已知向量a=(2sinx,√2cos(x-π/2)+1),向量b=(cosx,√2cos(x-π/2)-1),设f(x)=向量a·向量b,求f(x)最小正周期,
问题描述：
已知向量a=(2sinx,√2cos(x-π/2)+1),向量b=(cosx,√2cos(x-π/2)-1),设f(x)=向量a·向量b,求f(x)最小正周期,
答
f(x)=a·b
=2sinx*cosx+[√2cos(x-π/2)+1]*[√2cos(x-π/2)-1]
=2sinx*cosx+[-√2sinx+1]*[-√2cosx-1]
=sin2x+(√2sinx)^2-1
=sin2x-(1-2sin^2 x)
=sin2x-cos2x
=√2(sin2x*√2/2-cos2x*√2/2)
=√2sin(2x-π/4)
最小正周T=2π/w=2π/2=π再跪求f(x)的单调区间^ω^2kπ-π/2≤2x-π/4≤2kπ+π/22kπ-π/4≤2x≤2kπ+3π/4kπ-π/8≤2x≤kπ+3π/8(k∈Z)即递增区间为[kπ-π/8，kπ+3π/8 ](k∈Z)2kπ+π/2≤2x-π/4≤2kπ+3π/22kπ+3π/4≤2x≤2kπ+7π/4kπ+3π/8≤2x≤kπ+7π/8(k∈Z)即递减区间为[kπ+3π/8，kπ+7π/8](k∈Z)