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.
$\mathrm{sin}\theta $ is a constructible number;

3.
$\mathrm{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 $\{\mathrm{sin}\theta ,\mathrm{cos}\theta \}=\{0,1\}$. 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 $\mathrm{sin}\theta $ and $\mathrm{cos}\theta $, both of these are constructible numbers.
Now assume that $\mathrm{sin}\theta $ is a constructible number. At one endpoint of a line segment of length $\mathrm{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 $\mathrm{cos}\theta $ have been constructed.
A similar^{} procedure can be used given that $\mathrm{cos}\theta $ is a constructible number to prove the other two statements. ∎
Note that, if $\mathrm{cos}\theta \ne 0$, then any of the three statements thus implies that $\mathrm{tan}\theta $ is a constructible number. Moreover, if $\mathrm{tan}\theta $ is constructible, then a right triangle having a leg of length $1$ and another leg of length $\mathrm{tan}\theta $ is constructible, which implies that the three listed conditions are true.
Title  theorem on constructible angles 
Canonical name  TheoremOnConstructibleAngles 
Date of creation  20130322 17:15:59 
Last modified on  20130322 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 