2倍角公式
f(x)=cos^2(x-π/12)+sin^2(x+π/12)-1
=[cos(2x-π/6)+1]/2+[1-cos(2x+π/6)]/2-1
=1/2[cos(2x-π/6)-cos(2x+π/6)]
=1/2(√3/2cos2x+1/2sin2x-√3/2cos2x+1/2sin2x)
=1/2sin2x
周期是2π/2=π
f(x)=-f(-x)奇函数
2）
f（x）=1/2 sin2x
在（0,π/4）单调增,在（π/4,2π/3）单调减
f(0)=0
f(π/4)=1/2
f(2π/3)=1/2sin4π/3=-1/2sinπ/3=-根号3/4
最大值1/2,最小值-根号3/4