Showing that $\cos\left(\frac{\pi}{5}\right)=\frac{1}{2}\phi?$

Question

What is the usual way of proving things like

$$\cos\left(\frac{\pi}{5}\right)=\frac{1}{2}\phi?$$

I know that there is an identity which claims the above, but how was it derived? Are other identities used in the process? I am interested in knowing this.

For example, I could follow through as such:

$$\cos\left(\frac{\pi}{5}\right)=\cos\left(-\frac{\pi}{5}\right)=-\cos\left(\pi-\frac{\pi}{5}\right)=-\cos\left(\frac{4\pi}{5}\right)=1-2\cos^2\left(\frac{2\pi}{5}\right),$$

which gets me somewhat close to the golden ratio, but I must admit that I am stuck here.

Do you guys have any ideas?

Answer 1

I don't know about the "usual" way to do this, but here's a geometric proof: draw an isosceles triangle $ABC$ with angles $\pi/5$, $\pi/5$, and $3\pi/5$, this last angle being at $A$. Now draw $AD$ such that $D$ is on $BC$ and angle $BAD$ measures $\pi/5$. Notice both small triangles $DAB$ and $DAC$ are isosceles. Let $AB = AC = DC = 1$ and $BC = x$. Then since triangle $DAB$ is similar to triangle $ABC$, $BD/AB = CA/BC$, or $(x-1)/1 = 1/x$. Hence $x = \phi$. Now $\cos \pi/5 = \cos\angle B = \text{adj} / \text{hyp} = (BC/2) / AB = \phi /2$.

EDIT: darn, the image quality is atrocious.

Answer 2

You could continue on the track you've shown. You've got $\cos(\pi/5)=1-2\cos^2(2\pi/5)$, so $2\cos^2(2\pi/5)+\cos(\pi/5)-1=0$, and since $\cos(2\pi/5)=2\cos^2(\pi/5)-1$, then $8\cos^4(\pi/5)-8\cos^2(\pi/5)+\cos(\pi/5)+1=0$, so $\cos(\pi/5)$ is a root of $8x^4-8x^2+x+1$. In particular, it is a positive root. Some inspection of that polynomial shows that it has 1/2 and -1 as rational roots, and $\cos(\pi/5)$ can't be either of those (as $\cos(\pi/5)>\cos(\pi/3)=1/2$). Factoring those out leaves $4x^2-2x+1$, and the positive root of that one is $\phi/2$.

Of course, if you realize that $\phi/2$ is the unique positive root of $4x^2-2x+1$ and that $\cos(\pi/5)$ is positive, then you only need to check that $\cos(\pi/5)$ is a zero of $f(x)=4x^2-2x+1$. Here's one slick way to do that:

Let $x=\cos(\pi/5)$, and use the identity $2\cos^2\theta=\cos(2\theta)+1$ to give you $f(x)=2\cos(2\pi/5)-2\cos(\pi/5)+1$. As you've already noted, $\cos(\pi/5)=-\cos(4\pi/5)$, so that means $f(x)=2\cos(4\pi/5)+2\cos(2\pi/5)+1$. Now,

$$\begin{eqnarray*} f(x) & = & 1+\cos(2\pi/5)+\cos(4\pi/5)+\cos(-4\pi/5)+\cos(-2\pi/5)\\ & = & 1 + \cos(2\pi/5)+\cos(4\pi/5)+\cos(6\pi/5)+\cos(8\pi/5)\\ & = & \sum_{k=0}^4\cos(2k\pi/5) \end{eqnarray*}$$

Consider how the angles $2k\pi/5$ for $k=0,..,4$ are evenly spaced about the unit circle. If you set 5 equal weights at those spots on the edge of an ideal weightless disk (or any ideal disk perfectly balanced on a point at its center), the system of weights would be perfectly balanced on the point at the center of the disk. The same is then true if you were to balance the disk along any line through the disk's center, and in particular, the line corresponding to the $y$-axis. The law of the lever says that means the total signed distances of those weights from that line--that is, the sum of the cosines of those angles--is precisely 0. That is, $f(x)=0$.

Working with complex numbers, we can use a similar approach with much less work, but if you're unfamiliar with them, that won't help at all.

Answer 3

Here's a standard way of doing this with complex numbers:

Let $z=e^{\pi i/5}$, which is a primitive 10th root of unity. So it's a root of the 10th cyclotomic polynomial, which is $z^4-z^3+z^2-z+1$. In other words, $$e^{4\pi i/5}-e^{3\pi i/5}+e^{2\pi i/5}-e^{\pi i/5}+1=0.$$ So the real part of this is also equal to zero. But the real part of this is $$\cos\left(\frac{4\pi}5\right)-\cos\left(\frac{3\pi}5\right)+\cos\left(\frac{2\pi}5\right)-\cos\left(\frac{\pi}5\right)+1=2\cos\left(\frac{2\pi}5\right)-2\cos\left(\frac{\pi}5\right)+1.$$ Let $x=\cos\left(\frac{\pi}5\right)$. Then this can be written as \begin{align*}2(2x^2-1)-2x+1 &= 0\\ 4x^2-2x-1 &= 0,\end{align*} whose positive root is $\frac{1+\sqrt 5}4=\frac{\phi}2$.