I want to show that $$\cos\frac{2 \pi}{5}+\cos\frac{4\pi}{5}=-\frac12,\,\cos(2 \pi/5)\cos(4 \pi/5)=-1/4$$. I can't use the value of $\cos(2 \pi/5)$ without having proven it first.
It's a lot harder than I anticipated, so I thought why not ask here. Thanks in advance!
See the top answer here
if $z = e^{2 i \pi /5}$ or $e^{4 i \pi/5}$ then $\frac{z+z^{-1}}{2}$ is equal to the corresponding value of cos in either case
and $1 + z + z^2 + z^3 + z^4 = 0$ also in either case, so dividing through by $z^2$ shows that our values both satisfy $w^2 + \frac{w}{2} - \frac{1}{4} = 0$.
At which point we are done by Vieta's Formulas [suppose the roots are $\alpha$ and $\beta$ and expand $(x-\alpha)(x-\beta)$]
Hint: using roots of unity $z^5=1$ we get $z=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}, k=0,1,2,3,4$.
Using Vieta theorem, it follows that $1+\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{6\pi}{5}+\cos\frac{8\pi}{5}=0$.
Using the double angle identity,
\begin{align} \cos\frac{4\pi}5&=2\cos^2\frac{2\pi}5 - 1\tag1\\ \cos\frac{2\pi}5=\cos\frac{8\pi}5&=2\cos^2\frac{4\pi}5 - 1\tag2 \end{align}
$(1)-(2):$ $$\cos\frac{4\pi}5-\cos\frac{2\pi}5=2\left(\cos\frac{2\pi}5+\cos\frac{4\pi}5\right)\left(\cos\frac{2\pi}5-\cos\frac{4\pi}5\right)$$ $$\therefore\quad\cos\frac{2\pi}5+\cos\frac{4\pi}5=-\frac12$$
