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）=

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