe Jan 10 '17 at 01:36
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    How come your correct statement doesn't seem to have any relation to the question? Did you mean "algebraic angle" instead of "algebraic number"? In any case, suppose we extend the definition of algebraic angle to all of $\mathbb R$ (by removing the range restriction). Is it false that this periodic subset of $\mathbb R$ is a $\mathbb Q$-vector space? – Erick Wong Jan 10 '17 at 01:58
  • I see why it's not a vector space and I'll edit the post accordingly. – Mikhail Hogrefe Jan 10 '17 at 02:04
  • @Erick Wong : I might have typed algebraic number by accident, sorry about that. – Patrick Da Silva Jan 10 '17 at 02:05
  • @Mikhail Hogrefe : the problem is not in your argument, it's in your definition of algebraic angle. If you define algebraic angles as real numbers satisfying some property then they indeed form a $\mathbb Q$-vector space by your argument, but for some reason you insisted in defining them modulo $2\pi$. This changes things! – Patrick Da Silva Jan 10 '17 at 02:07
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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.

Marc van Leeuwen
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