f(x)=2sin(x/4)cos(x/4)+√3cos(x/2).求f(x)最小正周期.令g(x)=f(x+∏/3)判断g(x)奇偶性
来源:百度知道 编辑:UC知道 时间:2024/05/24 15:07:20
(x)=2sin(x/4)cos(x/4)+√3cos(x/2)=sin(x/2)+√3cos(x/2)=2sin(x/2+pi/3)
T=4pi
g(x)=2sin(x/2+pi/2)=2cos(x/2)
偶函数
f(x)=sin(x/2)+√3cos(x/2)
=2sin(x/2+arctan√3)
=2sin(x/2+π/3)
T=2π/(1/2)=4π
g(x)=f(x+π/3)=2sin[(x+π/3)/2+π/3]
=2sin(x/2+π/2)
=2cos(x/2)
g(-x)=2cos(-x/2)=2cos(x/2)=g(x)
且定义域x属于R,关于原点对称
所以是偶函数
2sin(x/4)cos(x/4)=sin(x/2)
然后提取2,得2(1/2sin(x/2)+根号3cos(x/2))=2sin(x/2+60)
T=4π
判断奇偶就带进去就好了
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