Question

2sin(pi-a)cos(pi/2-a)+3sin^2(pi/2-a)-2.

Answer 1

$2sin( pi - alpha )*cos( frac{ pi }{2}- alpha )+3 sin^{2}( frac{ pi }{2}- alpha )-2=$
$=2sin alpha *sin alpha +3( sin( frac{ pi }{2}- alpha ))^{2} -2=2 sin^{2} alpha + 3cos^{2} alpha -2=$
$(2 sin^{2} alpha +2 cos^{2} alpha )+ cos^{2} alpha -2=2*( sin^{2} alpha + cos^{2} alpha )+ cos^{2} alpha -2=$
$=2*1+ cos^{2} alpha -2= cos^{2} alpha$