theorem on constructible anglesTheoremOnConstructibleAngles2007-06-15 16:46:362008-01-13 21:58:23Theorem\usepackage{amssymb}
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\usepackage{amsthm}
\usepackage{xypic}
\newtheorem{thm*}{Theorem}\PMlinkescapeword{constructible}
\PMlinkescapeword{even}
\PMlinkescapeword{label}
\PMlinkescapeword{leg}
\PMlinkescapeword{measure}
\PMlinkescapeword{measures}
\PMlinkescapeword{perpendicular}
\PMlinkescapeword{similar}
\PMlinkescapeword{vertex}
\begin{thm*}
Let $\theta \in \mathbb{R}$. Then the following are equivalent:
\begin{enumerate}
\item An angle of \PMlinkname{measure}{AngleMeasure} $\theta$ is \PMlinkname{constructible}{Constructible2};
\item $\sin \theta$ is a constructible number;
\item $\cos \theta$ is a constructible number.
\end{enumerate}
\end{thm*}
\begin{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:
\begin{enumerate}
\item If an angle whose measure is $\theta$ is constructible, then so are angles whose measures are $\pi-\theta$, $\pi+\theta$, and $2\pi-\theta$;
\item If $x$ is a constructible number, then so is $|x|$.
\end{enumerate}
If $\theta \in \{0, \frac{\pi}{2} \}$, then clearly an angle of measure $\theta$ is constructible, and $\{\sin \theta, \cos \theta \}=\{0,1\}$. Thus, \PMlinkname{equivalence}{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 $0<\theta<\frac{\pi}{2}$.
Assume that an angle of measure $\theta$ is constructible. Construct such an angle and mark off a line segment of length $1$ from the \PMlinkname{vertex}{Vertex5} of the angle. Label the endpoint that is not the vertex of the angle as $A$.
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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.
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\rput[b](1.2,2.3){$A$}
\rput[a](0.7,-0.3){$\cos \theta$}
\rput[l](1.7,1.3){$\sin \theta$}
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\end{center}
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.
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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.
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\rput[b](2.3,1.5){$A$}
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\end{center}
A similar procedure can be used given that $\cos \theta$ is a constructible number to prove the other two statements.
\end{proof}
Note that, if $\cos \theta \neq 0$, 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.