Question

已知函数f(x)=2sin x/4 cos x/4 -2√3sin²x/4 +√3 求最小正周期及

已知函数f(x)=2sin x/4 cos x/4 -2√3sin²x/4 +√3
求最小正周期及最值
令g(x)=f(x+π/3)判断奇偶性说明理由

Answer 1

f(x)=2sin x/4 cos x/4 -2√3sin²x/4 +√3
= 2sin x/4 cos x/4 +√3(1-2sin²x/4)
= sinx/2+√3cosx/2
= 2sinx/2cosπ/3+cosx/2sinπ/3
= 2sin(x/2+π/3)
最小正周期T=2π/(1/2)=4π
最小值=-2；最大值=2

g(x) = f(x+π/3)
= 2sin[(x+π/3)/2+π/3]
= 2sin(x/2+π/2)
= 2cos(x/2)
偶函数

Answer 2

f(x)
=2sin(x/4).cos(x/4) -2√3[sin(x/4)]^2 +√3
=sin(x/2) -√3( 1- cos(x/2) ) + √3
=sin(x/2) +√3cos(x/2)
=2sin(x/2 + π/3)
最小正周期=4π
最大值=2
最小值=-2

g(x)
=f(x+π/3)
=2sin( (x+π/3) /2 + π/3)
=2sin( x/2 + π/2)
=2cos(x/2)
g(-x)=g(x)
g偶函数