Some months ago, I wrote about a method for finding $\cos(72º)$, or $\cos\br{\frac{2\pi}{5}}$ in proper units. Almost immediately, the good people of Twitter and Facebook – notably @ImMisterAl (Al) and @BuryMathsTutor (Mark)- suggested other ways of doing it.

Let’s start with Mark’s method, which he dissects in his book GCSE Maths Challenge^{1}.

Looking at Q16 here, we have three isosceles triangles. ACD and DCB are similar (they’re both isosceles, and base angle C is the same in both). If we call angle CAD $\theta$, then ABD is $180º – 2\theta$, so CBD and DCB are both $2\theta$.

That means the three angles in ACD add up to $5\theta$, which must be 180º; $\theta$ is therefore 36º.

Suppose the length BD (and therefore CD and BA) is one unit, and the length BC is $x$. We can do some trigonometry!

Applying the cosine rule to the big triangle ACD, we have $\cos(72º)=\frac{1 + (1+x)^2 – (1+x)^2}{2(1+x)} = \frac{1}{2(1+x)}$.

Doing the same to the smaller triangle CBD, we have $\cos(72º)=\frac{1+x^2-1}{2x} = \frac{x}{2}$.

Let $\cos(72º)=C$ for the purposes of algebra, and we have $2C=x$ from the second equation. Substituting this into the first gives $C=\frac{1}{2+4C}$.

Rearrange this to give $4C^2 + 2C – 1 = 0$, and $C$ drops out of the quadratic formula as $C=\frac{-1 \pm \sqrt{5}}{2}$; we know $\cos(72º)>0$, so only the positive branch makes sense.

Therefore $\cos(72º) = \frac{\sqrt{5}-1}{2}$.

I like this method a lot – it drops out super-neatly. However, it feels a bit like the triangles have appeared by magic; it’s easy to prove it once you have the scaffolding in place, but I wouldn’t have come up with the scaffolding on my own.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## Barney Maunder-Taylor

The other evening, I was trying to work out what shape triangles I needed to make a rectified pentagonal prism (as you do), and as part of the calculations needed cos(36deg).

Being too lazy to get out of bed and find a calculator, I observed that if a=36deg then cos(3a)=cos(180-2a)since both are cos(108deg). Expanding both sides down to single angle gave a cubic equation which easily factorised into a linear and a quadratic. Solving the quadratic gave cos(36deg)=(1+sqrt5)/4, whence it’s easy to get cos72 using double angle formula.

So there’s a “real-life” application of this, assuming wanting to build a rectified pentagonal dipyramid counts as “real-life”!!

Thanks as ever, Colin, for the blogs, keep ’em coming!

## Barney Maunder-Taylor

(Which I’ve just noticed is strikingly similar to Mark’s method).

## Colin

Neat!