Question

数学问题，求大神解答。。。。。。。

已知函数f(x)=cos(2x-π/3)+2sin(x-π/4sin(x+π/4）(1)求函数f(x)的最小正周期 和图像的对称轴方程(2)求函数f(x)在区间[一π/12，π/2]上的值域。

Answer 1

解：(1)f(x)＝cos(2x－π/3)＋2sin(x－π/4)sin(x＋π/4)＝1/2cos2x＋√3/2sin2x＋(sinx－cosx)(sinx＋cosx)＝1/2cos2x＋√3/2sin2x＋sin²x－cos²x＝1/2cos2x＋√3/2sin2x－cos2x＝sin(2x－π/6)∴最小正周期：T＝2π/2＝π由2x－π/6＝kπ(k∈Z)得x＝π/12+kπ/2(k∈Z)∴函数图象的对称中心为：(π/12+kπ/2,0)(k∈Z)(2)∵x∈[－π/12，π/2]∴2x－π/6∈[－π/3，5π/6]∵f(x)＝sin(2x－π/6)在区间[－π/12，π/3]上单调递增，在区间[π/3，π/2]上单调递减∴x＝π/3时，f(x)取最大值1又∵f(－π/12)＝－√3/2＜f(π/2)＝1/2当x＝－π/12时，f(x)取最小值－√3/2∴函数f(x)在区间[－π/12，π/12]上的值域是：[－√3/2，1]

Answer 2

1. f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4）
       =cos(2x-π/3)+cos[(x-π/4)-(x+π/4)]-cos[(x-π/4)+(x+π/4)]
       =cos(2x-π/3)-cos2x+cos(-π/2)
       =cos(2x-π/3)-cos2x
       =-2sin(2x-π/6)sin(-π/6)
       =sin(2x-π/6)
   故最小正周期为π
   令2x-π/6=kπ（k∈Z），解得：x=(k/2+1/12)π
   这就是图像的对称轴方程

2. 
   .由x∈[-π/12,π/2]得2x-π/6∈[-π/3,5π/6]
   由π/2∈[-π/3,5π/6]
   函数最大值为1
   最小值为sin(-π/3)=-√3/2

   所以函数的值域为[-√3/2,1]

Answer 3

（1）
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4）
　 =cos(2x-π/3)+cos[(x-π/4)-(x+π/4)]-cos[(x-π/4)+(x+π/4)]
=cos(2x-π/3)-cos2x+cos(-π/2)
=cos(2x-π/3)-cos2x
=-2sin(2x-π/6)sin(-π/6)
=sin(2x-π/6)
最小正周期T=2π/2=π
令2x-π/6=kπ（k∈Z）
　　x=(k/2+1/12)π
　　对称轴方程：x=(k/2+1/12)π
（2）
　　∵x∈[-π/12,π/2]
　　∴2x-π/6∈[-π/3,5π/6]
∵π/2∈[-π/3,5π/6]
∴函数最大值为sinπ/2=1
∴最小值为sin(-π/3)=-根号3 /2
　　∴函数的值域为[-根号3 /2,1]