1、若a和b∈(0,π/2).cos(a-b/2)=√3/2 sin(a/2-b)=-1/2,则cos(a+b)=2、已知a,b∈(3π/4,π),sin(a+b)=-3/5,sin(b-π/4)=12/13,则cos(a+π/4)=
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![1、若a和b∈(0,π/2).cos(a-b/2)=√3/2 sin(a/2-b)=-1/2,则cos(a+b)=2、已知a,b∈(3π/4,π),sin(a+b)=-3/5,sin(b-π/4)=12/13,则cos(a+π/4)=](/uploads/image/z/6112780-52-0.jpg?t=1%E3%80%81%E8%8B%A5a%E5%92%8Cb%E2%88%88%EF%BC%880%2C%CF%80%2F2%EF%BC%89.cos%EF%BC%88a-b%2F2%EF%BC%89%3D%E2%88%9A3%2F2+sin%EF%BC%88a%2F2-b%29%3D-1%2F2%2C%E5%88%99cos%EF%BC%88a%2Bb%29%3D2%E3%80%81%E5%B7%B2%E7%9F%A5a%2Cb%E2%88%88%283%CF%80%2F4%2C%CF%80%EF%BC%89%2Csin%EF%BC%88a%2Bb%29%3D-3%2F5%2Csin%28b-%CF%80%2F4%EF%BC%89%3D12%2F13%2C%E5%88%99cos%EF%BC%88a%2B%CF%80%2F4%EF%BC%89%3D)
1、若a和b∈(0,π/2).cos(a-b/2)=√3/2 sin(a/2-b)=-1/2,则cos(a+b)=2、已知a,b∈(3π/4,π),sin(a+b)=-3/5,sin(b-π/4)=12/13,则cos(a+π/4)=
1、若a和b∈(0,π/2).cos(a-b/2)=√3/2 sin(a/2-b)=-1/2,则cos(a+b)=
2、已知a,b∈(3π/4,π),sin(a+b)=-3/5,sin(b-π/4)=12/13,则cos(a+π/4)=
1、若a和b∈(0,π/2).cos(a-b/2)=√3/2 sin(a/2-b)=-1/2,则cos(a+b)=2、已知a,b∈(3π/4,π),sin(a+b)=-3/5,sin(b-π/4)=12/13,则cos(a+π/4)=
a,b属于(0,π/2),a/2,b/2属于(0,π/4),-π/4<a-b/2<π/2,-π/2<a/2-b<π/4,cos(a-b/2)=√3/2,a-b/2=π/6或a-b/2=-π/6,sin(a/2-b)=-1/2,a/2-b=-π/6,当a-b/2=π/6,a/2-b=-π/6时得a=π/3,b=π/3;当a-b/2=-π/6,a/2-b=-π/6时得a=-π/3,b=-π/3,不属于(0,π/2),故舍去;则cos(a+b)=cos(2π/3)=-1/2