TITLE: Why $\sin(x)+\sin(\pi x)$ is not periodic?
QUESTION [6 upvotes]: Why $\sin(x)+\sin(\pi x)$ is not periodic?
There is an answer here which tries to explain it, but I somehow do not get it. 
If we assume that $T>0$ is a period of $\sin(x)+\sin(\pi x)$, then 
$$\sin(x)+\sin(\pi x)=\sin(x+T)+\sin(\pi (x +T))$$
Apparently one needs to differentiate the equation above two times to get:
$$\sin(x)+\pi^2 \sin(\pi x)=\sin(x+T)+ \pi^2 \sin(\pi (x +T))$$
and then what?

REPLY [3 votes]: Suppose $f(x)=\sin(x)+\sin(\pi x)$ is periodic with period $T$. Then its derivative $f'(x)=\cos(x)+\pi\cos(\pi x)$ has period $T$ as well and the same for $f''(x)=-\sin(x)-\pi^2\sin(\pi x)$
Thus you have, evaluating $f(0)=f(T)$ and $f''(0)=f''(T)$,
$$
\begin{cases}
\sin(T)+\sin(\pi T)=0\\[4px]
\sin(T)+\pi^2\sin(\pi T)=0
\end{cases}
$$
which entails $\sin(\pi T)=0$ and $\sin(T)=0$. Since $\pi$ is irrational, this is impossible.