Question

函数y=cos²(x-π/12)+sin²(x+π/12)-1的最小正周期

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
以上回答你满意么？

Answer 2

y=cos²(x-π/12)+sin²(x+π/12)-1
=1/2+1/2cos(2x-π/6)+1/2-1/2cos(2x+π/6)-1
=1/2[cos(2x)cos(π/6)+sin(2x)sin(π/6)]-1/2[cos(2x)cos(π/6)-sin(2x)sin(π/6)]
=1/2sin(2x)
T=2π/2=π