已知sinα+sinβ+sinγ=0,cosα+cosβ+cosγ=0,求cos(β-γ).

sinα=-(sinβ+sinγ),sinα^2=(sinβ+sinγ)^2=sinβ^2+2sinγsinβ+sinγ^2cosα=-(cosβ+cosγ),cosα^2=(cosβ+cosγ)^2=cosβ^2+2cosγcosβ+cosγ^2加起来有1=2+2(sinγsinβ+cosγcosβ)=2+2cos(β-γ).所以c...