Question

Sin^2xcos^2x=1-cos4x/8

Answer 1

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