Question

问几道高一数学题！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

Answer 1

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)≤1
0≤1/2cos(α-β)+1/4≤3/4

y=sinαsinβ的最大值是3/4\最小值是0.
第二题sin（A+B）=sinAcosB+sinBcosA=3/5...(1)
sin（A-B）=sinAcosB-sinBcosA=1/5...(2)
(1)=3*(2)
sinAcosB+sinBcosA=3sinAcosB-3sinBcosA
2sinBcosA=sinAcosB
tanA/tanB =2......(3)
sin（A+B）=3/5,所以sinC=3/5,cosC=4/5,tanC=3/4,所以tan(A+B)=-3/4=(tanA+tanB)/(1-tanAtanB),代入(3)式
所以tanA=2+根号6,tanB =1+根号6/2
设高为h,画图知:AB=h/tanA+h/tanB
所以h=2tanB*AB/3=2+根号6

Answer 2

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