Review question

Ref: R7021

## Solution

$ABCDE$ 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

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.

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.

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.