化简sin^2α+sin^2β+2sinαsinβcos(α+β)

来源：学生作业帮助网编辑：作业帮时间：2024/04/29 01:29:18

化简sin^2α+sin^2β+2sinαsinβcos(α+β)
化简sin^2α+sin^2β+2sinαsinβcos(α+β)

化简sin^2α+sin^2β+2sinαsinβcos(α+β)
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)

化简sin^2α+sin^2β+2sinαsinβcos(α+β) 化简sin(α+β)+sin(α-β)+2sinαsin(3π/2-β)= 化简sin(α+β)-2sinαsinβ/2sinαsinβ+cos(α+β) sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β) 求证:sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα 求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα 求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα 求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα 证明：sin(2α+β)/sinα - 2cos(α+β)=sinβ/sinα sin(α+β)sin(α-β)=sin^2α-sin^2β的推导过程若sin^2β-sin^2α=m,则sin(α+β)sin(α-β) sin（α+β）-2sinαcosβ/2sinαsinβ+cos（α+β）化简化简：sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β) 化简sin(α+β)-2sinαcosβ 除以 2sinαsinβ+cos(α+β）化简 sin(α+β)-2sinαcosβ/1sinαsinβ+cos(α+β) 化简:(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+cos2αcos2β 化简sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β 化简sin^2α+sin^2β-sin^2cos^2β-cos^2αsin^2β