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/2 4)]

Answer 2

化解得f(x）=2sin(π/3+4x)
周期T=2π/4=π/2
得递减区间为（kπ/2+π/24，kπ/2+7π/24）

Answer 3

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)]