Question

已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1

（1）求函数的最小正周期 (2）求f（x）在区间｛-π/4，π/4｝上的最大值和最小值，求详细步骤

Answer 1

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
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

Answer 2

f(x)=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x（sin(α±β)=sinα·cosβ±cosα·sinβ ，cos(2α)==2cos^2α-1）
=sin2x+cos2x
=√2*（sin（2x+π/4)）
所以 周期是π
因，x属于［-π/4,π/4］
所以，2x+π/4属于［-π/4，3π/4］
所以，√2*sin(2x+π/4)属于［-1，√2］