We need
#sin(2x)=2sinxcosx#
#cos(2x)=1-2sin^2x#
#sinx=cos(90-x)#
#sin^2x+cos^2x=1#
Let's start
#sin(72)=sin(2*36)#
#=2sin36cos36#
#=2sin(2*18)cos(2*18)#
#=2sin(2sin18cos18)(1-2sin^2 18)#
#=4sin18cos18(1-2sin^2 18)#
Therefore,
#sin72=cos(90-72)=4sin18cos18(1-2sin^2 18)#
#cos18=4sin18cos18(1-2sin^2 18)#
Divide both sides by #cos18#
#1=4sin18(1-2sin^2 18)#
#8sin^3 18-4sin18+1=0#
Factorise
#(2sin18-1)(4sin^2 18+ 2sin18-1)=0#
Therefore,
#sin18=1/2# which is #O/#
The solution to the quadratic equation is
#sin18=(-2+-sqrt(2^2-4*4*(-1)))/(2*4)#
#=(-2+-sqrt20)/8#
#=(-2+-2sqrt5)/8#
#=(-1+-sqrt5)/4#
Discard #sin18=(-1-sqrt5)/4# as this is #<0#
We keep only #sin18=(-1+sqrt5)/4#
So,
#cos72=(-1+sqrt5)/4#
#sin72=sqrt(1-cos^2 72)#
#=sqrt(1-((-1+sqrt5)/4)^2)#
#=sqrt((16-(1+5-2sqrt5)))/4#
#=sqrt(10+2sqrt5)/4#
