4cos(2θ)sin(4θ)

We notice that this has a product of a sine and a cosine
and we remember that such a product appears in the formulas: 

sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
sin(A-B) = sin(A)cos(B) - cos(A)sin(B)

So we substitute A=4θ and B=2θ in those

sin(4θ+2θ) = sin(4θ)cos(2θ) + cos(4θ)sin(2θ)
sin(4θ-2θ) = sin(4θ)cos(2θ) - cos(4θ)sin(2θ)

or

sin(6θ) = sin(4θ)cos(2θ) + cos(4θ)sin(2θ)
sin(2θ) = sin(4θ)cos(2θ) - cos(4θ)sin(2θ)

If we add those equations together, equal to equals,
the last terms cancel and we get

sin(6θ) + sin(2θ) = 2sin(4θ)cos(2θ)

So our problem

4cos(2θ)sin(4θ) = 2[2sin(4θ)cos(2θ)] = 2[sin(6θ) + sin(2θ)]

or

2sin(6θ) + 2sin(2θ)

Edwin