证明$f(x)=sinx^2$不是周期函数.
反证:假设是周期函数，周期为$T,T>0$.
$$f(0)=f(T)\Rightarrow sinT^2=0\Rightarrow T^2=k_1\pi,k_1\in N^{*}$$
$$f(\sqrt{2}T)=f(\sqrt{2}T+T)\Rightarrow sin2T^2=sin(\sqrt{2}T+T)^2$$
$$\Rightarrow 0=sin2k_1\pi=sin(\sqrt{2}T+T)^2$$
$$\Rightarrow(\sqrt{2}T+T)^2=k_2\pi,k_2\in N^{*}$$
$$\Rightarrow (\sqrt{2}+1)^2=\frac{k_2}{k_1}$$
等式左边为无理数$\ne$等式右边为有理数，矛盾，故假设不成立。$\therefore f(x)=sinx^2$不是周期函数.
评：此类证明非周期的题，套路基本都是反证，取一些特殊值，得出矛盾.