已知f(x）=cosx^²(x-π/12)+sin^²(x+π/12)-1

题目

已知f(x）=cosx^²(x-π/12)+sin^²(x+π/12)-1
1,求f（x)的最小正周期
2若x∈[0,2π/3]求f(x)的最值

答案

f(x)=cos²(x-π/12)+sin²(x+π/12)-1
=[1+cos(2x-π/6)]/2 +[1-cos(2x+π/6)]/2 -1
=[2+cos(2x-π/6)-cos(2x+π/6)]/2 -1
=[2-2sin(2x)sin(-π/6)]/2 -1
=(1/2)sin(2x)
所以周期是T=2π/2=π
若x∈[0,2π/3]
则2x∈[0,4π/3]
sin(2x)∈[-√3/2,1]
∴f(x)∈[-√3/4,1/2]
最大值是1/2,最小值是-√3/4