<s>[INST] How to evaluate cos(2$\pi$/5)? I understand evaluation for cos($\pi$/8), but I cannot figure it out how to solve cos(2$\pi$/5). 
Can anyone please show it?
 [/INST] Note that:
$$\cos\left(\frac{4\pi}5\right)=\cos\left(2\pi-\frac{4\pi}5\right)=\cos\left(\frac{6\pi}5\right)\implies\cos 2x =\cos 3x$$
$$\implies 2\cos^2x-1=4\cos^3x-3\cos x\implies 4y^3-2y^2-3y+1=0$$
$$\implies (y-1)(4y^2+2y-1)=0 \implies 4y^2+2y-1=0 \implies y=\frac{-2\pm\sqrt{20}}{8} $$
since $y>0$
$$\cos\left(\frac{4\pi}5\right)=\frac{-1+\sqrt{5}}{4}$$
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