Question

已知函数f(x)=2cosxsin(x+π/3)-√3(sinx)^2+sinxcosx

（1）求函数f(x)的最小正周期；
（2）若x∈[-π/2,π/2]时，求f(x)的单调递减区间。

Answer 1

(1)
2cosxsin(x+π/3)=sin(2x+π/3)+sinπ/3,(和差化积)
-√3sin²x=( -√3/2)(1-cos2x),
sinxcosx=½sin2x.
∴f(x)=sin(2x+π/3)+sinπ/3-√3/2+√3/2cos2x+½sin2x
=sin(2x+π/3)+[√3/2cos2x+½sin2x]
=sin(2x+π/3)+sin(2x+π/3) (辅助角公式)
=2sin(2x+π/3).
(2)由2x+π/3∈[π/2+2kπ,3π/2+2kπ]得 x∈[π/12+kπ,7π/12+kπ]. (k∈Z)
又x∈[-π/2,π/2].∴x∈[-π/2,-5π/12]∪[π/12,2/π] .此即为所求递减区间.