Question

sin(pi-b)cos(pi/2-b)-sin(pi/2+b)cos(pi-b)

Answer 1

Sin(pi-b)cos(pi/2-b)-sin(pi/2+b)cos(pi-b)=sin(b)*sin(b)-cos(b)*(-cos(b)) =sin(b)*sin(b)+cos(b)*cos(b)=sin^2(b)+cos^2(b)=1