y=(sin^2)x是否为周期函数 求证 sin(x^2+x) 不是周期函数
\u5982\u4f55\u8bc1\u660ey=x sin^2 x \u4e0d\u662f\u5468\u671f\u51fd\u6570\uff1f\u7531\u4e09\u89d2\u51fd\u6570\u7684\u4e8c\u500d\u89d2\u516c\u5f0f\u6709\uff1a
cos2x=1\uff0d2sin²x\u3002\u6545
sin²x=(1\uff0dcos2x)/2
y=xsin²x=x/2\uff0dxcos2x/2\u3002
\u7531\u4e8ex/2\u4e0d\u662f\u5468\u671f\u51fd\u6570\uff0c\u540e\u9762\u4e00\u9879\u4e5f\u4e0d\u662f\u5468\u671f\u51fd\u6570\u3002\u56e0\u6b64\u8be5\u51fd\u6570\u4e0d\u662f\u5468\u671f\u51fd\u6570\u3002
\u5047\u8bbey= sin(x^2+x) \u662f\u5468\u671f\u51fd\u6570
\u90a3\u4e48\u5fc5\u7136\u5b58\u5728\u4e00\u4e2aT>0\uff0c\u4f7f\u5f97\u5bf9\u4efb\u610f\u7684x\uff0c\u90fd\u6709 sin(x^2+x) = sin(\uff08x+T\uff09^2+\uff08x+T\uff09)
\u90a3\u4e48\u7531\u4e8e y=sin(x^2+x) \u7684\u53ef\u5bfc\u6027\u8d28\u53ef\u4ee5\u77e5\u9053
y ' = \uff082x+1\uff09cos\uff08x^2 +x\uff09\u4e5f\u662f\u4ee5T\u4e3a\u5468\u671f\u7684\u51fd\u6570\u3002
\u90a3\u4e48y \u2019\u7684\u96f6\u70b9\u96c6\u5408\u91cc\u5fc5\u7136\u6709\u5b50\u5e8f\u5217\uff5bx\u3002+nT\uff0cn\u662f\u6574\u6570\uff5d\uff08\u5176\u4e2dx\u3002\u53ef\u4ee5\u662fx^2+x=\u03c0/2 \u4e00\u4e2a\u6839\uff09\uff0c\u4e5f\u5c31\u662f\u8be5\u5b50\u5e8f\u5217\u7684\u6bcf\u9879\u90fd\u662fx^2+x=\u03c0/2 +k\u03c0\u7684\u6839\uff0c\u7b80\u5355\u8ba1\u7b97\u4e00\u4e0b\u5f88\u5bb9\u6613\u9a8c\u8bc1\u8fd9\u663e\u7136\u662f\u4e0d\u53ef\u80fd\u7684\u3002
=[1-cos(2x)]/2
=1/2-cos(2x)/2
所以周期为π
y=xcosx不是周期函数
因为y'=cosx-xsinx
可见Y的单调性不具有周期性
y=(sinx)^2=(1-cos2x)/2
是周期函数，最小正周期是t=2π/2=π
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