根据和角公式和倍角公式,得:
sin(π/4+a)·sin(π/4-a)
=[sin(π/4)·cosa+cos(π/4)·sina]·[sin(π/4)·cosa-cos(π/4)·sina]
=[sqrt(2)/2·(cosa+sina)]·[sqrt(2)/2·(cosa-sina)]
=(1/2)·(cosa+sina)·(cosa-sina)
=(1/2)·(cos²a-sin²a)
=(1/2)·cos(2a)
=1/6
∴cos(2a)=1/3
∵a∈(π/2,π)
∴2a∈(π,2π)
∵cos(2a)=1/3>0
∴2a∈(3π/2,2π)
∴sin(2a)