<s>[INST] Showing that $\cos\left(\frac{\pi}{5}\right)=\frac{1}{2}\phi?$ 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?
 [/INST] 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 $\frac{1+\sqrt 5}4=\frac{\phi}2$.
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