Question

已知函数f(x)=2cosx(sinx-cos)+1,x∈R.(1)求函数f(x)的最小正周期及单调递增区间;(2)求函数f(x)在区间[π/8

已知函数f(x)=2cosx(sinx-cos)+1,x∈R.
(1)求函数f(x)的最小正周期及单调递增区间;
(2)求函数f(x)在区间[π/8,3π/4]上的最小值和最大值.

Answer 1

f(x)=2cosx(sinx-cos)+1
=2sinxcosx-2cos²x+1
=sin2x-cos2x
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
所以：最小正周期=2π/2=π

π/8≤x≤3π/4
π/4≤2x≤3π/2
0≤2x-π/4≤5π/4
所以：
当2x-π/4=π/2时，函数有最大值=√2
当2x-π/4=5π/4时，函数有最小值=√2×(-√2/2)=-1

Answer 2

f(x)=2cosx(sinx-cosx)+1
=2sinxcosx-2cos²x+1
=sin2x-cos2x
=√2[(√2/2)sin2x-(√2/2)cos2x]
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
(1)所以：最小正周期T=2π/2=π

(2) π/8≤x≤3π/4
π/4≤2x≤3π/2
0≤2x-π/4≤5π/4
所以：
当2x-π/4=π/2时，函数有最大值=√2
当2x-π/4=5π/4时，函数有最小值=√2×(-√2/2)=-1

Answer 3

同意