Question

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

Answer 1

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)=<x<=(kπ/2)+(7π/24)
递减区间是[(kπ/2)+(π/24)，(kπ/2)+(7π/24)]

Answer 2

sin(PI/3+4X)=cos((PI/2)-(PI/3+4X))=cos(-4X+PI/6)=cos(4X-PI/6);在自己算！