求函数f(x)=sin(π/3+4x)+cos(4x-π/6)的最小正周期和递减区间

题目

求函数f(x)=sin(π/3+4x)+cos(4x-π/6)的最小正周期和递减区间

答案

sin(π/3+4x)=cos[π/2-(π/3+4x)]=cos(π/6-4x)
f(x)=sin(π/3+4x)+cos(4x-π/6)=2cos(4x-π/6)
最小正周期是：T=2π/4=π/2
2kπ=<4x-π/6<=2kπ+π
2kπ+π/6=<4x<=2kπ+7π/6
(kπ/2)+(π/24)=递减区间是[(kπ/2)+(π/24),(kπ/2)+(7π/24)]