Question

若θ∈(0,),a＞b＞0,求f(θ)=的最小值.

Answer 1

解法一:(化弦為切)

f(θ)=

=a²(1+tan²θ)+b²(1+)

=a²+b²+(a²tan²θ+)

≥a²+b²+2ab=(a+b)².

當(dāng)且僅當(dāng)a²tan²θ=,即tan²θ=,tanθ=時“=”成立.

故f(θ)的最小值為(a+b)².

解法二:(利用cos²θ+sin²θ=1)

(cos²θ+sin²θ)()=a²+b²+

≥a²+b²+2ab=(a+b)²,

當(dāng)且僅當(dāng),即tanθ=時“=”成立.

∴f(θ)的最小值為(a+b)².