<s>[INST] complex numbers exponential form I wish to show that $\cos^2(\frac{\pi}{5})+\cos^2(\frac{3\pi}{5})=\frac{3}{4}$
I know the solutions to $z^5+1=0$ are $-1$, $e^{i\frac{\pi}{5}}$, $e^{-i\frac{\pi}{5}}$, $e^{i\frac{3\pi}{5}}$, $e^{i\frac{-3\pi}{5}}$ and that
$z^5+1=(z+1)(z^4-z^3+z^2-z+1)$
which means that the solutions to $z^4-z^3+z^2-z+1=0$ are $e^{i\frac{\pi}{5}}$, $e^{-i\frac{\pi}{5}}$, $e^{i\frac{3\pi}{5}}$, $e^{i\frac{-3\pi}{5}}$
and so $z^4-z^3+z^2-z+1=(z-e^{i\frac{\pi}{5}})$$(z-e^{-i\frac{\pi}{5}})$ $(z-e^{i\frac{3\pi}{5}})$$(z-e^{i\frac{-3\pi}{5}})$
but I am not sure where to go from there.
 [/INST] Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
Divide both sides by $z^2$ and replace $2u=z+\dfrac1z$
The roots of $$(2u)^2-2-2u+1=0$$ are $$\cos\frac\pi5,\cos\frac{3\pi}5$$
Now use https://mathworld.wolfram.com/VietasFormulas.html
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