Question

三角函数求过程

设函数f(x)=2cosx(sinx-cosx)，求最小正周期与单调递减区间

Answer 1

解：f(x)=2cosx(sinx-cosx)
=2cosxsinx-2cosxcosx(利用二倍角公式化简）
=sin2x-(1+cos2x)
=sin2x-cos2x-1
=根号2sin(2x-π/4）-1
T=2π/2=π
当2kπ+π/2小于等于2x-π/4小于等于2kπ+3π/2
解得单调递减区间是：x属于【kπ+3π/8，kπ+7π/8】

Answer 2

2cosx(sinx-cosx)+1
=sin2x-2(cos2x)^2+1=√2sin(2x-π/4)
则f(x)=2cosx(sinx-cosx)=√2sin(2x-π/4) -1
则2kπ-2/π<2x-π/4<2kπ+2/π
kπ-π/8<x<kπ+3π/8增
kπ+3π/8<x<kπ+7π/8减
周期是 T=2π/2=π