How do you prove #sin^4 X = (cos^2 2X - 2 cos 2x + 1) / 4#?
1 Answer
Apr 30, 2016
Using the following:
#a^2-2ab+b^2 = (a-b)^2# #cos(2x) = cos^2(x)-sin^2(x)# #1 = cos^2(x)+sin^2(x)#
we have: