函数f(x)=sin²(x+π/4)+cos²(x-π/4)-1,f(x)是周期为多少?
f(x)=sin²(x+π/4)-[1-cos²(x-π/4)]=sin²(x+π/4)-sin²(x-π/4)
=[sin(x+π/4)+sin(x-π/4)][sin(x+π/4)-sin(x-π/4)]
=[2sinxcos(π/4)][2cosxsin(π/4)]
=4(√2/2)²sinxcosx=2sinxcosx=sin2x
故f(x)是一个最小正周期T=π的周期函数.
函数f(x)=sin²(x+π/4)+cos²(x-π/4)-1,f(x)是周期为多少?
f(x)=sin²(x+π/4)-[1-cos²(x-π/4)]=sin²(x+π/4)-sin²(x-π/4)
=[sin(x+π/4)+sin(x-π/4)][sin(x+π/4)-sin(x-π/4)]
=[2sinxcos(π/4)][2cosxsin(π/4)]
=4(√2/2)²sinxcosx=2sinxcosx=sin2x
故f(x)是一个最小正周期T=π的周期函数.