Question

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

Answer 1

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