# Ask Uncle Colin: an unexpected Golden Ratio

Dear Uncle Colin,

Why is $\cos(36º)$ equal to $\frac{\phi}{2}$? I find trigonometry difficult and, uh, let’s say I have some demons to banish.

Puzzled Educator Noticed That A Golden Ratio Appears Magically

Hi, PENTAGRAM, and thanks for your message!

I would imagine there are several ways to demonstrate this, but I’m going with the first one to jump into my mind: trigonometric identities.

**Identity number 1**: $\cos(xº) \equiv -\cos(180º-xº)$ - and in particular, $\cos(36º) = -\cos(144º)$. Because $\frac{144}{36} = 4$, this means 36º is a solution to $\cos(x) + \cos(4x) = 0$ [*].

**Identity number 2**: $\cos(2\theta) = 2\cos^2(\theta) -1$. And, by extension, $\cos(4x) = 2\cos^2(2x) - 1$. By further extension, $\cos(4x) = 2(2\cos^2(x)-1)^2 - 1$.

That means, if I let $c = \cos(x)$, equation [*] becomes $2\left(4c^4 - 4c^2 + 1\right) + c - 1= 0$, or $8c^4 - 8c^2 + c + 1 = 0$.

### Solving the quartic

A “fairly obvious” solution to this is $c = -1$ - the first pair of terms factorise as a difference of two squares, so $(c+1)$ is a factor of both ‘halves’.

Factoring that out gives $(c+1)\left(8c^3 - 8c^2 + 1\right) = 0$.

That’s *less* obvious, but $c=\frac{1}{2}$ is also a solution to this, so $(2c-1)$ is a factor. Dividing it out gives $(c+1)(2c-1)(4c^2 - 2c - 1)= 0$.

Now, we know that $\cos(36º)$ isn’t -1 or 0.5 – yes yes, thank you sensei, a bit more than 0.8, we know – so we’ll need to solve the quadratic to find the number we want.

Let’s make completing the square easy by multiplying everything by 4: $16c^2 - 8c - 4 = 0$. That’s $(4c-1)^2 = 5$, so $4c-1 = \pm \sqrt{5}$ and $c = \frac{ 1 \pm \sqrt{5}}{4}$.

$c$ is positive, so the positive root is the one we want - and it’s $\frac{\phi}{4}$.

### A digression into the other solutions

Clearly, 36º is not the only $x$ that satisfies the equation $\cos(4x) + \cos(x) = 0$. What are the others?

**Identity number 3** is $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$.

In the case where $A=4x$ and $B=x$, this gives $2\cos\left(\frac{5}{2}x\right)\cos\left(\frac{3}{2}x\right) = 0$.

This is satisfied when $\cos\left(\frac{5}{2}x\right) = (2k+1)90º$ or $\cos\left(\frac{3}{2}x\right) = (2k+1)90º$.

The first of those gives $x = (2k+1)36º$, so $x$ can be 36º, 108º (which gives the negative root), 180º (-1), 324º (the positive root), and so on modulo 360º.

The second gives $x = (2k+1)60º$, so $x$ can also be 60º (0.5), 180º (-1), 300º (0.5), and so on modulo 360º.

I hope that helps!

- Uncle Colin