Question

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

Answer 1

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

Answer 2

sinα+cosβ=-sinγ
cosα+sinβ=-cosγ
两边平方再两式相加
得2（sinα*cosβ+cosα*sinβ）=2sin(α+β)=-1
所以sin(α+β)=-0.5