Question

一道数学题，拜托了，请留下详细的解题过程，谢谢~

Answer 1

∵α-β=π/6
∴sinαsinβ=1/2[cos(α-β)-cos(α+β)]=1/2[cos(π/6)-(cos(2α-π/6)]=√3/4-1/2*cos(2α-π/6)
∵0<α<π/2, 0<β<π/2
∴0<β=α-π/6<π/2
∴π/6<α<π/2
∴π/3<2α<π
∴π/6<2α-π/6<5π/6
∴cos(2α-π/6)的取值范围是(-√3/2,√3/2)
∴sinαsinβ的取值范围是(0,√3/2)

Answer 2

∵α-β=π/6
∴sinαsinβ=-1/2(cos(α+β)-cos(α-β))=-1/2(cos(2α-π/6)-cos(π/6))=-1/2*cos(2α-π/6)+√3/4
∵α、β是锐角,α-β=π/6
π/6＜α＜π/2
π/6＜2α-π/6＜5π/6
cos(2α-π/6)的取值范围是(-√3/2,√3/2)
∴sinαsinβ的取值范围是(0,√3/2)