y=cos^2(x-π/12)+sin^2(x+π/12)-1 =cos^2(x-π/12)+sin^2(x+π/12)-[sin^2(x+π/12)+cos^2(x+π/12)] =cos^2(x-π/12)-cos^2(x+π/12) =[cos(x-π/12)-cos(x+π/12)][cos(x-π/12)+cos(x+π/12)] =[sinxsin(π/12)][cosxcos(π/12)] =(sinxcosx)[sin(π/12)cos(π/12)] =(1/4)sin2xsin(π/6) =(1/8)sin2x 所以：T=2π/2=π,是奇函数。
问：求函数y=sin(π/3-2x)+sin2x的最小正周期
答：sin(π/3-2x)+sin2x =sin(π/3+2x)所以最小正周期为π详情>>
答：详情>>
