Question

已知函数f（x）=2cos^2(x-π/6)+2sin(x-π/4)sin(x+π/4)-1 求函数的最小正周期及单调递增区间。

Answer 1

f（x）=2cos^2(x-π/6)+2sin(x-π/4)sin(x+π/4)-1
=cos(2x-π/3)-sin(2x+π/2)
=1/2*cos2x+根号3/2*sin2x+cos2x
=3/2*cos2x+根号3/2*sin2x
=根号3(cos2x-π/6)
所以函数的最小正周斯是π

函数在2kπ-π/2<=2x-π/6<=2kπ单调递增(k是任意整数)
kπ-π/6<=x<=kπ+π/12
所以函数的单调递增区间是[kπ-π/6,kπ+π/12]

Answer 2

解：（1） f(x)=2√3sin (π/4+x)+2cos x-√3 =√3[1-cos(π最小正周期：π 2x+π/6的单调递增区间是 2kπ-π/2<=2x+π/6<=2k,