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

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&#