Answer :
from cos(2x) = 2cos^2(x) -1
cos(4x) = 2[cos(2x)]^2 -1
= 2[2cos(x) -1]^2 -1
= 2[4cos^2(x) - 4cos(x) + 1] -1
= 8cos^2(x) - 8cos(x) + 1
cos(4x) = 2[cos(2x)]^2 -1
= 2[2cos(x) -1]^2 -1
= 2[4cos^2(x) - 4cos(x) + 1] -1
= 8cos^2(x) - 8cos(x) + 1