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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.

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:

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 $\pi.$

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.

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 $$ \color{magenta}{ \alpha + x \beta \notin E ?} $$

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 $\pi.$