Theorem   Let $\theta \in \mathbb{R}$ . Then the following are equivalent:

1. An angle of measure $\theta$ is constructible;
2. $\sin \theta$ is a constructible number;
3. $\cos \theta$ is a constructible number.

Proof. First of all, due to periodicity, we can restrict our attention to the interval $0 \le \theta <2\pi$ . Even better, we can further restrict our attention to the interval $0 \le \theta \le \frac{\pi}{2}$ for the following reasons:

1. If an angle whose measure is $\theta$ is constructible, then so are angles whose measures are $\pi-\theta$ , $\pi+\theta$ , and $2\pi-\theta$ ;
2. If $x$ is a constructible number, then so is $|x|$ .

If $\theta \in \{0, \frac{\pi}{2} \}$ , then clearly an angle of measure $\theta$ is constructible, and $\{\sin \theta, \cos \theta \}=\{0,1\}$ . Thus, equivalence has been established in the case that $\theta \in \{0,\frac{\pi}{2}\}$ . Therefore, we can restrict our attention even further to the interval $0<\theta<\frac{\pi}{2}$ .

Assume that an angle of measure $\theta$ is constructible. Construct such an angle and mark off a line segment of length $1$ from the vertex of the angle. Label the endpoint that is not the vertex of the angle as $A$ .

Drop the perpendicular from $A$ to the other ray of the angle. Since the legs of the triangle are of lengths $\sin \theta$ and $\cos \theta$ , both of these are constructible numbers.

Now assume that $\sin \theta$ is a constructible number. At one endpoint of a line segment of length $\sin \theta$ , erect the perpendicular to the line segment.

From the other endpoint of the given line segment, draw an arc of a circle with radius $1$ so that it intersects the erected perpendicular. Label this point of intersection as $A$ . Connect $A$ to the endpoint of the line segment which was used to draw the arc. Then an angle of measure $\theta$ and a line segment of length $\cos \theta$ have been constructed.

A similar procedure can be used given that $\cos \theta$ is a constructible number to prove the other two statements.