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This is about compass and straightedge constructions, although I suspect nothing changes if we expand to real numbers algebraic over the rationals.

Anyway, let $E$ be the "constructible numbers," meaning the smallest subfield of the reals such that, if $x > 0, x \in E,$ then $\sqrt x \in E.$ So the elements are towers of square roots, add and mix those together.

Next, suppose we have real numbers $\alpha, \beta$ such that $\sin \alpha \in E,\sin \beta \in E. $ These are constructible angles written in radians.

It is easy to show with Hermite-Lindemann that, for $m,n \in \mathbb Z,$ that $$ m \alpha + n \beta \notin E $$ unless $$ m \alpha + n \beta = 0.$$ Divide though by another integer, we get the same statement for rational coefficients. Divide by the first, we have the statement for $\alpha + r \beta$ with $r \in \mathbb Q.$

Question: if $x \in E$ but $x \notin \mathbb Q,$ also $\sin \alpha \in E,\sin \beta \in E $ and $\alpha, \beta \neq 0,$ can we conclude that $$ \color{magenta}{ \alpha + x \beta \notin E ?} $$

Caution: it is all nice if $\alpha, \beta$ are rational or algebraic multiples of $\pi.$ However, perfectly good angles such as $\arctan 2$ are involved, this being a transcendental multiple of $\pi.$

Motivation: if we cannot conclude this, the proof by Chebotarev and his student about the lunes of Hippocrates is incomplete, in the sense that it places restrictions on the pairs of angles considered.

The Pacific Loon:

enter image description here

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  • $\begingroup$ For those keeping score at home, there is an informed suggestion: taking my $e^{i \alpha}$ and $e^{i \beta}$ which are nonzero algebraic, it is a "standard consequence" of Baker on linear forms in logarithms that either $\alpha, \beta $ are linearly dependent over $\mathbb Q$ or linearly independent over $\bar{\mathbb Q},$ the algebraics. If anyone knows how to fill in the details I'd be happy to hear about it. $\endgroup$
    – Will Jagy
    Commented Dec 13, 2013 at 0:05
  • $\begingroup$ Is the Loon of Hippocrates too rare in this day for you to get a photograph of one? $\endgroup$
    – Michael Hardy
    Commented Dec 13, 2013 at 2:00
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    $\begingroup$ @MichaelHardy, yes, that's exactly it. $\endgroup$
    – Will Jagy
    Commented Dec 13, 2013 at 2:08

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This is from a variant of Gelfond-Schneider, which was mentioned as well, and in Baker's book. I just couldn't see it. Theorem 10.2 on page 135 in Niven, Irrational Numbers reads: If $\alpha$ and $\gamma$ are non-zero algebraic numbers, and if $\alpha \neq 1, $ then $(\log \gamma)/(\log \alpha)$ is either rational or transcendental.

In my case, I am taking two angles, rename them $\theta, \omega,$ such that $e^{i \theta}$ and $e^{i \omega}$ are algebraic, while $\theta / \omega$ is irrational. As a result, $\theta / \omega = (i \theta) / ( i \omega)$ is transcendental.

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