sin^2xcos^2x=1-cos4x/8
\(\sin^2x\cos^2x=\left(\dfrac{1-\cos2x}2\right)\left(\dfrac{1+\cos2x}2\right)=\dfrac{1-\cos^22x}4=\dfrac{1-\dfrac{1+\cos4x}2}4=\dfrac{2-(1+\cos4x)}8=\dfrac{1-\cos4x}8\)
