Question

函数f(x)=-√2(sin2x+π/4)+6 sin x cos x-2cos²x+1

1）求f(x)的最小正周期；2)求f(x)在区间[0，π/2]上的最大值和最小值

Answer 1

解：f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1
=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1
=-√2(√2/2·sin2x+√2/2·cos2x)+3sin2x-cos2x
=-sin2x-cos2x+3sin2x-cos2x
=2sin2x-2cos2x
=√[2²+(-2)²]sin(2x-π/4)
=2√2sin(2x-π/4)
1)最小正周期=2π/2=π
2)-π/2+2kπ≤2x-π/4≤π/2+2kπ k属于Z
-π/8+kπ≤x≤3π/8+kπ
递增区间[-π/8+kπ，3π/8+kπ]
∴f(x)在区间[0,π/2]上最大值为f(3π/8)=2√2,最小值为f(0)=-2