We are to evalute
#cos((4pi)/5)=cos(pi-pi/5)=-cos(pi/5)#
This problem of evaluating #cos(pi/5)#can be solved by evaluating #sin(pi/10)# in the following way.
Let #A=pi/10#
#=>5A=pi/2#
#=>3A=pi/2-2A#
#:.cos(3A)=cos(pi/2-2A)#
#=>4cos^3A-3cosA=sin2A=2sinAcosA#
#=>4cos^2A-3=2sinA#
#=>4-4sin^2A-3=2sinA#
#=>4sin^2+2sinA-1=0#
#=>sinA=(-2+sqrt(2^2-4*4*
(-1)))/(2*4)#
#=(-2+sqrt20)/8=(sqrt5-1)/4#
#:.sin(pi/10)=(sqrt5-1)/4#,
Now
#cos((4pi)/5)=cos(pi-pi/5)=-cos(pi/5)#
#=-(1-2sin^2(pi/10))#
#=2sin^2(pi/10)-1#
#=2*((sqrt5-1)/4)^2-1#
#=2/16(5+1-2sqrt5)-1#
#=1/8*2*(3-sqrt5)-1#
#=(3-sqrt5)/4-1#
#=(3-sqrt5-4)/4#
#=-(sqrt5+1)/4#