Question

已知函数f(x)=sin^2 x+2根号3sinxcosx+sin(x+π/4)sin(x-π/4),x属于R，求f(x)的最小正周期和值域

若x=x0 0≤x0小于等于π/2为f(x)的一个零点，求sin2x0的值

Answer 1

解：f(x)=sin^2 x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)
=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)
=2-2cos2x+√3sin2x+(1/2)sin(2x-π/2)
=2+√3sin2x-3/2cos2x
=2+√3(1/2sin2x-√3/2cos2x)
=2+√3sin(2x-π/3)
由最小正周期公式得：T=2π/w=2π/2=π
f(x)(max)=2+√3
f(x)(min)=2-√3
所以f(x)的值域为【2-√3,2+√3】.

Answer 2

f(x)=(1-cos2x)/2+√3sin2x+cos(x-π/4)sin(x-π/4)
=1/2-(1/2)sin2x+√3sin2x+(1/2)sin(2x-π/2)
=1/2-(1/2)sin2x+√3sin2x - (1/2)cos2x
=√3sin2x -cos2x+1/2
=2sin(2x-π/6)+1/2
由最小正周期公式得：T=2π/w=2π/2=π
f(x)(MAX)=2+1/2=5/2
f(x)(min)=-2+1/2=-3/2
所以f(x)的值域为【-3/2,5/2】

设x0是f(x)的一个在【0，π/2】内的一个零点， ==>sin(2x-π/6)+1/2=0

sin(2x-π/6)=-1/4
2x0-π/6=arcsin(-1/4)=-arcsin1/4
2x0=π/6-arcsin1/4
sin2x0=sin[π/6-arcsin1/4]=(1/2)cos(arcsin1/4)-(√3/2)sin(arcsin1/4)
=(1/2)√[1-sin^2(arcsin1/4)]-(√3/2)(1/4)
=(√15-√3)/8

Answer 3

1、f(x)=(sinx)^2+√3sinxcosx

=(1-cos2x)/2+√3/2sin2x

=1/2+sin(2x-π/6)

其最小正周期为π

2、若x∈[-π/4，π/4]则2x-π/6∈[-2π/3，π/3]

从而sin(2x-π/6)∈[-1，√3/2]

y=f(x)的值域为：[-1/2，(1+√3)/2]