If an angle $\theta$ is not constructable, then $\cos\theta$ is not constructable.

Question

Claim: if an angle $\theta$ is not constructable, then $\cos\theta$ is not constructable.

I am at the very end of a proof, and I suspect this claim is true and it would help me very much in this proof. Does anyone have any insight on this? I've been trying for a bit, drawing it out, and I cannot think of a method to prove it. Any guidance is appreciated.

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Edit: I have since tried something that seems to make sense to me. I use a previous result that the square roots of positive constructable numbers are constructable and that the constructable numbers form a number field. So I instead prove the contrapositive. I assume $\cos\theta$ is constructable. I draw a line and mark $\cos\theta$ on it. Since $\cos\theta$ is constructable, it follows that $\sqrt{1 - \cos^2\theta}$ is constructable. I construct a bisector at one end of $\cos\theta$ and mark $\sqrt{1 - \cos^2\theta}$ on it so that they are end-to-end. Then, using straightedge, connect their opposite ends. So the non-right angle which is next to $\cos\theta$ must be $\theta$.

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Edit: Thanks for the help. I understand now.

Answer 1

Prove the contrapositive. Suppose $\cos\theta$ is constructible. Then in right triangle $ABC$ with $\angle B$ right, $AC=1$ and $BC=\cos\theta$, $\angle C=\theta$.

Answer 2

If you could construct $\cos \theta$ then to get theta all that's needed is to intersect a vertical line with unit circle.

Answer 3

If you have a unit circle and the $x$ axis (certainly both constructable) then $\cos \theta$ will be the length along the $x$ axis where if you dropped perpendicular (constructable) from the point where the $\angle \theta$ intersects to circle.

Likewise if you raised a perpendicular (constructable) from $\cos \theta$ you'd get the point of intersection. And it's a matter of constructing a line from the point of intersection to the origin (constructable) to get $\angle \theta$.

So $\angle \theta$ is constructable $\iff \cos \theta$ is constructable.