后面的sin應(yīng)該有平方
(1)
f (x )=√3sin（2X－π/6）+2sin²(x-π/12)
= √3sin（2X－π/6）+1-cos(2x-π/6)
=3sin（2X－π/6）-cos(2x-π/6) +1
=2[√3/2sin（2X－π/6）-1/2cos(2x-π/6) ]+1
=2sin(2x-π/6-π/6)+1
=2sin(2x-π/3)+1
f(x)最小正周期T=2π/2=π
(2)
函數(shù)f (x )取得最大值為2+1=3
由2x-π/3=2kπ+π/2,k∈Z
得 x=kπ+5π/12,k∈Z
f (x )取得最大值的X的集合為
｛x| x=kπ+5π/12,k∈Z}用的是輔助角公式asinα+bcosα=√(a²+b²)[a/√(a²+b²)*sinα+b/√(a²+b²)*cosα]=√(a²+b²)*sin(α+φ)本題中α=2x-π/6提取√[(√3)²+1²]=2