Question

已知函数f（x）=sinxsin（x+兀/6）

已知函数f（x）=sinxsin（x+兀/6）求最小正周期
2、当x属于0到兀/2时。函数范围？
|

Answer 1

1、
f(x)=sinx(sinxcosπ/6+cosxsinπ/6)
=√3/2*sin²x+1/2*sinxcosx
=√3/2*(1-cos2x)/2+1/2*(sin2x)/2
=√3/4+1/2*(1/2*sin2x-√3/2*cos2x)
=√3/4+(sin2xcosπ/3-cos2xsinπ/3)
=√3/4+sin(2x-π/3)
所以T=2π/2=π

2、
0≤x≤π/2
-π/3≤2x-π/3≤2π/3
-√3/2≤sin(2x-π/3)≤1
-√3/4≤√3/4+sin(2x-π/3)≤√3/4+1
所以f(x)∈[-√3/4，√3/4+1]

Answer 2

(1)
f(x) = sinx. sin(x+π/6)
=sinx .[ (√3/2)sinx + (1/2)cosx ]
=(√3/2)(sinx)^2 + (1/2)sinx.cosx
= (√3/2)(1-cos2x)/2 + (1/4)sin2x
= (1/4)sin2x - (√3/4)cos2x +√3/4
= (1/2)sin(2x-π/3) +√3/4
最小正周期 =π
(2)
x∈[0,π/2]
max f(x)= 1/2 +√3/4
min f(x)
= f(0)
=(1/2)(-√3/2) +√3/4
=0

0≤ f(x) ≤ 1/2 +√3/4