已知θ∈(0,π/2),f(θ)=(sin2θ+1)2/sin2θ,求f(θ)的最小值及相应的θ值
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已知θ∈(0,π/2),f(θ)=(sin2θ+1)2/sin2θ,求f(θ)的最小值及相应的θ值
(sin2θ+1)2后一个2为平方
(sin2θ+1)2后一个2为平方
![已知θ∈(0,π/2),f(θ)=(sin2θ+1)2/sin2θ,求f(θ)的最小值及相应的θ值](/uploads/image/z/8645182-70-2.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B8%E2%88%88%280%2C%CF%80%2F2%29%2Cf%28%CE%B8%29%3D%28sin2%CE%B8%2B1%292%2Fsin2%CE%B8%2C%E6%B1%82f%28%CE%B8%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E5%8F%8A%E7%9B%B8%E5%BA%94%E7%9A%84%CE%B8%E5%80%BC)
θ∈(0,π/2)
2θ∈(0,π)
f(θ)=(sin2θ+1)^2/sin2θ
=[(sin2θ)^2+2sin2θ+1]/sin2θ
=(sin2θ)^2/sin2θ+2sin2θ/sin2θ+1/sin2θ
=sin2θ+2+1/sin2θ
=sin2θ+1/sin2θ+2
>=2+2
=4
f(θ)的最小值为:4
sin2θ=1/sin2θ
(sin2θ)^2=1
2θ=π/2
θ=π/4
2θ∈(0,π)
f(θ)=(sin2θ+1)^2/sin2θ
=[(sin2θ)^2+2sin2θ+1]/sin2θ
=(sin2θ)^2/sin2θ+2sin2θ/sin2θ+1/sin2θ
=sin2θ+2+1/sin2θ
=sin2θ+1/sin2θ+2
>=2+2
=4
f(θ)的最小值为:4
sin2θ=1/sin2θ
(sin2θ)^2=1
2θ=π/2
θ=π/4
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