Determine the roots of $z^5=1$, and deduce that $$z^5-1=(z-1)(z^2-2z\cos \dfrac{2\pi}{5}+1)(z^2-2z\cos \dfrac{4\pi}{5}+1).$$ Deduce that $$\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}=-\dfrac{1}{2},\quad \cos \dfrac{2\pi}{5}\cos \dfrac{4\pi}{5}=-\dfrac{1}{4},$$ and hence show that $$\cos \dfrac{\pi}{5}=\dfrac{\sqrt{5}+1}{4},\quad \cos \dfrac{2\pi}{5}=\dfrac{\sqrt{5}-1}{4}.$$
Note that $$z^5=1 \Rightarrow (e^{2\pi ki/5})^5=1,\quad k=0,1,2,3,4.$$ We know the roots to $z^5-1=0$ are $1,e^{2\pi i/5},e^{4\pi i/5},e^{-2\pi i/5},e^{-4\pi i/5}$. We thus have $$(z-1)(z-e^{2\pi i/5})(z-e^{-2\pi i/5})(z-e^{4\pi i/5})(z-e^{-4\pi i/5})=0,$$ or $$(z-1)(z^2-z(e^{2\pi i/5}+e^{-2\pi i/5})+1)(z^2-z(e^{4\pi i/5}+e^{-4\pi i/5})+1)=0,$$ which is $$(z-1)(z^2-2z\cos{\dfrac{2\pi i}{5}}+1)(z^2-2z\cos{\dfrac{4\pi i}{5}}+1)=0.$$ Note also that $$z^5-1=(z-1)(z^4+z^3+z^2+z+1).$$ Now expand the expression $$(z^2-2z\cos{\dfrac{2\pi i}{5}}+1)(z^2-2z\cos{\dfrac{4\pi i}{5}}+1)\\=z^4-2z^3(\cos \dfrac{4\pi}{5}+\cos \dfrac{2\pi}{5})+z^2(4\cos \dfrac{2\pi}{5}\cos \dfrac{4\pi}{5}+2)-2z(\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5})+1.$$ Comparing the two equations, we have $$\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}=-\dfrac{1}{2},\quad \cos \dfrac{2\pi}{5} \cos \dfrac{4\pi}{5}=-\dfrac{1}{4}.$$ Treating $\cos \dfrac{2\pi}{5}$ as $a$ and $\cos \dfrac{4\pi}{5}$ as $b$, it remains to solve the linear system: $$\begin{cases}a+b=-\dfrac{1}{2}\\ ab=-\dfrac{1}{4}.\end{cases}$$ Using quadratic formula, we have $$a=\cos \dfrac{2\pi}{5}=\dfrac{\sqrt{5}-1}{4},\quad b=\cos \dfrac{4\pi}{5}=\dfrac{-\sqrt{5}-1}{4}.$$ Therefore $$\cos \dfrac{\pi}{5}=\cos (\pi-\dfrac{4\pi}{5})=-\dfrac{-\sqrt{5}-1}{4}=\dfrac{\sqrt{5}+1}{4}.$$
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