I’ve recently proved a theorem that is probably well-known to anyone that studied number theory. I didn’t know it before, though, and it was a lot of fun, so I decided to write it up here in case there are other people around that like number theory a bit but not enough to actually learn it.

The question I wanted to answer was which beautiful angles have beautiful cosines? My completely objective definitions are that a beautiful angle is of the form $\frac{p}{q}\pi$ for small integers $p,q$, and a beautiful cosine is of the form $\frac{a}{b}+\sqrt{\frac{c}{d}}$, again for small integers. Surprisingly, the question can be answered completely, and it is not even difficult if you know number theory (I don’t).

The first thing to notice is that any algebraic number uniquely defines its minimal polynomial, which is the polynomial with rational coefficients (and leading coefficient one) of least degree that has this algebraic number as a root. So we only need to compute the minimal polynomial of $\alpha$ and see if it matches the minimal polynomial of $\cos(\frac{p}{q}\pi)$ for any $p,q$. Which doesn’t really help, because what the hell is the minimal polynomial of $\cos(\frac{p}{q}\pi)$?

Luckily though the minimal polynomial of $e^{i\pi\frac{p}{q}}$ instead is very well characterized: it is known as a cyclotomic polynomial, and it is given simply by $\Phi_n(x) := \prod_k (x-e^{i2\pi\frac{k}{n}})$, over all $k$ that are coprime to $n$.

Since $\cos(\frac{p}{q}\pi) = \alpha$ iff $e^{i\frac{p}{q}\pi} = \alpha \pm i\sqrt{1-\alpha^2}$ what we do instead is to compute the minimal polynomial of $\alpha + i\sqrt{1-\alpha^2}$ (which is equal to the minimal polynomial of $\alpha – i\sqrt{1-\alpha^2}$), and see if it matches some cyclotomic polynomial.

Let’s start by disregarding the trivial cases where $\frac{a}{b}+\sqrt{\frac{c}{d}}$ is $1,0,$ or $-1$. We know that $\cos(0),\cos(\pi/2),\cos(-\pi/2),$ and $\cos(\pi)$ give them, there’s no need for the big guns. We’ll assume from here on that $\frac{a}{b}+\sqrt{\frac{c}{d}}$ is never an integer.

The first interesting case is when $c=0$, so we’re just asking which beautiful angles have rational cosines. To answer that we need the minimal polynomial of $z=\frac{a}{b}+i\sqrt{1-\frac{a}{b}^2}$. It’s clear that $(x-z)(x-z^*)$ works, and we can’t get rid of the root $z^*$, so that’s the minimal one. Explicitly it is

\[ x^2 -2 \frac{a}{b} x +1, \]so it can only match cyclotomic polynomials of degree 2. Well, the degree of a cyclotomic polynomial $\Phi_n(x)$ is simply the number of integers $k \le n$ that are coprime to $n$, so we need to find which integers $n$ have exactly 2 smaller integers that are coprime to it.

This is kind of a strange question, but the reverse is rather straightforward: the number of integers coprime to $n$ is the definition of Euler’s totient function $\varphi(n)$. It has the simple lower bound of $\varphi(n) \ge \sqrt{n/2}$, which allows us to brute-force our problem: we just need to look at $n$ up to 8 and we’ll have all integers with exactly 2 coprimes. They are 3,4, and 6, so we just need to check whether our minimal polynomial matches $(x-e^{i2\pi\frac{1}{3}})(x-e^{i2\pi\frac{2}{3}})$, $(x-e^{i2\pi\frac{1}{4}})(x-e^{i2\pi\frac{3}{4}})$, or $(x-e^{i2\pi\frac{1}{6}})(x-e^{i2\pi\frac{5}{6}})$. Which is kind of overkill, because we already have the angles, so we can simply check instead which of them have rational cosines, which turn out to be $\cos(\frac{2}{3}\pi) = \cos(\frac{4}{3}\pi) = -\frac{1}{2}$, and $\cos(\frac{1}{3}\pi) = \cos(\frac{5}{3}\pi) = \frac{1}{2}$. And these are the only ones.

