若sinαsinβ+cosαcosβ=0,则sinαcosα+sinβcosβ= .
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/03 00:01:52
若sinαsinβ+cosαcosβ=0,则sinαcosα+sinβcosβ= .
1若sinαsinβ+cosαcosβ=0,则sinαcosα+sinβcosβ=
2已知sin(α+β)=1,则cos(α+2β)+sin(2α+β)=
急,
1若sinαsinβ+cosαcosβ=0,则sinαcosα+sinβcosβ=
2已知sin(α+β)=1,则cos(α+2β)+sin(2α+β)=
急,
∵sinαsinβ+cosαcosβ=0
∴cos(α-β)=0
∴α-β=kπ+π/2(k∈Z),α=β+kπ+π/2(k∈Z)
∴2α=2β+2kπ+π (k∈Z)
∴sin2α=sin(2β+2kπ+π)=sin(2β+π)=-sin2β (k∈Z)
sinαcosα+sinβcosβ=1/2sin2α+1/2sin2β=1/2(sin2α+sin2β)=0
∵ sin(a+b)=1
∴cos(α+β)=0,α+β=π/2+2kπ(k∈Z)
cos(α+2β)+sin(2α+β)
=cos(π/2+2kπ +β)+sin(α+π/2+2kπ)(k∈Z)
=cos(π/2 +β)+sin(α+π/2) (因为cos(2kπ +β)=cosβ,sin(α+2kπ)=sinα)
=cosα-sinβ (因为sin(α+π/2)=cosα,cos(π/2 +β)=-sinβ)
=cos(π/2+2kπ -β)-sinβ
=sinβ-sinβ=0
∴cos(α-β)=0
∴α-β=kπ+π/2(k∈Z),α=β+kπ+π/2(k∈Z)
∴2α=2β+2kπ+π (k∈Z)
∴sin2α=sin(2β+2kπ+π)=sin(2β+π)=-sin2β (k∈Z)
sinαcosα+sinβcosβ=1/2sin2α+1/2sin2β=1/2(sin2α+sin2β)=0
∵ sin(a+b)=1
∴cos(α+β)=0,α+β=π/2+2kπ(k∈Z)
cos(α+2β)+sin(2α+β)
=cos(π/2+2kπ +β)+sin(α+π/2+2kπ)(k∈Z)
=cos(π/2 +β)+sin(α+π/2) (因为cos(2kπ +β)=cosβ,sin(α+2kπ)=sinα)
=cosα-sinβ (因为sin(α+π/2)=cosα,cos(π/2 +β)=-sinβ)
=cos(π/2+2kπ -β)-sinβ
=sinβ-sinβ=0
若sinαsinβ+cosαcosβ=0,则sinαcosα+sinβcosβ= .
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