## Precalculus (6th Edition) Blitzer

$\frac{1}{2}[sin(10x)+sin(4x)]$
Using the Product-to-Sum Formula $sin(u)cos(v)=\frac{1}{2}[sin(u+v)+sin(u-v)]$, we have $sin(7x)cos(3x)=\frac{1}{2}[[sin(7x+3x)+sin(7x-3x)]=\frac{1}{2}[sin(10x)+sin(4x)]$