(2008江西文数)函数f(x)=sinx/(sinx+2sin(x/2))为什么周期是4pi,偶函数?
人气:237 ℃ 时间:2020-06-12 05:05:35
解答
f(x)=sinx/[sinx+2sin(x/2)]
f(-x)=sin(-x)/[sin(-x)+2sin(-x/2)]
=-sinx/[-(sinx+2sin(x/2))]
=sinx/[sinx+2sin(x/2)]
所以是偶函数
f(x)=sinx/(sinx+2sin(x/2))
=(2sin(x/2)cos(x/2))/[2sin(x/2)cos(x/2)+2sin(x/2)]
=cos(x/2)/(1+cos(x/2))
=1-1/(1+cos(x/2))
而cos(x/2)周期2π/(1/2)=4π
所以f(x)周期4π
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