How do you find the exact values of cos 2pi/5?

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Answer 1 · 2016-03-16T22:03:54

#cos(2pi/5)=(-1+sqrt(5))/4#

Here the most elegant solution I found in:

#cos(4pi/5)=cos(2pi-4pi/5)=cos(6pi/5)#

So if # x=2pi/5#:

#cos(2x)=cos(3x)#

Replacing the cos(2x) and cos(3x) by their general formulae:

#color(red)(cos(2x)=2cos^2x-1 and cos(3x)=4cos^3x-3cosx)#,

we get:

#2cos^2x-1=4cos^3x-3cosx#

Replacing #cosx# by #y#:

#4y^3-2y^2-3y-1=0#

#(y-1)(4y^2+2y-1)=0#

We know that #y!=1#, so we have to solve the quadratic part:

#y=(-2+-sqrt(2^2-4*4*(-1)))/(2*4)#

#y=(-2+-sqrt(20))/8#

since #y>0#, #y=cos(2pi/5)=(-1+sqrt(5))/4#