## Precalculus (6th Edition) Blitzer

a. $y=cos(x)$ b. see explanations.
a. The graph appears to be the same as $y=cos(x)$ b. Using the addition and double-angle formulas, we have $y=\frac{sin(x)+sin(2x+x)}{2sin(2x)}=\frac{sin(x)+sin(2x)cos(x)+cos(2x)sin(x)}{4sin(x)cos(x)}=\frac{sin(x)+2sin(x)cos^2(x)+cos(2x)sin(x)}{4sin(x)cos(x)}=\frac{1+2cos^2(x)+cos(2x)}{4cos(x)}=\frac{1+2cos^2(x)+2cos^(x)-1}{4cos(x)}=cos(x)$