Using angles in a pentagon...

Using angles in a pentagon...

Question

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

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

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

Suggestion

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

Let’s assume that the side lengths of the pentagon are $1$.

Can we draw a large, clear diagram of the pentagon $ABCDE$, with $AB$ horizontally at the bottom?

What are the angles inside a regular pentagon, in radians?

If we now drop perpendiculars from $D$ and $E$ to the horizontal, can we start to find lengths and angles?

Using the first part, are there any trigonometrical identities we could use? To get this just in terms of $\cos\dfrac{\pi}{5}$?

What’s the sum of $\dfrac{3\pi}{5}$ and $\dfrac{2\pi}{5}$? How does this connect $\cos\dfrac{3\pi}{5}$ and $\cos\dfrac{2\pi}{5}$?

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

$$\begin{equation*} \cos\frac{\pi}{5}-\cos\frac{2\pi}{5}=\frac{1}{2}. \qquad \tag{$*$} \end{equation*}$$

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 $\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{5} = \dfrac{2\phi^2-1}{2\phi^2}- \dfrac{1}{2\phi} = \dfrac{2\phi^2-1}{2\phi^2}- \dfrac{\phi}{2\phi^2}= \dfrac{2\phi^2-\phi-1}{2\phi^2} = \dfrac{\phi^2}{2\phi^2} = \dfrac{1}{2}.$

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

$(star)$

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: $\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}.$

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.