Question

2sin^2 2 альфа + 2cos(п:2 - альфа) + 2 cos^2 2 альфа

Answer 1

$2sin^22 alpha +2cos(frac{pi}{2}- alpha )+2cos^22 alpha =\\=2(sin^22 alpha +cos^22 alpha )+2sin alpha =2cdot 1+2sin alpha =2(1+sin alpha ); ;\\ili\\2sin^22 alpha +2cos(frac{pi}{2}- alpha )+2cos^22 alpha =\\=2(sin^22 alpha +cos^22 alpha )+2cos(frac{pi}{2}- alpha )=2(1+cos(frac{pi}{2}- alpha ))=\\=[, cos^2x=frac{1+cos2x}{2}, ]=2cdot 2cos^2(frac{pi}{4}-frac{ alpha }{2})=4cos^2(frac{pi}{4}-frac{ alpha }{2}); ;$