Question

已知f(x)=sin²wx+(根号3/2)sin2wx-(1/2)(x∈R,w>0)若f(x)的最小正周期为2π

已知f(x)=sin²wx+(根号3/2)sin2wx-(1/2)(x∈R,w>0)若f(x)的最小正周期为2π (1)求f(x)的表达式和f(x)的单调递增区间（2）求f(x)在区间【-（π/6），（5π/6）】的最大值和最小值

Answer 1

（1）
f(x)=sin²wx+(根号3/2)sin2wx-(1/2)
=1/2(1-cos2wx)+√3/2sin2wx-1/2
=√3/2sin2wx-1/2cos2wx
=sin(2wx-π/6)
∵f(x)的最小正周期为2π
∴2π/(2w)=2π
∴w=1/2
∴f(x)=sin(x-π/6)
由2kπ-π/2≤x-π/6≤2kπ+π/2,k∈Z
得2kπ-π/3≤x≤2kπ+2π/3,k∈Z
∴f(x)的单调递增区间是
[2kπ-π/3,2kπ+2π/3],k∈Z
(2)
∵x∈【-（π/6），（5π/6）】
∴x-π/6∈[-π/3,2π/3]
∴x-π/6=-π/3,x-π/6时，f(x)取得最小值-√3/2
x-π/6=π/2，x=2π/3时，f(x)取得最大值1

Answer 2

f(x)=sin²wx+(√3/2)sin2wx-(1/2)=(1-cos2wx)/2+(√3/2)sin2wx-(1/2)
=+(√3/2)sin2wx-(1/2)cos2wx=sin(2wx-π/3)
T=2π/2W=2π, W=1/2
f(x)=sin(x-π/3)
递增区间 2kπ-π/2<x-π/3<2kπ+π/2 , 2kπ-π/6<x<2kπ+5π/6
递减区间；2kπ+5π/6<x<2kπ+11π/6
f(5π/6)=1最大值
f(-π/6)=-1最小值

追问

你的递增区间还有最小值都错了诶~

追答

f(x)=sin(x-π/3)这个错了，使后面都错了应该是f(x)=sin(x-π/6)

Answer 3

=(√3/2)sin2wx-(1/2)cos2wx+1 =sin(2wx-π/6)+1 已知最小正周期为TT 则T=2π/2w=π w=1 所以f（x）的解析式为 f(x)=sin(2x-π/6