已知α,β属于(3π/4,π),sin(α+β)=-3/5sin(β-π/4)=12/13求cos(α+π/4)

已知α,β属于(3π/4,π),sin(α+β)=-3/5sin(β-π/4)=12/13求cos(α+π/4)
已知α,β属于(3π/4,π),sin(α+β)=-3/5sin(β-π/4)=12/13求cos(α+π/4)

已知α,β属于(3π/4,π),sin(α+β)=-3/5sin(β-π/4)=12/13求cos(α+π/4)
α,β属于(3π/4,π),
α+β属于(3π/2,2π),
sin(α+β)=-3/5
所以
cos(α+β)=4/5
β-π/4∈（π/2,3π/4）
sin(β-π/4)=12/13
cos(β-π/4)=-5/13
所以
cos(α+π/4)
=cos[(α+β)-(β-π/4)]
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
=4/5*(-5/13)+(-3/5)*12/13
=-56/65