急 一直1.已知α+β=2π/3,且0≤α≤π/2,求y=sinαsinβ的最大值和最小值2.已知锐角△ABC中,sin（A+B)=3/5,sin(A-B)=1/5,（1）求证：tanA=2TANB (2)设AB=3,求AB边上的高
问题描述：
急 一直
1.已知α+β=2π/3,且0≤α≤π/2,求y=sinαsinβ的最大值和最小值
2.已知锐角△ABC中,sin（A+B)=3/5,sin(A-B)=1/5,（1）求证：tanA=2TANB (2)设AB=3,求AB边上的高
答
y=sinαsinβ
=sinαsin(2π/3-α)
=sinα(sin2π/3*cosα-cos2π/3sinα)
=√3/2sinαcosα+1/2(sinα)2
=√3/4sin2α+1/4(1-cos2α)
=1/2(√3/2sin2α-1/2cos2α)
= 1/2sin(2α-π/6)+1/4
因为0≤α≤π/2 所以-π/6≤2α-π/6≤5π/6
从图象上看当α=0时，有最小值y=0
当α=2π/3时，有最大值y=3/4
答
第一题y=sinαsinβ=1/2[cos(α-β)-cos(α+β)]=1/2[cos(α-β)-cos(2π/3)]=1/2[cos(α-β)+1/2]=1/2cos(α-β)+1/4又因为α+β=2π/3,且0≤α≤π/2,知-2π/3≤α-β≤π/2,-1/2≤cos(α-β)≤10≤1/2cos(α-β)+...