Question

函数f(x)=根号3sin(2x-π/6）+2sin²（x-π/12），f(x)的最小正周期。

Answer 1

y=√3sin(2x-π/6)+2(sin(x-π/12))^2=√3sin(2x-π/6)+1-cos(2x-π/6)
=1+√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)
所以该函数的周期T=2π/2=π.

Answer 2

由cos2α=1-2sin²α，把2sin^2(x-π/12）表示成1-cos2(x-π/12)=1-cos(2x-π/6)，这样f(x)=根号3sin(2x-6/π)-cos(2x-π/6）+1.然后用辅助角公式把sinα和cosα结合起来就可以了。f(x)=2cos2x+1.所以最小正周期是π