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

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 $\pi$.

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You don't need the Lindemann theorem for the forward direction; de Moivre's formula yields algebraicity of rational cosines fairly directly. –  Henning Makholm Sep 24 '11 at 6:17
Indeed! Used a cannon to kill a fly... –  Mateus Araújo Sep 24 '11 at 6:26
Peering at the simple continued fraction expansion of $\arccos\,\varphi^{-1}$, it doesn't seem rational to me... –  Ｊ. Ｍ. Sep 24 '11 at 6:39
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 $\pi$... –  Henning Makholm Sep 24 '11 at 6:40
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. Sep 24 '11 at 6:45

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 $n\ge 1$

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

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 $\pi$.

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