更新时间2021-06-15 07:02:15
那个TT是派
f(x) = cos( x - π/6 ) - sinx
= cosxcos(π/6) + sinxsin(π/6) - sinx
= cosxcos(π/6) - (1/2)sinx
= cosxcos(π/6) - sinxsin(π/6)
= cos( x - π/6 )；
T = 2π；
f(x)最小点，x - π/6 = -π，x = -5π/6 ；
f(x)最大点，x = π - 5π/6 = π/6；
故 f(x) 单调增区间为 ( 2kπ - 5π/6，2kπ + π/6 )，k是整数 。
f(x)=cos(x-π/6)-sinx
=cos(x-π/6)-cos(π/2-x)
=-2sin[(x-π/6)/2+(π/2-x)/2]sin[(x-π/6)/2-(π/2-x)/2]
=-2sin(π/6)sin(x-π/3)
=-sin(x-π/3)
f(x)的单调递增区间是sin(x-π/3)的单调递减区间
2πk+π/2<x-π/3<2πk+3π/2【k是整数】
2πk+5π/6<x<2πk+11π/6
爱问答