Friday, August 29, 2014

Cosine of 72 degrees (and constructing a regular pentagon)

By using (say) DeMoivre's theorem, we have that \( \left( \cos\frac{2 \pi}{5}+ i \sin\frac{2\pi}{5} \right)^5=1\)
Expanding the left side as the fifth power of a binomial, equating the imaginary parts on both sides of the equation, and then replacing  \( \sin^2 72^\circ\) with  \(1- \cos^2 72^\circ \)
 \( i\sin 72^\circ \left(5 \cos^4 72^\circ+10 \cos^2 72^\circ i^2 \sin ^2 72 ^\circ+ i^4 \sin^4 72^\circ \right)=0i\)
 \(16 \cos^4 72^\circ - 12 \cos^2 72^\circ  + 1=0\)
Solving this quadratic in \( \cos^2 72^\circ \), we get
 \[ \cos^2 72^\circ  = \frac{12 \pm  \sqrt{80} } {32}   \]
 \[ \cos^2 72^\circ  = \frac{6 \pm 2 \sqrt{5} } {16}=\frac{\left( \sqrt{5}\pm 1 \right)^2}{4^2}   \]
 \[ \cos 72^\circ  = \pm \frac{\sqrt{5} \pm 1}{4}  \]
where we can choose the correct value of the four possible values by noting that, because 72° is between 45° and 90°,  \( \cos 72^\circ \) must lie between  \(1 / \sqrt{2}\) and 0. Because  \( \cos 72^\circ \) is positive, we choose the "+" before the fraction, and because  \( \cos 72^\circ \) is less than \( 1 / \sqrt{2}\), which in turn is less than \(\frac{\sqrt{5}+1}{4}\), we choose the "-" in the numerator:
\[ \cos 72^\circ = \frac{\sqrt{5}-1}{4} \]

Constructing a regular pentagon

So we can construct \( \cos 72^\circ\). For example, the diagonal of a 1-by-2 rectangle is \(\sqrt{5}\). We could cut off one unit from a segment of length \(\sqrt{5}\), then divide the segment of length \(\sqrt{5}-1\) into four pieces of length \( \frac{\sqrt{5} -1} {4} \). (Or we could construct the appropriate solution to the equation \( 4x^2 +2x -1 = 0 \). See my post on Solving quadratic equations via geometric construction.)

Construct a unit circle centered at O, and construct a radius \(\overline{OA}\).  Construct the point B on \(\overline{OA}\) so that \(\overline{OB}\) has length  \( \cos 72^\circ\). If C is a point on the circle so that \(\overline{BC}\) is perpendicular to \(\overline{OA}\), then \(\angle COA\) is a 72° angle, and both A and C are vertices of a regular pentagon inscribed in the circle.

1 comment:

Bruce Yoshiwara said...

This method of evaluating cos(2pi/5) was shown to me by my then-UCLA office mate Larry Miller in my first year of grad school.