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已知0<θ<π/2,求函数F(θ)=[2sin^2(θ)+1]/sin2θ的最小值

f(θ)=[2sin^2(θ)+1]/sin2θ =[3sin^2(θ)+cos^2(θ)]/sin2θ =3sin^2(θ)/sin2θ+cos^2(θ)/2sin(2θ) =3sin^2(θ)/2sinθcosθ+cos^2(θ)/2sinθcosθ =3sinθ/2cosθ+cosθ/2sinθ >=√3所以最小值为√3

f(θ)=(sin2θ+2)^2/sin2θ = sin2θ+4+4/sin2θ sin2θ=t(04所以g'(t)=1-4/t^2 评论0 0 0

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=4f(θ)的最小值为:4sin2θ=1/sin2θ(sin2θ)^2=12θ=π/2θ=π/4

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=4f(θ)的最小值为:4sin2θ=1/sin2θ(sin2θ)^2=12θ=π/2θ=π/4

你好希望下面我的回答对你有帮助tanθ+cotθ=sinθ/cosθ+cosθ/sinθ=(sinθ+cosθ)/sinθcosθ=2/sin(2θ)又0<θ<π/2,则0<2θ<πsin(2θ)在此区间内恒为正所以当sin(2θ)为最大值时,tanθ+cotθ可取到最小值在0<2θ<π区间内,sin(2θ)可取到最大值1所以tanθ+cotθ的最小值为2/1=2

f(x)=1/sinθ+2/cosθ=(sinθ+cosθ)/sinθ +2(sinθ+cosθ)/cosθ=1+cotθ+2+2tanθ≥3+2√【(cotθ)*2tanθ】=3+2√2当且仅当(cotθ)=2tanθ,tanθ=(1/2)^(1/4),即θ=arctan2^(-1/4)时,取等号.故最小值为3+2√2.

f(θ)=(sin2θ+2)^2/sin2θ = sin2θ+4+4/sin2θ sin2θ=t(0<t<=1);g(t)=t+4/t+4;g'(t)=1-4/t^2; -1=<t<=1 所以 t^2<=1 所以 4/t^2>4所以g'(t)=1-4/t^2<0所以g(t)=t+4/t+4;在(0,1]上是减函数所以最小值为 t=1时g(t)min=1+4+4=9f(θ)min= 9 此时 sin2θ=1 所以 θ=π/4

f(θ)=2sin^2(π/4+θ)-cos(2θ+π/6) =-cos(π/2+2θ)-cos(2θ+π/6)+1 =sin2θ-(√3/2)cos2θ+(1/2)sin2θ+1 =√3[(√3/2)sin2θ-(1/2)cos2θ]+1 =√3sin(2θ-π/6)+1π/4&lt;=θ&lt;=π/2,则π/3&lt;=2θ-π/6&lt;=5π/6,1/2&lt;=sin(2θ-π/6)&lt;=1.当2θ-π/6=π/2,即θ=π/3时,f(θ)=√3sin(2θ-π/6)+1取得最大值f(π/3)=√3+1.

0<a<π/2 0<2a<π 0<sin2a<=1f(a)=(sin2a+2)^2/sin2a 令 sin2a=t y=(t+2)^2/t=(t^2+4t+4)/t=t+4/t+4 函数在(0,2)上是减函数,所以t=1 ymin=9sin2a=1 2a=π/2a=π/4

解:1.f(θ)=-1/2+(sin5θ/2)/(2sinθ/2)=-1/2+[sin(2θ)cos(θ/2)+cos(2θ)sin(θ/2)]/[2sin(θ/2)] =-1/2+cos(2θ)/2+[2sinθcosθcos(θ/2)]/[2sin(θ/2)] =-1/2+cos(2θ)/2+2cosθ[cos(θ/2)]^2=-1/2+[2(cosθ)^2-1]/2+cosθ(cosθ+1) =2(cosθ)^2+cosθ-1 (0&lt;θ&lt;π)2.y=f(θ)=

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