## Precalculus (6th Edition) Blitzer

$\frac{1}{16}(5-7cos(2x)+3cos(4x)-cos(4x)cos(2x))$
Step 1. Using the formula $sin^2x=\frac{1-cos(2x)}{2}$, we have $sin^6x=(sin^2x)^3=(\frac{1-cos(2x)}{2})^3=\frac{1}{8}(1-3cos(2x)+3cos^2(2x)-cos^3(2x))$ Step 2. Using the formula $cos^2(2x)=\frac{1+cos(4x)}{2}$, we have $sin^6x=\frac{1}{8}(1-3cos(2x)+3(\frac{1+cos(4x)}{2})-(\frac{1+cos(4x)}{2})cos(2x))=\frac{1}{16}(2-6cos(2x)+3(1+cos(4x))-(1+cos(4x))cos(2x))=\frac{1}{16}(5-7cos(2x)+3cos(4x)-cos(4x)cos(2x))$