f（x）=sin（2x-π/4）-2√2sin²x的最小正周期

题目

f（x）=sin（2x-π/4）-2√2sin²x的最小正周期
注：2√2sin²x,sin²x不在根号下

答案

f(x) = sin(2x-π/4) - 2√2sin^2x= sin2xcosπ/4 - cos2xsinπ/4 - √2*(2sin^2x)= √2/2 sin2x - √2/2 cos2x - √2*(1-cos2x)= √2/2 sin2x - √2/2 cos2x - √2 + √2cos2x= √2/2 sin2x + √2/2 cos2x - √2= s...