cosα=4/5,α∈(3π/2,2π),=>sinα=-√[1-cos²α]=-√[1-(4/5)^2]=-3/5tanβ=3/4,β∈(0,π) => β∈(0,π/2)=> sinβ=3/5,cosβ=4/5cos(α-β)=cosαcosβ+sinαsinβ=(4/5)*(4/5)+(-3/5)*(3/5)=7/25
