已知f（x）=Sin（2x+π⼀6）+cos（2x-π⼀3）

(1)利用公式cosα=sin(π/2+α)
cos(2x-π/3)=sin(2x-π//3+π/2)=sin(2x+π/6)代入：
f(x)=sin²(2x+π/6）=1/2-（1/2）cos（4x+π/3）；
T=2π/ω=2π/4=π/2

(2)
f（x）=Sin（2x+π/6）+cos（2x-π/3）
g（x）=f（x+π/3）
=Sin（2（x+π/3）+π/6）+cos（2（x+π/3）-π/3）
=sin(2x+5π/6)+cos(2x+π/3)
=√2sin(2x+π/6+π/4)
=√2sin(2x+5π/12) （公式AsinX+BcosX=√(A^2+B^2) sin(X+ARGTAN（B/A）
在[-π/4，π/4]值域

2x+5π/12 属于[-π/12，11π/12]
最大值√2，最小值-√2sin(π/12)
值域[-√2sin(π/12)，√2]

f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1
=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x
=2sin2xcosπ/3+cos2x
=sin2x+cos2x
=√2*(√2/2*sin2x+√2/2*cos2x)
=√2*(sin2xcosπ/4+cos2xsinπ/4)
=√2*sin(2x+π/4)
t=2π/2=π
sin函数单调性：
在[-(π/2)+2kπ,(π/2)+2kπ]，k∈z上是增函数
在[(π/2)+2kπ,(3π/2)+2kπ]，k∈z上是减函数
2x+π/4代入求
x∈[-π/4,π/4]
2x∈[-π/2,π/2]
2x+π/4∈[-π/4,3π/4]
-1<=√2*sin(2x+π/4)<=√2
f(x)在区间[-π/4，π/4]上的最大值为：√2
f(x)在区间[-π/4，π/4]上的最小值为：-1

答：
f(x)=sin(2x-π/3)-cos(2x+π/6)
=cos(π/3)sin2x-sin(π/3)cos2x-cos(π/6)cos2x+sin(π/6)sin2x
=sin2x-√3cos2x
=2*[(1/2)sin2x-(√3/2)cos2x]
=2sin(2x-π/3)
1）
图像向右平移m个单位得到：f(x)=2sin(2x-2m-π/3)
经过原点：f(0)=2sin(-2m-π/3)=0
所以：2m+π/3=kπ
所以：m=kπ/2-π/6>0
所以：k=1时，m的最小值为π/3
2）
f(x)=2sin(2x-π/3)=-8/5
所以：sin(2x-π/3)=-4/5
所以：sin(2x+π/6-π/2)=-sin[π/2-(2x+π/6)]=-cos(2x+π/6)=-4/5
所以：cos(2x+π/6)=4/5
因为：
-π/12<=x<=5π/12
-π/6<=2x<=5π/12
0<=2x+π/6<=π
所以：
0<=2x+π/6<=π/2
结合
恒等式
(cosa)^2+(sina)^2=1解得：
sin(2x+π/6)=3/5
所以：
f(x+π/4)
=2sin(2x+π/2-π/3)
=2sin(2x+π/6)
=2*(3/5)
=6/5