f(x)=2sin^2+sin2x-1
=2sin²x-1+sin2x
=-cos2x+sin2x
=√2sin(2x-π/4)
函数的最小正周期为 2π/2=π
函数的最大值为 √2 最小值为-√2
最小值时
2x-π/4=2kπ-π/2
x=kπ-π/8
取得最小值x的集合为{x|x=kπ-π/8,k∈Z}
最大值时
2x-π/4=2kπ+π/2
x=kπ+3π/8
取得最大值时x的集合为{x|x=kπ+3π/8,k∈Z}=-cos2x+sin2x =√2sin(2x-π/4)这步怎么算的.?=-cos2x+sin2x=sin2x-cos2x=√2[(√2/2)sin2x-(√2/2)cos2x]因为 cos(π/4)=√2/2 sin(π/4)=√2/2=√2sin(2x-π/4)