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

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 四分之派的最大值和最小值

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

Answer 2

f(x)=sin²x-sin²(x-π/6)
= {sinx+sin(x-π/6)}{sinx-sin(x-π/6)}
= 2sin[(x+x-π/6)/2]cos[(x-x+π/6)/2] * 2cos[(x+x-π/6)/2]sin[(x-x+π/6)/2]
= 2sin[x-π/12]cos[π/12] * 2cos[x-π/12]sin[π/12]
= 2sin[x-π/12]cos[x-π/12] * 2sin[π/12]cos[π/12]
= sin(2x-π/6)*sin(π/6)
= (1/2) sin(2x-π/6)

f(x)最小正周期 T = 2π/2 = π

在【-π/3，π/4】
2x∈【-2π/3，π/2】
2x-π/6∈【-5π/6，π/3】

2x-π/6=-5π/6时，最小值 = (1/2)×sin(-5π/6) = -1/4
2x-π/6=π/6时，最大值 = (1/2)×sin(π/3) = √3/4