Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 15

$\frac{1}{128}(3-4cos4x+cos8x)$

Use $sin2x=2sin(x)cos(x), sin^2x=\frac{1-cos2x}{2}, cos^2x=\frac{1+cos2x}{2}$, we have $cos^4xsin^4x=\frac{1}{16}(2sin(x)cos(x))^4=\frac{1}{16}(sin2x)^4 =\frac{1}{16}(\frac{1-cos4x}{2})^2 =\frac{1}{64}(1-2cos4x+ cos^24x) =\frac{1}{64}(1-2cos4x+\frac{1+cos8x}{2}) =\frac{1}{128}(3-4cos4x+cos8x)$