已知sinα+cosβ=3/5,cosα+sinβ=4/5,求sin(α+β)

(sinα+cosβ)^2=(sinα)^2+(cosβ)^2+2sinαcosβ=9/25
(cosα+sinβ)^2=(cosα)^2+(sinβ)^2+2cosαsinβ=16/25
两式相加：
(sinα)^2+(cosα)^2+(sinβ)^2+(cosβ)^2+2cosαsinβ+2sinαcosβ
=1+1+2(cosαsinβ+sinαcosβ)
=2+2(cosαsinβ+sinαcosβ)=1
sin(α+β)=cosαsinβ+sinαcosβ=-1/2

sinα+cosβ=3/5
cosα+sinβ=4/5

so 1=(sinα+cosβ)^2+(cosα+sinβ)^2=2+2(sinacosb+cosasinb)=2+2sin(a+b)

so sin(a+b)=(1-2)/2=-1/2