Question

设函数f（x）＝2cosx（sinx－cosx）（1）求f（x）的最小正周期。（2）求f（x）的

单

Answer 1

解：f(x)=2cosx(sinx－cosx)
=2sinxcosx-2cos²x
=sin2x-cos2x-1
=√2sin(2x-π/4)-1
1)T=2π/2=π
2)2kπ-π/2<2x-π/4<2kπ+π/2
kπ-π/8<x<kπ+3π/8
单增区间：(kπ-π/8,kπ+3π/8)(k∈Z)
单减区间：(kπ+3π/8,kπ+7π/8)(k∈Z)