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

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

Answer 1

T=π/2 将式子展开f(x)=2sin（4x+π/3）
递减区间【π/24+2kπ，7π/24+2kπ】

Answer 2

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)]