0所以π/2<π/2+a<π,π/4<π/4-b/2<3π/4又cos(π/2+a)=1/3,cos(π/4-b/2)=根号3/3,所以sin(π/2+a)=2根2/3,sin(π/4-b/2)=-根6/3cos(a+b/2)=cos[(π/2+a)-(π/4-b/2)] =cos(π/2+a)cos(π/4-b/2)+sin(π/2+a)sin(π/4-b/2) =..........=-根3/3记得采纳哦