已知函数f(x)=2sin(x+5π/24)cos(x+5π/24)-2cos²(x+5π/24)+1

问题描述:

已知函数f(x)=2sin(x+5π/24)cos(x+5π/24)-2cos²(x+5π/24)+1
(1)求f(x)的最小正周期(2)求函数f(x)的单调递增区间

f(x)=sin(2x+5π/12)-cos(2x+5π/12)
=√2sin(2x+5π/12-π/4)
=√2sin(2x+π/6)
所以,最小正周期T=2π/2=π
递增区间:
-π/2+2kπ