证明:由sinα+sinβ+sinγ=0,有sinα+sinγ=-sinβ,两边平方,有(sinα)^2+(sinγ)^2+2sinαsinγ=(sinβ)^2.(1)由cosα+cosβ+cosγ=0,有cosα+cosγ=-cosβ,两边平方,有(cosα)^2+(cosγ)^2+2cosαcosγ=(cosβ)^2.(2)将(1)(2)两式相加,有(sinα)^2+(cosα)^2+(sinγ)^2+(cosγ)^2+2sinαsinγ+2cosαcosγ=(sinβ)^2+(cosβ)^2.即1+1+2(cosαcosγ+sinαsinγ)=1,所以cosαcosγ+sinαsinγ=cos(α-γ)=-1/2
求采纳