∵0<α<π/2
∴sin(π/4+а)=√[1-cos²(π/4+а)]=2√2/3
∵-π/2<в<0
∴sin(π/4-в/2)=√[1-cos²(π/4-в/2)]=√6/3
∴cos(а+в/2)=cos[(π/4+а)-(π/4-в/2)]
=cos(π/4+а)cos(π/4-в/2)+sin(π/4+а)sin(π/4-в/2)
=(1/3)(√3/3)+(2√2/3)(√6/3)
=√3/9+4√3/9
=5√3/9
希望能帮到你，祝学习进步O(∩_∩)O
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