Question

函数y=cos²(x－ π/12) + sin²(x＋ π/12)－1 的最小正周期是？

Answer 1

y=cos²(x-π/12) + sin²(x＋π/12)－1
=(1+cos(2x-π/6))/2+(1-cos(2x+π/6))/2－1
= cos(2x-π/6)/2-cos(2x+π/6)/2
=1/2[cos(2x-π/6)- cos(2x+π/6)]
=1/2[cos2x cosπ/6+sin2x sinπ/6-( cos2x cosπ/6-sin2x sinπ/6)]
=1/2[2sin2x sinπ/6]
= sin2x sinπ/6
=1/2 sin2x
∴函数的最小正周期是π.