Written by Colin+ in ask uncle colin, trigonometry.

Dear Uncle Colin,

How would you calculate $\cos(72º)$ by hand?

– Pointless Historical Inquiry

Hi, PHI, and thanks for your message. There seems to be an awful lot of degree use around at the moment, and I’m not very happy about it. But still, in the spirit of answering what the question asks, I’ll roll with it.

How would I do it by hand? I’d probably start by figuring out I can work out.

For a start, I know $X = \cos(72º) = \sin(18º)$, which is good.

I can also say that $\sin(36º) = 2\sin(18º)\cos(18º)$, which I can express as $2X\sqrt{1-X^2}$, in case that helps. Also, $\cos(36º) = 1 – 2\sin^2(18º) = 1 – 2X^2$

Meanwhile, $\sin(72º) = 2\sin(36º)\cos(36º)$, which is $4X\sqrt{1-X^2}\br{1-2X^2}$.

The square of that, plus $X^2$, must equal 1!^{1}

We then have $16X^2 \br{1-X^2}\br{1-2X^2}^2 + X^2 = 1$.

The first thing I notice is that if I subtract $X^2$ from both sides, there’s a common factor of $1-X^2$ that I can simply divide out – I know $0 < X < 1$, because of common sense.

That gives $16X^2 \br{1-2X^2}^2 = 1$

That’s a cubic in $X^2$. If I let $s = X^2$, it becomes: $16s \br{1 – 4s + 4s^2} – 1 = 0$, or

$64s^3 – 64s^2 + 16s – 1 = 0$.

Now, I don’t know about you, but I can’t generally solve cubics in my head. I know there’s a method, but it’s not one of the tools I have at my disposal. However, I might make an inspired spot: $s = \frac{1}{4}$ is a solution to this, so $(4s-1)$ is a factor.

Dividing through gives $16s^2 -12s + 1=0$ for the other solutions, which we can solve: if $a=16$, $b=-12$ and $c=1$, then $s = \frac{12 \pm \sqrt{144 – 4\times 16}}{32} = \frac{3 \pm \sqrt{5}}{8}$.

OK! We’re nearly there. Remember that $s = X^2$, so we need the square root of this: $X = \sqrt{\frac{3 \pm \sqrt{5}}{8}}$. (Strictly, we need to consider all four possibilities: however, common sense tells us $X > 0$, still.)

Can we narrow it down to one or the other? $\frac{3 + \sqrt{5}}{8}$ is rather more than $\frac{5}{8}$, and its square root is larger still – but $\sin(18º) < \sin(30º) = \frac{1}{2}$. That means we need the negative branch.

Can we find $\sqrt{\frac{3 – \sqrt{5}}{8}}$? Indeed we can.

Suppose $(a +b\sqrt{5})^2 = \frac{3}{8} – \frac{\sqrt{5}}{8}$.

Then $a^2 + 2ab\sqrt{5} + 5b^2 = \frac{3}{8} – \frac{\sqrt{5}}{8}$.

Splitting out the rational and irrational parts gives: $a^2 + 5b^2 = \frac{3}{8}$ and $2ab = – \frac{1}{8}$.

This remains a mess.

Let $a = -\frac{1}{16b}$, from the second equation, and substitute into the first to give $ \frac{1}{256b^2} + 5b^2 = \frac{3}{8}$. Multiplying through by $256b^2$ gives $1280b^4 – 96b^2 +1 = 0$.

Spot a factorisation there? Good, $1280 = 80 \times 16$, so we can make it $(80b^2 -1)(16b^2 -1)=0$.

The whole point of this is getting rational values for $b$, so we can ignore the $80b^2$ bracket and say instead that $b= \frac{1}{4}$ and $a=-\frac{1}{4}$.

This gives us a final value of $\frac{1}{4}\br{\sqrt{5}-1}$ – which, after all of that, is the correct answer.

Hope that helps!

– Uncle Colin

- Surprise factorial that changes nothing! [↩]

## Barney Maunder-Taylor

Good God! Much applause etc, nicely done!! – especially dealing with the degrees (must send you an “Imperial Measures” question soon hehe).

## Colin

Thanks, Barney – there are more posts on the same theme coming 🙂