Question

高中数学，第四题…

Answer 1

y=cos²(x-π/12)+sin²(x+π/12)-1
=½[1+cos(2x-π/6)]+½[1-cos(2x+π/6)]-1
=½[cos(2x-π/6)-cos(2x+π/6)]
=-½ * 2 sin[(2x-π/6+2x+π/6)/2] sin[(2x-π/6-2x-π/6)/2]
=sin2x sin(π/6)
=½sin2x
=½sin(2x+2kπ)
=½sin2(x+kπ) k为整数
最小正周期为π

Answer 2

选C，很简单嘛，用余弦的二倍角公式先化简，然后最后得y=1/2*sin(2x),所以周期为π