Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 13

$\frac{1}{16}(1-cos2x-cos4x+cos2x\cdot cos4x)$

Use $sin2x=2sin(x)cos(x), sin^2x=\frac{1-cos2x}{2}$, we have $cos^2xsin^4x=\frac{1}{4}(2sin(x)cos(x))^2sin^2x=\frac{1}{4}(sin2x)^2sin^2x=\frac{1}{4}\frac{1-cos4x}{2}\frac{1-cos2x}{2}=\frac{1}{16}(1-cos2x-cos4x+cos2x\cdot cos4x)$