Question

函数y=sin(π/2x)cos[π/2(x-1)]的最小正周期是

Answer 1

y=sin(π/2x)cos[π/2(x-1)]
=sin(π/2x)cos(π/2x-π/2)
=sin(π/2x)cos(π/2-π/2x)
=sin(π/2x)*sin(π/2x)
=[sin(π/2x)]^2=(1/2)(1-cosπx)
所以函数y=sin(π/2x)cos[π/2(x-1)]的最小正周期是：2.

Answer 2

先化简：cos[π/2(x-1)]=cos(π/2x-π/2)=sin(π/2x)(诱导公式), 则y=sin(π/2x)cos[π/2(x-1)]=[sin(π/2x)]^2=1/2×[1-cos(πx)](逆用二倍角公式),∵y=1/2cos(πx)的最小正周期为T=2，而y=1/2×[1-cos(πx)]的图像仅仅由y=1/2cos(πx)的图像先向下平移1/2个单位、再做关于x轴的对称变换得到，因此所求函数的最小正周期还是T=2.

Answer 3

y=sin((π/2)x)*cos((π/2)(x-1))=sin((π/2)x)*cos((π/2)x-π/2)=sin((π/2)x)*cos(π/2-(π/2)x)=sin((π/2)x)*sin((π/2)x)=sin²（πx/2)=（1-cosπx）/2，所以ω=π，则周期T=2π/ω=2，最小正周期为2。