求（2+sinA）/(2+cosA)的最大值?

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若60度<=∠A<=120度，求(2+SinA)/(2+CosA)的最大值?Sol∠A=2∠B30度<=∠B<=60度SinA=Sin(2B)=2SinBCosB=2SinBCosB/(Sin^2 A+Cos^2 A)=2TanB/(1+Tan^2 B)CosA=Cos(2B)
=Cos^2 B－Sin^2 B
=(1－Tan ^2B)/(1+Tan^2 B)
p=(2+SinA)/(2+CosA)
=[2+2TanB/(1+Tan^2 B)]/[2+(1－Tan ^2 B)/(1+Tan^2 B)]
=(2+2Tan^2 B+2TanB)/(2+2Tan^2 B+1－Tan^2 B)
=(2Tan^2 B+2TanB+2)/(Tan^2 B+3)
x=TanB
√3/3<=x<=√3
y=(2x^2+2x+2)/(x^2+3)
y(x^2+3)=2x^2+2x+2
y’(x^2+3)+2xy=4x+2
y’(x^2+3)+2x(2x^2+2x+2)/(x^2+3)=4x+2
y’(x^2+3)=4x-(4x^3+4x+4)/(x^2+3)
y’(x^2+3)=(4x^3+12x-4x^3-4x-4)/(x^2+3)
y’(x^2+3)=(8x－4)/(x^2+3)
y’=(8x－4)/(x^2+3)^2
Set y’=0
x=1/2
1/2<√3/3
當x=√3/3
(2*1/3+2*√3/3+2)/(1/3+3)
=(2+2*√3+6)/(1+9)
=(4+√3)/5
當x=√3
(2*3+2*√3+2)/(3+3)
=(4+√3)/3
最大值=(4+√3)/3

考慮切線斜率即可得.

若 60° ≤ A ≤ 120°，30° < A/2 < 60°。

令 t = tan (A/2)，
那 tan 30° < tan (A/2) < tan 60°
即 1/√3 < t < √3。

2015-05-10 19:06:39 補充：
使用複角公式可知
sin A
= sin[2(A/2)]
= 2 sin(A/2) cos(A/2)
= 2 sin(A/2) cos(A/2) / [cos²(A/2) + sin²(A/2)]
= {[2 sin(A/2) cos(A/2)]/cos²(A/2)} / {[cos²(A/2) + sin²(A/2)]/cos²(A/2)}
= 2 tan(A/2) / (1 + tan²(A/2))
= 2t/(1 + t²)

2015-05-10 19:06:46 補充：
cos A
= cos[2(A/2)]
= cos²(A/2) - sin²(A/2)
= [cos²(A/2) - sin²(A/2)] / [cos²(A/2) + sin²(A/2)]
= {[cos²(A/2) - sin²(A/2)]/cos²(A/2)} / {[cos²(A/2) + sin²(A/2)]/cos²(A/2)}
= [1 - tan²(A/2)] / (1 + tan²(A/2))
= (1 - t²)/(1 + t²)

2015-05-10 19:06:52 補充：
因此，
(2 + sin A)/(2 + cos A)
= [2 + 2t/(1 + t²)]/[2 + (1 - t²)/(1 + t²)]
= [2(1 + t²) + 2t]/[2(1 + t²) + (1 - t²)]
= (2t² + 2t + 2)/(t² + 3)
= 2(t² + t + 1)/(t² + 3)
= 2(t² + 3 + t - 2)/(t² + 3)
= 2[1 + (t - 2)/(t² + 3)]