Question

已知fx=cos(2x-π/3)+2sin（x-π/4）xsin（x+π/4） 1求fx的最小正周期及单调递增区间;

2：求函数fx在区间[-π/12,π/2]上的值域：

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Answer 1

f(x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)
=cos(2x-π/3)+sin(2x-π/2)
=cos(2x-π/3)-cos2x
=2sin(π/6)sin(2x-π/6)
=sin(2x-π/6)

1. 最小正周期T=2π/2=π
单增区间2x-π/6∈[2kπ-π/2, 2kπ+π/2]
x∈[kπ-π/6, kπ+π/3]
2. x∈[-π/12,π/2]
2x-π/6∈[-π/3,5π/6]
sin(2x-π/6)∈[-√3/2, 1]
∴值域为f(x)∈[-√3/2，1]
希望能帮到你，祝学习进步O(∩_∩)O

Answer 2

sin(x+π/4)=sin[(x-π/4)+π/2]=cos(x-π/4)
∴f(x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)
=cos(2x-π/3)+sin(2x-π/2)
=cos(2x-π/3)-cos2x
=(1/2)cos2x + (√3/2)sin2x-cos2x
= (√3/2)sin2x-(1/2)cos2x
=sin(2x-π/6)

1.
T=2π/2=1
-π/2+2kπ≤2x-π/6≤π/2+2kπ
∴-π/6+kπ≤x≤π/3+kπ
∴单调递增区间为[-π/6+kπ，π/3+kπ]

2.
x∈[-π/12,π/2]
2x∈[-π/6,π]
2x-π/6∈[-π/3,5π/6]
∴f(x)∈[-√3/2，1]
即：值域为[-√3/2，1]