求函数f(x)=cos^2(x-pai/12)+sin^2(x+pai/12)-1的周期

cos²x=(1+cos2x)/2 sin²x=(1-cos2x)/2 所以f(x)=[1+cos(2x-π/6)]/2+[1-cos(2x+π/6)]/2-1 =1/2[cos(2x-π/6)-cos(2x+π/6)] =1/2[(cos2xcosπ/6+sin2xsinπ/6)-(cos2xcosπ/6-sin2xsinπ/6)] =sin2xsinπ/