求 1/[2sinθcosθ+2√3cos^2(θ)]極小值

sin(2θ+α)
=sin(2θ)cos(α)+sin(α)cos(2θ)
∵sin(2θ+α)=(1/2)*(sin2θ+√3cos2θ)
sin(2θ)cos(α)+sin(α)cos(2θ)=(sin2θ)/2+(√3cos2θ)/2
∴cos(α)=1/2 and sin(α)=√3/2
α=60°

1/[2sinθcosθ+2√3cos^2(θ)]
=1/{sin2θ+2√3[cos(2θ)+1]/2}
=1/[sin2θ+√3cos(2θ) +√3]
=1/[2sin(2θ+α) +√3]
=1/[2sin(2θ+60°) +√3]
當分母變為最大時,1/[2sinθcosθ+2√3cos^2(θ)]的ans便是極小值
∵-1≦sin(2θ+60°)≦1
∴sin(2θ+60°)=1

∴極小值=1/[2sin(2θ+60°) +√3]=1/(2x1+√3)=1/(2+√3)

2014-01-09 18:31:26 補充：
(1/2)*(sin2θ+√3cos2θ)
=1/2*sin2θ+√3/2*cos2θ
=sin2θcos60°+sin60°cos2θ
=sin(2θ+60°)
∵sin(2θ+α)=(1/2)*(sin2θ+√3cos2θ)
sin(2θ+α)=sin(2θ+60°)
α=60°

已知對任意θ,Sin(2θ+α)=(1/2)*(Sin2θ+√3Cos2θ),求α，0°≦α≦90°
Sol
Sin(2θ+α)=Sin2θCosα+Cos2θSinα
Sin2θCosα+Cos2θSinα=(1/2)*(Sin2θ+√3Cos2θ)
2Sin2θCosα+2Cos2θSinα=Sin2θ+√3Cos2θ
Sin2θ(2Cosα－1)+Cos2θ(2Sinα－√3)=0
對任意θ成立
2Cosα－1=0，2Sinα－√3=0
Cosα=1/2，Sinα=√3/2
0°<=α<=90°
α=60°
由此求 1/[2SinθCosθ+2√3Cos^2θ]極小值
2SinθCosθ+2√3Cos^2 θ
=Sin2θ+√3*(1+Cos2θ)
=Sin2θ+√3Cos2θ+√3=2*[(1/2)*Sin2θ+√3Cos2θ]+√3
=2Sin(2θ+60°)+√3
-1<=Sin(2θ+60°)<=1
-2<=2Sin(2θ+60°)<=2
-2+√3<=2Sin(2θ+60°)+√3<=2+√3
Since －2+√3<0<2+√3
1/[2Sinθcosθ+2√3Cos^2θ]極小值不存在