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
问题描述：
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
答
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