f(X)=sin(2x-π/4)-2*2^1/2*sin^2(x)的最小正周期是

问题描述:

f(X)=sin(2x-π/4)-2*2^1/2*sin^2(x)的最小正周期是

f(x)=sin(2x-π/4)-2√2sin²x
=sin2xcos(π/4)-cos2xsin(π/4)-√2(1-cos2x)
=sin2xcos(π/4)+cos2xsin(π/4)-√2
=sin(2x+π/4)-√2
所以 最小正周期是T=2π/2=π第二步是怎么来的??1.降次公式:2sin²x=1-cos2x2.sin2xcos(π/4)-cos2xsin(π/4)-√2(1-cos2x)=sin2xcos(π/4)-(√2/2)cos2x +√2cos2x - √2=sin2xcos(π/4)+cos2xsin(π/4)-√2