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
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![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](/uploads/image/z/122158-46-8.jpg?t=%E6%80%A5+%E4%B8%80%E7%9B%B41.%E5%B7%B2%E7%9F%A5%CE%B1%2B%CE%B2%3D2%CF%80%2F3%2C%E4%B8%940%E2%89%A4%CE%B1%E2%89%A4%CF%80%2F2%2C%E6%B1%82y%3Dsin%CE%B1sin%CE%B2%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC2.%E5%B7%B2%E7%9F%A5%E9%94%90%E8%A7%92%E2%96%B3ABC%E4%B8%AD%2Csin%EF%BC%88A%2BB%29%3D3%2F5%2Csin%28A-B%29%3D1%2F5%2C%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9AtanA%3D2TANB+%282%29%E8%AE%BEAB%3D3%2C%E6%B1%82AB%E8%BE%B9%E4%B8%8A%E7%9A%84%E9%AB%98)
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
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
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
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
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 ...
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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
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