已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,则cos(α+Π/4)=?

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已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,则cos(α+Π/4)=?
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已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,则cos(α+Π/4)=?
已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,则cos(α+Π/4)=?

已知α,β∈(3π/4,π),sin(α+β)=-3/5,sin(β-π/4)=12/13,则cos(α+Π/4)=?
3π/43π/43π/2cos(a+b)=4/5
3π/4π/2cos(b-π/4)=-5/13
cos(a+π/4)=cos(a+b+π/4-b)=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
=(4/5)*(-5/13)+(-3/5)*(12/13)
=-56/65

cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
=cos(a+b)cos(b-π/4)-36/65;

3π/43π/43π/2所以:cos(a+b)=4/5;

3π/4π/2所以:cos(b-π/4)=-5/13.

所以:
cos(a+π/4)=-20/65-36/65=-56/65.