题目

化简sin^2α+sin^2β+2sinαsinβcos(α+β)

提问时间：2020-07-27

答案

sin^2α+sin^2β+2sinαsinβcos(α+β)
=1/2(1-cos2α)+1/2(1-cos2β)+
2sinαsinβ(cosαcosβ-sinαsinβ)
=1/2(1-cos2α)+1/2(1-cos2β)+
2sinαcosαsinβcosβ-2sinαsinβsinαsinβ=1/2(1-cos2α)+1/2(1-cos2β)+
1/2sin2αsin2β-2sin^2αsin^2β=1-1/2cos2α-1/2cos2β+1/2sin2αsin2β-1/2(1-cos2α)(1-cos2β)=1-1/2cos2α-1/2cos2β+1/2sin2αsin2β-1/2+1/2cos2α+1/2cos2β-1/2cos2αcos2β=1/2-1/2(cos2αcos2β-sin2αsin2β)=1/2-1/2cos(2α+2β)