a,b∈(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos(a+pi/4)=
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a,b∈(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos(a+pi/4)=a,b∈(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)
a,b∈(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos(a+pi/4)=
a,b∈(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos(a+pi/4)=
a,b∈(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13,则cos(a+pi/4)=
解cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
由a,b∈(3pi/4,pi),sin(a+b)=-3/5,sin(b-pi/4)=12/13
知a+b属于(3π/2,2π),b-π/4属于(π/2,3π/4)
cos(a+b)=4/5,cos(b-π/4)=5/13
故
cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
=4/5*5/13+(-3/5)*(12/13)
=20/65-36/65
=-16/65
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