已知sinα=1/3，α∈（π/2,π）,cosβ=-3/5,β∈（π,3π/2),求sin（α+β）、cos（α-β）的值

Question

如题所述

Answer 1

解：∵α∈(π/2,π),β∈(π,3π/2)
∴cosα<0，sinβ<0
∵sinα=1/3，cosβ=-3/5
∴cosα=-√(1-sin²α)=-2√2/3
sinβ=-√(1-cos²β)=-4/5
故sin(α+β)=sinαcosβ+cosαsinβ
=(1/3)(-3/5)+(-2√2/3)(-4/5)
=(8√2-3)/15
cos(α-β)=cosαcosβ+sinαsinβ
=(-2√2/3)(-3/5)+(1/3)(-4/5)
=(6√2-4)/15