函数f(x)=2cos²(x-π/4)-1 的周期
问题描述：
函数f(x)=2cos²(x-π/4)-1 的周期
答
f(x)=2cos²(x-π/4)-1
= cos[2(x-π/4)]
= cos(π/2-2x)
= sin2x
周期=π
答
f(x)=2cos²(x-π/4)-1
=cos[2(x-π/4)]
=cos(2x-π/2)
=sin2x
所以T=2π/2=π