Question

求证: Sin2α＋sin2β－Sin2α×sin2β＋cos2α× cos2β＝1

Answer 1

以上的2都是平方吧
解：sin2α+sin2β-sin2αsin2β+cos2αcos2β
=sin2α（1-sin2β）+sin2β+cos2αcos2β
=sin2α•cos2β+sin2β+cos2αcos2β
=cos2β（sin2α+cos2α）+sin2β
=cos2β+sin2β=1

Answer 2

Sin2α＋sin2β－Sin2α×sin2β＋cos2α× cos2β
=Sin2α＋sin2β-4Sinαcosαsinβ cosβ+(cos^α-Sin^α)× (cos^β-Sin^β)
=Sin2α+sin2β+(cosαcosβ-SinαSinβ)^-(cosαSinβ+Sinαcosβ)^
=Sin2α+sin2β+cos^(α+β)-Sin^(α+β)
=1+Sin2α+sin2β-2Sin^(α+β)
=1+2(Sinαcosα+Sinβcosβ-Sinαcosα-Sinβcosβ)
=1