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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