Question 703828
Recall that sin(2θ) = 2sin(θ)cos(θ) => sin(2θ)/2 = sin(θ)cos(θ)

sin^2(θ) * cos^2(θ) = (sin(2θ)/2)^2 = 1/4*sin^2(2θ)

Simplifying that further and remembering that sin^2(θ) = (1 - cos(2θ))/2

1/4*sin^2(2θ) = 1/8 * (1 - cos(4θ))