化简sin^2α+sin^2β+2sinαsinβcos(α+β)
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/30 02:24:14
化简sin^2α+sin^2β+2sinαsinβcos(α+β)
![化简sin^2α+sin^2β+2sinαsinβcos(α+β)](/uploads/image/z/18325381-13-1.jpg?t=%E5%8C%96%E7%AE%80sin%5E2%CE%B1%2Bsin%5E2%CE%B2%2B2sin%CE%B1sin%CE%B2cos%28%CE%B1%2B%CE%B2%29)
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)
=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:(1)sin(α+β)−2sinαcosβ2sinαsinβ+cos(α+β)
化简:sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β)
化简sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β
求证sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β=1
求证sinα-sinβ=2cos(α+β)/2sin(α-β)/2
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求证:[sin(2α+β)/2sinα]-cos(α+β)=sinβ/2sinα
求证:sin(2α+β)sinα
sinα^2+sinβ^2+sinγ^2=1,那么cosαcosβcosγ最大值等于
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