Question

已知函数f(x)=4sin²x+2sin2x-2,x∈R

（1）求f(x)的最小正周期及f(x)取得最大值时的集合;
(2)求证：函数f(x)的图像关于直线x=－π/8对称.

Answer 1

f(x)=4sin²x+2sin2x-2
=2(1-cos2x)+2sin2x-2
=2(sin2x-cos2x)
=2根号2(sin2xcosπ/2-cos2xsinπ/2)
=2根号2sin(2x-π/4)

最小正周期：2π/2=π
(2x+π/4)=2kπ-π/2，k∈Z 时，sin(2x-π/4)=1，f(x)有最大值2根号2
此时x=kπ-3π/8，k∈Z

f(-π/8-x)=2根号2sin[2(-π/8-x)-π/4]=2根号2sin[-2x-π/2]=2根号2sin[π-(-2x-π/2)]=2根号2sin[2x+3π/2]=2根号2sin[(2x+3π/2)-2π]=2根号2sin[2x-π/2]
f(-π/8+x)=2根号2sin[2(-π/8+x)-π/4]=2根号2sin[2x-π/2]
f(-π/8-x) = f(-π/8+x)，得证