已知sinα+cosβ+sinγ=0,且cosα+sinβ+cosγ=0,求sin(α+β)

sinα+cosβ+sinγ=0
sinα+cosβ=-sinγ
(sinα+cosβ)^2=(sinγ)^2
(sinα)^2+2sinαcosβ+(cosβ)^2=(sinγ)^2.1
cosα+sinβ+cosγ=0
cosα+sinβ=-cosγ
(sinβ+cosα)^2=(cosγ)^2
(sinβ)^2+2sinβcosα+(cosα)^2=(cosγ)^2.2
1式+2式得
2+2sinαcosβ+2sinβcosα=1
2(sinαcosβ+sinβcosα)=-1
2sin(α+β)=-1
sin(α+β)=-1/2