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

 \[ \cos^2 72^\circ  = \frac{12 \pm  \sqrt{80} } {32}   \]

so

 \[ \cos^2 72^\circ  = \frac{6 \pm 2 \sqrt{5} } {16}=\frac{\left( \sqrt{5}\pm 1 \right)^2}{4^2}   \]

 \[ \cos 72^\circ  = \pm \frac{\sqrt{5} \pm 1}{4}  \]

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.

Solving quadratic equations via geometric construction

We can solve a quadratic equation of the form x² - sx + p = 0, , using the standard construction tools of compass and straightedge.  The method has been attributed to critic Thomas Carlyle.

Construct the circle in the Cartesian plane with center and passing through A(0,1).  By symmetry, the circle also passes through and .

Because the circle has center   and passes through A(0,1), the equation of the circle is

This reduces to

x² - sx + y² - (p + 1)y + p = 0

So the x-intercepts of the circle are the solutions to  x² - sx + p = 0.

Alternate justification:

The segment joining the x-intercepts has a length , hence x₁ + x₂ = s.

intercepts the arc AX₁X₂, and  intercepts the opposite arc, hence the two angles are supplementary.  But is also supplementary with , so is congruent to , which in turn impies that  .  Hence

 ,

which implies that , so x₁ x₂ = p.

Thus (x -  x₁)(x - x₂) = x² - sx + p, and the solutions to x² - sx + p = 0 are x₁ and x₂ .