How to evaluate cos(2$\pi$/5)?

Question

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?

Answer 1

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 $y>0$

$$\cos\left(\frac{4\pi}5\right)=\frac{-1+\sqrt{5}}{4}$$