Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 14

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

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