Question

已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+cos2x (1)求函数的最小正周期

(2)将函数f(x)的图像沿向量m=(-3π/8,2)平移得到函数g(x)的图像，求函数g(x)的单调区间

Answer 1

解析：
f(x)=sin(2x+π/3)+sin(2x-π/3)+cos2x
=sin2x*cos(π/3)+cos2x*sin(π/3)+sin2x*cos(π/3)-cos2x*sin(π/3)+cos2x
=2sin2x*cos(π/3)+cos2x
=sin2x+cos2x
=√2*(sin2x *√2/2 +cos2x *√2/2)
=√2*[sin2x *cos(π/4) +cos2x *sin(π/4)]
=√2*sin(2x + π/4)
(1) 可知函数f(x)的最小正周期T=2π/2=π
(2) 将函数f(x)的图像沿向量m=(-3π/8,2)平移得到函数g(x)的图像，那么：
函数g(x)=√2*sin[2(x+ 3π/8) + π/4] +2
=√2*sin(2x+ 3π/4 + π/4) +2
=√2*sin(2x+ π) +2
=-√2*sin2x +2
则可知：
当2kπ-π/2≤2x≤2kπ+π/2即：kπ-π/4≤ x ≤kπ+π/4，k∈Z时，函数g(x)是减函数，即对应的单调减区间为[kπ-π/4，kπ+π/4]，k∈Z；
当2kπ+π/2≤2x≤2kπ+3π/2即：kπ+π/4≤ x ≤kπ+3π/4，k∈Z时，函数g(x)是增函数，即对应的单调增区间为[kπ+ π/4，kπ+ 3π/4]，k∈Z。