三角函数 (14 14:54:32)
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三角函数 (14 14:54:32)
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)=
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.
=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.