\(ABCDE\) is a regular pentagon. By projecting the broken line \(AED\) on the line \(AB\), or otherwise, show that \[\cos\frac{\pi}{5}-\cos\frac{2\pi}{5}=\frac{1}{2}.\]

Method 1

Regular pentagon ABCDE with base AB and sides labelled clockwise. D-dash is the point on AB directly below D, and P is the point on DD-dash such that PE is horizontal. E-dash is directly below E such that BAE-dash is a straight line. Angles are as follows: E-dash A E = 2 pi over 5, DEP = pi over 5

The sum of angles inside a \(n\)-sided polygon is \((n-2)\pi\), so each angle inside our regular pentagon is \(\dfrac{3\pi}{5}\). For simplicity we choose each side to be of length \(1\).

Let \(D'\) and \(E'\) be the points \(D\) and \(E\) projected onto the line \(AB\) as shown. Since \(ABCDE\) is a regular pentagon, \[D'A =\frac{1}{2}.\] From the right-angled triangle \(AEE'\) we see that \[AE' =\cos\frac{2\pi}{5},\] and from the triangle \(DPE\) we find that \[D'E' = PE =\cos\frac{\pi}{5}.\] Since \[D'E' = D'A + AE',\] we get the result \[\begin{equation*} \cos\frac{\pi}{5}-\cos\frac{2\pi}{5}=\frac{1}{2}. \qquad \label{eq:star}\tag{$*$} \end{equation*}\]

Method 2

The question says ‘or otherwise’, so we’re free to pick another method if we wish.

Regular pentagon ABCDE with base AB. Triangle ABD is drawn. Angles are  D A B = 2 pi over 5, A D B = pi over 5. Sides are phi, phi and 1.

The regular pentagon is intimately connected with \(\phi\), the Golden Ratio \(= 1.61803...\), which is the positive solution to \(x^2 - x - 1 = 0\).

Using the Cosine Rule in \(ABD\) one way, \(\cos \dfrac{2\pi}{5} = \dfrac{1}{2\phi}\). Using the Cosine Rule in \(ABD\) another way, \(\cos \dfrac{\pi}{5} = \dfrac{2\phi^2-1}{2\phi^2}\).

Thus \[\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{5} = \dfrac{2\phi^2-1}{2\phi^2}- \dfrac{1}{2\phi} = \dfrac{2\phi^2-1}{2\phi^2}- \dfrac{\phi}{2\phi^2}= \dfrac{2\phi^2-\phi-1}{2\phi^2} = \dfrac{\phi^2}{2\phi^2} = \dfrac{1}{2}.\]

Hence, or otherwise, show that \(\cos\dfrac{\pi}{5} = \dfrac{\sqrt{5}+1}{4}\).

Let \[\gamma=\cos\frac{\pi}{5}.\] Using the cosine double angle formula \[\cos 2\theta=2\cos^2\theta-1\] with \(\theta=\dfrac{\pi}{5},\) we can use \(\eqref{eq:star}\) to get \[\begin{align*} \gamma-(2\gamma^2-1) &= \frac{1}{2} \\ \iff \quad 4\gamma^2-2\gamma-1 &= 0. \end{align*}\]

We then use the quadratic formula to solve for \(\gamma\): we have \[\gamma = \cos\frac{\pi}{5}=\frac{2\pm\sqrt{4-4\times 4\times (-1)}}{2\times 4}=\frac{1\pm\sqrt{5}}{4}.\] We know that \(\dfrac{\pi}{5} < \dfrac{\pi}{2}\) so \(\cos\dfrac{\pi}{5}>0\) and thus we choose the positive root: \[\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}.\]

Show further that \(\cos\dfrac{3\pi}{5} = -\dfrac{\sqrt{5}-1}{4}\).

We use the fact that \[\cos(\pi-\theta)=-\cos\theta,\] which we can see from the diagram.

Graph of cos x and cos (pi minus x). One is a reflection of the other in the x axis.
This gives \[\cos\frac{3\pi}{5}=-\cos\frac{2\pi}{5}\] so \[\begin{align*} \cos\frac{3\pi}{5} &= -\cos\frac{2\pi}{5}=\frac{1}{2}-\cos\frac{\pi}{5}\\ &= \frac{2-(\sqrt{5}+1)}{4}=-\frac{\sqrt{5}-1}{4} \end{align*}\]

as required.