Question

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

要完整解题过程

Answer 1

3π/4<a<π
3π/4<b<π
3π/2<a+b<2π
cos(a+b)=4/5
3π/4<b<π
π/2<b-π/4<3π/4
cos(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

Answer 2

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<a<π
3π/4<b<π
3π/2<a+b<2π
所以：cos(a+b)=4/5;

3π/4<b<π;
π/2<b-π/4<3π/4.
所以：cos(b-π/4)=-5/13.

所以：
cos(a+π/4)=-20/65-36/65=-56/65.