假设(xcos\,x)有周期,依据周期函数的规律,可得
[egin{aligned}
xcos\,x & = (x+T)cos\,(x+T) \
& = (x+T)cos\,xcos\,T - sin\,xsin\,T \
& = xcos\,xcos\,T - xsin\,xsin\,T + Tcos\,xcos\,T - Tsin\,xsin\,T \
end{aligned}
]
上式需要成立,则(cos\,T = 1并且Tcos\,xcos\,T-Tsin\,xsin\,T-xsin\,xsin\,T=0)
假设(cos\,T=1)成立,则(sin\,T=1-cos^2\,T=0),则(Tcos\,xcos\,T-Tsin\,xsin\,T-xsin\,xsin\,T=Tcos\,x=0)
(Tcos\,x=0)发现只有(T=0)时,(Tcos\,x=0)条件才成立,因此(xcos\,x)函数没有周期