作业帮已知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/4
3π/4
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/4
3π/4
所以：
cos(a+π/4)=-20/65-36/65=-56/65.
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