Question

2sin32°cos2°+2sin²28+ 1/2

Answer 1

$2sin32cos2+2sin^228+frac{1}{2}=2cdotfrac{1}{2}(sin(32+2)+sin(32-2))+2cdotfrac{1}{2}(cos(28-28)-cos(28+28))+frac{1}{2}=sin34+sin30+cos0-cos56+frac{1}{2}=sin34+sin30+cos0-cos56+frac{1}{2}=cos(90-34)+frac{1}{2}+1-cos56+frac{1}{2}=cos56-cos56+2=2$