问题标题:
函数f(x)=sinx•cos(x-π/2)的最小正周期是()A.π/2B.π函数f(x)=sinx•cos(x-π/2)的最小正周期是()A.π/2B.πx05C.3π/2D.2π
问题描述:
函数f(x)=sinx•cos(x-π/2)的最小正周期是()A.π/2B.π
函数f(x)=sinx•cos(x-π/2)的最小正周期是()
A.π/2B.πx05C.3π/2D.2π
刘林源回答:
f(x)=sinx•cos(x-π/2)
=sinx•(cosx.cosπ/2+sinxsinπ/2)
=sinx(0+sinx)
=sinxsinx
=1-cos²x
最小正周期是2π.
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