Recall that, #1+cos2theta=2cos^2theta.#
In ther words, #cos2theta=2cos^2theta-1..................(1).#
Replacing #theta" by "4x,# we get,
#1+cos(2*4x)=2cos^2 4x," i.e., "1+cos8x=2cos^2 4x.#
#:. (1+cos8x)/2=cos^2 4x.#
Now, by #(1), cos4x=2cos^2 2x-1,#
#=2(cos2x)^2-1,#
#=2(2cos^2x-1)^2-1.....[because, (1)],#
#=2(4cos^4x-4cos^2x+1)-1,#
#=8cos^4x-8cos^2x+1.#
#:. cos^2 4x=(8cos^4x-8cos^2x+1)^2,#
#=64cos^8x+64cos^4x+1-128cos^6x-16cos^2x+16cos^4x,#
#=64cos^8x-128cos^6x+80cos^4x-16cos^2x+1.#
#:.(1+cos8x)/2=64cos^8x-128cos^6x+80cos^4x-16cos^2x+1.#
Enjoy Maths.!
