cos^x sin^x = A + B cos 2x + C cos 4x + D cos 2x cos 4x

Question

I have no idea how to solve this problem. 

Answer 1

You will need to know these formulas

cos²x = (1 + cos 2x)/2

sin²x = (1 − cos 2x)/2

 

For this problem

cos²xsin⁴x = [(1 + cos 2x)/2][(1 − cos 2x)/2]²

                 = (1/8)(1 + cos 2x)(1 − 2 cos 2x + cos²2x)

                 = (1/8)(1 + cos 2x)[1 − 2 cos 2x + ½(1 + cos 4x)]

                 = (1/8)(1 + cos 2x)(3/2 − 2 cos 2x + ½ cos 4x)

                 = (1/8)(3/2 − 2 cos 2x + ½ cos 4x + 3/2 cos 2x − 2 cos²2x + ½ cos 2x cos 4x)

                 = (1/8)(3/2 − ½ cos 2x + ½ cos 4x  − 1 − cos 4x + ½ cos 2x cos 4x)

                 = (1/8)(½ − ½ cos 2x − ½ cos 4x + ½ cos 2x cos 4x)

                 = (1/16)(1 − cos 2x − cos 4x + cos 2x cos 4x)

 

So A = D = 1/16; B = C = −1/16.