Moving on to the next interesting case, we set $a=0$, so that we are asking which beautiful angles have cosines equal to $\sqrt{\frac{c}{d}}$. We need then to figure out the minimal polynomial of $z = \sqrt{\frac{c}{d}} +i\sqrt{1-\frac{c}{d}}$. Now the polynomial $(x-z)(x-z^*)$ doesn’t work anymore, as it has irrational coefficients, but this is easy to fix by considering instead $(x-z)(x-z^*)(x+z)(x+z^*)$, which does work and is also clearly the minimal polynomial. Working it out, we get

\[x^4 + \left(2-4\frac{c}{d}\right)x^2 + 1.\] Calculating which integers have exactly four coprimes, like before, we see that they are 5, 8, 10, and 12, which is way too much work to check by hand, so we’ll do something else instead: another nice property of cyclotomic polynomials is that they only have integer coefficients, so we only need to look for $c,d$ such that $4\frac{c}{d}$ is integer. The only nontrivial cases are $c=1,d=2$ and $c=3,d=4$, which tells us that the only beautiful cosines that are square roots of rationals are $\pm\sqrt{1/2}$ and $\pm\sqrt{3/4}$. The angles that give these cosines are the well-known $\pi/4,7\pi/4,3\pi/4,5\pi/4$ and $\pi/6,11\pi/6,5\pi/6,5\pi/6$.

To finalize let’s consider cosines of the form $\frac{a}{b}+\sqrt{\frac{c}{d}}$ for $a \neq 0$ and $c \neq 0$. We need the minimal polynomial for the hideous thing

\[z = \frac{a}{b}+\sqrt{\frac{c}{d}}+i\sqrt{1-\left(\frac{a}{b}+\sqrt{\frac{c}{d}}\right)^2}.\] The previous case does not work, but if you think about it, it is easy to see that with

\[w = \frac{a}{b}-\sqrt{\frac{c}{d}}+i\sqrt{1-\left(\frac{a}{b}-\sqrt{\frac{c}{d}}\right)^2}\] the polynomial $(x-z)(x-z^*)(x-w)(x-w^*)$ will do the job. Working it out with Mathematica, because that’s already above what I can do by hand, we get

\[x^4 – 4\frac{a}{b}x^3 + \left(2+4\frac{a^2}{b^2}-4\frac{c}{d}\right)x^2 – 4\frac{a}{b}x + 1.\] At this point it’s better to just look up what are the cyclotomic polynomials for $n=5,8,10,12$, which as we said before are the only ones with degree 4. They are

\begin{align*}

\Phi_5(x) &= x^4 + x^3 + x^2 + x +1 \\

\Phi_8(x) &= x^4 + 1 \\

\Phi_{10}(x) &= x^4 – x^3 + x^2 – x + 1 \\

\Phi_{12}(x) &= x^4 – x^2 + 1,

\end{align*}which is not that bad. $\Phi_8(x)$ and $\Phi_{12}(x)$ are right out because they require $a=0$, which we analysed before. To make our minimal polynomial equal to $\Phi_5(x)$ we need $a=-1$ and $b=4$, which tells us that $c/d = 5/16$. We have a match! It implies that the cosines of the angles $2\pi/5, 4\pi/5, 6\pi/5, 8\pi/5$ must be $-\frac{1}{4}+\sqrt{\frac{5}{16}}$ or $-\frac{1}{4}-\sqrt{\frac{5}{16}}$. A similar matching occurs with $\Phi_{10}(x)$, the only difference is that $a=1$ instead of $-1$. It implies that the cosines of the angles $\pi/5, 3\pi/5, 7\pi/5, 9\pi/5$ must be $\frac{1}{4}+\sqrt{\frac{5}{16}}$ or $\frac{1}{4}-\sqrt{\frac{5}{16}}$. And that’s it! There can be no more.

To summarize, we found out that the only possible beautiful cosines of beautiful angles are $0, \pm1, \pm\frac{1}{2},\pm\frac{1}{\sqrt{2}},\pm\frac{\sqrt{3}}{2},\frac{1\pm\sqrt{5}}{4},\frac{-1\pm\sqrt{5}}{4}$. And the beautiful angles with these cosines are all the integer multiples of $\pi,\pi/2,\pi/3,\pi/4,\pi/5,\pi/6$. But be careful, this holds really only for cosines, the sines of integer multiples of $\pi/5$ are never beautiful.1

Funnily enough, these are precisely the angles and cosines we learn in school, except for $\pi/5$. Usually we only learn a couple of examples that are good enough, and leave behind an incredible wealth of interesting cases. But here no, out teachers taught us all the angles and cosines that were worth knowing! $\pi/5$ is an understandable exception, it is not as beautiful as the others, and we should anyway demand both the sines and cosines to be beautiful.