Use the formulas for lowering powers to rewrite cos4 x sin4

Use the formulas for lowering powers to rewrite cos^4 x sin^4 x in terms of the first power of cosine

Solution

I imagine you\'re supposed to use the Pythagorean theorem, sin^2(x) = 1-cos^2(x), and the double angle formula (rearranged), [cos(2x) + 1]/2 = cos^2(x). Then...

cos^4(x) = (cos^2(x))^2
= [cos(2x) + 1]^2 / 4
= [cos^2(2x) + 2cos(2x) + 1] / 4
= [(cos(4x) + 1)/2 + 2cos(2x) + 1] / 4
= [cos(4x) + 1 + 4cos(2x) + 2] / 8
= [cos(4x) + 4cos(2x) + 3] / 8

sin^4(x) = (sin^2(x))^2
= [1-cos^2(x)]^2
= [1 - (cos(2x) + 1)/2]^2
= [2 - cos(2x) - 1]^2 / 4
= [1 - cos(2x)]^2 / 4
= [1 - 2cos(2x) + cos^2(2x)] / 4
= [1 - 2cos(2x) + (cos(4x) + 1)/2] / 4
= [2 - 4cos(2x) + cos(4x) + 1] / 8
= [3 - 4cos(2x) + cos(4x)] / 8


Multiply these two expressions to get
cos^4(x) sin^4(x)
= [cos(4x) + 4cos(2x) + 3] / 8 * [3 - 4cos(2x) + cos(4x)] / 8
= ...expand and simplify...
= [cos^2(4x) + 6cos(4x) - 16cos^2(2x) + 9] / 64
= [(cos(8x) + 1)/2 + 6cos(4x) - 16(cos(4x) + 1)/2 + 9]/64
= [cos(8x) + 1 + 12cos(4x) - 16cos(4x) - 16 + 18] / 128
= [cos(8x) - 4cos(4x) + 3] / 128