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
1、若a和b∈(0,π/2).cos(a-b/2)=√3/2 sin(a/2-b)=-1/2,则cos(a+b)=
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