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已知向量a=(2sinx,√2cos(x-π/2)+1),向量b=(cosx,√2cos(x-π/2)-1),设f(x)

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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,
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
再问: 再跪求f(x)的单调区间^ω^
再答: 2kπ-π/2≤2x-π/4≤2kπ+π/2 2kπ-π/4≤2x≤2kπ+3π/4 kπ-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/2 2kπ+3π/4≤2x≤2kπ+7π/4 kπ+3π/8≤2x≤kπ+7π/8 (k∈Z) 即递减区间为[kπ+3π/8,kπ+7π/8 ] (k∈Z)