Reciproc of the Lindemann theorem and the arc cosine of the golden ratio

Question

Via the Lindemann theorem it is easy to prove that the cosine of any rational multiple of $\pi$ is an algebraic number; however its contrapositive only tells us that the arc cosine of an algebraic number is a transcendental number, not necessarily linearly dependent from $\pi$ over the rationals.

I ask, then, is it true via any other theorem? Is there a nice counterexample?

I came to this question specifically as I've found in my research an angle $\theta$ such that $$\cos\theta = \frac{\sqrt{5}-1}{2} =: \varphi^{-1},$$ and I wanted to find out if $\theta$ has any nice representation, such as a rational multiple of $\pi$.

You don't need the Lindemann theorem for the forward direction; de Moivre's formula yields algebraicity of rational cosines fairly directly. — hmakholm left over Monica, Sep 24 '11 at 06:17
Peering at the simple continued fraction expansion of $\arccos,\varphi^{-1}$, it doesn't seem rational to me... — J. M. ain't a mathematician, Sep 24 '11 at 06:39
If the arccosine of an algebraic number were always a rational multiple of $\pi$, I would expect to see a lot more single-term Machin-like formulas for $\pi$... — hmakholm left over Monica, Sep 24 '11 at 06:40
Whoops, I forgot a $1/\pi$ factor in my previous comment. On the other hand, Mathematica tells me that the number is very nearly equal to one of the roots of $x^{11}-7x^{10}+5x^9-x^8+4x^7-4x^6+x^5-6x^4-3x^3+3x^2+3x-1$... — J. M. ain't a mathematician, Sep 24 '11 at 06:45

score 2 · Answer 1 · answered Sep 24 '11 at 14:18

2

No. As a concrete counterexample, $\cos^{-1}(0.6)$ is not a rational multiple of $\pi$.

$\cos^{-1}(0.6)$ is a rational multiple of $\pi$ if and only if $0.6+0.8i$ is a root of unity. But if $(0.6+0.8i)^n = 1$ then $(3+4i)^n$ must be real. And this is impossible because $(3+4i)^n \equiv 3+4i \pmod 5$ for all $n\ge 1$

answered Sep 24 '11 at 14:18

hmakholm left over Monica

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1

Nice argument. Unfortunately it does not seem to apply to $\varphi^{-1}$. – Mateus Araújo Sep 24 '11 at 17:36

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

No.

As Henning said, $\cos^{-1} \alpha$ is a rational multiple of $\pi$ if and only if $z = \alpha+i\sqrt{1-\alpha^2}$ is a root of unity. Note that as a sum of algebraic numbers, $z$ is itself an algebraic number; and to find out for which $F \in \mathbb{Q}[x]$ is true that $F(z)=0$ we only need the resolvent $F(z) = \rho(q(y),p(z-x))$, where $q \in \mathbb{Q}[y]$ is the polynomial for which $q(\alpha)=0$, and $p\in\mathbb{Q}[x]$ is the polynomial for which $p(i\sqrt{1-\alpha^2})=0$.

But using this answer we can easily decide whether $z$ is a root of unity: An algebraic $z$ of modulus one is a root of unity if and only if its (irreducible) $F$ is a cyclotomic polynomial. Let $N = \deg F$; there are finitely many cyclotomic polynomials $\Phi_M$ such that $\deg\Phi_M = \varphi(M) = N$, where $\varphi$ is Euler's totient function, so we only need to compare them.

For example, in the case of $z = 3/5 + i4/5$, $F(x) = x^2-\frac{6}{5}x+1$, which can never be a cyclotomic polynomial, since they have only integer coefficients, as roots of unity are algebraic integers.

In the case of $\varphi^{-1}$, $F(x) = x^4 + 2x^3 - 2x^2 + 2x + 1$, which we can compare to the $4$ cyclotomic polynomials of degree $4$ and see that it matches none, so $\cos^{-1}(\varphi^{-1})$ is not a rational multiple of $\pi$.

score 0 · Answer 3 · answered Feb 29 '24 at 21:54

To generalize Makholm's answer: Niven's theorem says that if $\theta$ is any rational multiple of $\pi$ other than an integer multiple of $\pi/2$ or of $\pi/3$, then $\cos \theta$ is irrational. Thus if $a$, $b$ and $c$ are a Pythagorean triple, that is if they are integers and $a^2+b^2=c^2$, then $\arccos(a/c)=\arcsin(b/c)$ and $\arcsin(a/c)=\arccos(b/c)$ are not rational multiples of $\pi$.

Reciproc of the Lindemann theorem and the arc cosine of the golden ratio

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