Question

已知函数f(x)=sin²x-sin²(x-π/6) 求f(x)最小正周期 求f在负三分之派 四分之派

Answer 1

解：
f(x)=sin²x-sin²(x-π/6)
=½[1-cos(2x)]-½[1-cos(2x-π/3)]
=-½[cos(2x)-cos(2x-π/3)]
=-½[cos(2x)-cos(2x)cos(π/3)-sin(2x)sin(π/3)]
=-½[cos(2x)-½cos(2x)-sin(2x)sin(π/3)]
=-½[½cos(2x)-sin(2x)sin(π/3)]
=-½[cos(2x)cos(π/3)-sin(2x)sin(π/3)]
=-½cos(2x+π/3)
最小正周期T=2π/2=π
x∈[-π/3，π/4]
-π/3≤2x+π/3≤5π/6
-√3/2≤cos(2x+π/3)≤1
-½≤-½cos(2x+π/3)≤√3/4
-½≤f(x)≤√3/4
函数的最大值为√3/4，最小值为-½