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三角函数式的取值范围已知3(sinα)^2+2(sinβ)^2=

2006-07-26 16:19:36N***
已知3(sinα)^2+2(sinβ)^2=2sinα 求(sinα)^2+(sinβ)^2的取值范围。三角函数式的取值范围已知3(sinα)^2+2(sinβ)^2=2sinα求(sinα)^2+(sinβ)^2的取值范围。：解:由3(sinα)^2+2(sin?

最佳回答

  • 解:由3(sinα)^2+2(sinβ)^2=2sinα得,3(sinα)^2-2sinα=-2(sinβ)^2 -2<=3(sinα)^2-2sinα<=0,0<=sinα<=2/3 (sinα)^2+(sinβ)^2=(sinα)^2-[3(sinα)^2-2sinα]/2 = -1/2(sinα)^2^2+sinα=(-1/2)[sinα-1]^2+1/2 0<=(sinα)^2+(sinβ)^2<=4/9
    2006-07-26 16:34:47
  • 3(sinα)^2+2(sinβ)^2=2sinα 所以(sinβ)^2=[2sinα-3(sinα)^2]/2 (sinα)^2+(sinβ)^2=sinα-(sinα)^2/2=-(sinα-1)^2/2+1/2 又由3(sinα)^2+2(sinβ)^2=2sinα得:2sinα-3(sinα)^2=2(sinβ)^2 因为0小于等于2(sinβ)^2小于等于2,所以0小于等于2sinα-3(sinα)^2小于等于2.所以0小于等于sinα小于等于2/3 所以0小于等于(sinα)^2+(sinβ)^2小于等于4/9
    2006-07-26 16:41:07
  • 同意不厌不倦之人的回答
    2006-07-26 16:39:42
  • sinα)^2+(sinβ)^2的取值范围。 -9/2到0之间
    2006-07-26 16:37:39
  • 不妨设S=(sina)^2+(sinb)^2 则2S=-(sina)^2+2sina=-(sina-1)^2+1 -1≤sina≤1,-2≤sina-1≤0,0≤(sina-1)^2≤4 -3≤1-(sina-1)^2≤1 -3≤2S≤1,显然0≤S，所以0≤S≤1/2 即(sinα)^2+(sinβ)^2的取值范围[0,1/2]
    2006-07-26 16:32:18
  • (sinα)^2是什么意思？平方还是2倍？
    2006-07-26 16:26:53
    • 很赞哦！ (75)

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