theorem on constructible angles

Theorem 1.

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

1. An angle of measure (http://planetmath.org/AngleMeasure) $\theta$ is constructible (http://planetmath.org/Constructible2);

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 . 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 (http://planetmath.org/Equivalent3) has been established in the case that $\theta \in \{0,\frac{\pi }{2}\}$. Therefore, we can restrict our attention even further to the interval .

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 (http://planetmath.org/Vertex5) 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. ∎

Note that, if $\cos \theta \ne 0$, then any of the three statements thus implies that $\tan \theta$ is a constructible number. Moreover, if $\tan \theta$ is constructible, then a right triangle having a leg of length $1$ and another leg of length $\tan \theta$ is constructible, which implies that the three listed conditions are true.

Title	theorem on constructible angles
Canonical name	TheoremOnConstructibleAngles
Date of creation	2013-03-22 17:15:59
Last modified on	2013-03-22 17:15:59
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	13
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 33B10
Classification	msc 51M15
Classification	msc 12D15
Related topic	ConstructibleNumbers
Related topic	CompassAndStraightedgeConstruction
Related topic	ConstructibleAnglesWithIntegerValuesInDegrees
Related topic	ExactTrigonometryTables
Related topic	ClassicalProblemsOfConstructibility
Related topic	CriterionForConstructibilityOfRegularPolygon