I'm writing a geometric software library and I came up with a useful concept.
Let's call a real number α an algebraic angle if α∈[0,2π) and cosα 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 2π. Then algebraic angles are closed under negation and addition (since cos(α+β)=cosαcosβ−sinαsinβ).
- Define multiplication of an algebraic angle by a rational number to also wrap around at 2π. Multiplying an algebraic angle by a rational yields another algebraic angle! This can be proved from the identity cos(nα)=Tn(cosα), which I learned about here. Tn is a Chebyshev polynomial of the first kind. In particular, it is a polynomial with integer coefficients.
- Since π is an algebraic angle, so is any rational multiple of π in [0,2π).
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(yα) is not necessarily equal to (xy)α. For example, (1/2)(2⋅π) is 0 but (1/2⋅2)π is π.
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(−15√5).
Let me elaborate on why an algebraic angle divided by an integer yields an algebraic angle: Suppose that cosα is algebraic, so that there is some polynomial P(x) with integer coefficients such that P(cosα)=0. Let n be a positive integer. By the Chebyshev identity above, P(Tn(cos(α/n)))=0, so cos(α/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 α↦eiα. 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.
No comments:
Post a Comment