Chapter 6 - Analytic Trigonometry - Section 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 519: 41

See below.

$sin^4\theta=(sin^2\theta)^2=(\frac{1-cos(2\theta)}{2})^2=\frac{1}{4}(1-2cos(2\theta)+cos^2(2\theta))=\frac{1}{4}(1-2cos(2\theta)+\frac{1+cos(4\theta)}{2})=\frac{1}{4}+\frac{1}{2}-\frac{1}{2}cos(2\theta)+\frac{1}{8}cos(4\theta) =\frac{3}{8}-\frac{1}{2}cos(2\theta)+\frac{1}{8}cos(4\theta)$