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