cos(2x-π/3)=cos2xcosπ/3+sin2xsinπ/3=cos2x×1/2+sin2x×√3/2
2sin²x-1=-cos2x (兩倍角公式)
∴f(x)=cos2x×1/2+sin2x×√3/2-cos2x
=-1/2cos2x+√3/2sin2x
=sin(-π/6)cos2x+cos(-π/6)sin2x
=sin(2x-π/6)
①T=2π/2=π
②2x-π/6=π/2+kπ 解得 x=π/3+kπ/2
對(duì)稱(chēng)軸方程為 x=π/3+kπ/2 k∈z
③x∈[-π/12,π/12] 則 2x-π/6∈[-π/3,0] sin(2x-π/6)∈[-√3/2,0]
f(x)值域?yàn)閇-√3/2,0]