已知sina+sinβ=1,cosa+cosβ=0,则cos(a+β）=?

/>sinα+sinβ=1 sinα=1-sinβ
cosα+cosβ=0 cosα=-cosβ
sin²α+cos²α=1
(1-sinβ)²+(-cosβ)²=1
2-2sinβ=1
sinβ=1/2 cos²β=1-sin²β=1-1/4=3/4
cos(α+β)=cosα-sinαsinβ
=-cos²β-(1-sinβ)sinβ
=-(3/4) -(1-1/2)(1/2)
=-3/4 -1/4
=-1