Question

设函数f(x)=cos(x+2/3π)+2cos^2(x/2),x属于R,记三角形内角ABC对边为abc,若f(B)=1

Answer 1

1) f(x)=cos(x+2π/3)+2cos²(x/2)=-(cosx)/2-(√3sinx)/2+1+cosx=1-[(√3sinx)/2-(cosx)/2]=1-[sin(x-π/6)]/2, ∴ 1/2≤f(x)≤3/2, 值域[1/2,3/2].由2kπ-π/2≤x-π/6≤2kπ+π/2,得2kπ-π/3≤x≤2kπ+2π/3,由2kπ+π/2≤x-π/6≤2kπ+3π/2,得2kπ+π/3≤x≤2kπ+5π/3, ∴ 减区间[2kπ-π/3,2kπ+2π/3];增区间[2kπ+π/32,kπ+5π/3]

(k∈Z)

(2) 利用你的结论:a²+c²-√3ac=1---->1≥(2-√3)ac---->

ac≤2+√3, ∵ S=0.5acsin30°=√3ac/4, ∴ 4S/√3≤2+√3,

∴ 0