作业帮已知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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![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：](/uploads/image/z/7629207-15-7.jpg?t=%E5%B7%B2%E7%9F%A5fx%3Dcos%282x-%CF%80%2F3%29%2B2sin%EF%BC%88x-%CF%80%2F4%EF%BC%89xsin%EF%BC%88x%2B%CF%80%2F4%EF%BC%89+1%E6%B1%82fx%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%8F%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%3B2%EF%BC%9A%E6%B1%82%E5%87%BD%E6%95%B0fx%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-%CF%80%2F12%2C%CF%80%2F2%5D%E4%B8%8A%E7%9A%84%E5%80%BC%E5%9F%9F%EF%BC%9A)
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：
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：
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：
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]
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<...
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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]
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