已知3sin^2 α+2sin^2 β=2sinα,求sin^2α+cos^2β的取值范围 要有具体过程
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![已知3sin^2 α+2sin^2 β=2sinα,求sin^2α+cos^2β的取值范围 要有具体过程](/uploads/image/z/3984535-55-5.jpg?t=%E5%B7%B2%E7%9F%A53sin%5E2+%CE%B1%2B2sin%5E2+%CE%B2%3D2sin%CE%B1%2C%E6%B1%82sin%5E2%CE%B1%2Bcos%5E2%CE%B2%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4+%E8%A6%81%E6%9C%89%E5%85%B7%E4%BD%93%E8%BF%87%E7%A8%8B)
已知3sin^2 α+2sin^2 β=2sinα,求sin^2α+cos^2β的取值范围 要有具体过程
已知3sin^2 α+2sin^2 β=2sinα,求sin^2α+cos^2β的取值范围
要有具体过程
已知3sin^2 α+2sin^2 β=2sinα,求sin^2α+cos^2β的取值范围 要有具体过程
解 因为 3sin²α+2cos²β=2sinα 2cos²β=2sinα-3sin²α=sinα(2-3sinα)≥0
0≤sinα≤2/3
所以 sin²α+cos²β=sinα-½sin²α
=-½(sin²α-2sinα)
=-½(sin²α-2sinα+1-1)
=-½(sinα-1)²+½
因为 0≤sinα≤2/3
所以 -1≤sinα-1≤-1/3
1/9≤(sinα-1)²≤1
0≤-½(sinα-1)²+½≤4/9
即 0≤sin²α+cos²β≤4/9
不懂的欢迎追问,如有帮助请采纳,谢谢!
2sin^2 β=-3sin^2 α+2sin α∈[0,2]
sin α∈[0,2/3]
sin^2 α+cos^2 β=sin^2 α+1-sin^2 β=sin^2 α+1+3/2*sin^2 α-sin α=5/2*sin^2 α-sin α+1
=5/2*(sin α-1/5)^2+9/10
∈[9/10,13/9]
cos^ α+cos^2 β=...
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2sin^2 β=-3sin^2 α+2sin α∈[0,2]
sin α∈[0,2/3]
sin^2 α+cos^2 β=sin^2 α+1-sin^2 β=sin^2 α+1+3/2*sin^2 α-sin α=5/2*sin^2 α-sin α+1
=5/2*(sin α-1/5)^2+9/10
∈[9/10,13/9]
cos^ α+cos^2 β=1-sin^2 α+1-sin^2 β=2-sin^2 α+3/2*sin^2 α-sin α=1/2*sin^2 α-sin α+2
=1/2*(sin^2 α-1)^2+3/2
∈[16/9,2]
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