Is $(\sin{x})(\sin{\pi x})$ periodic? I'm wondering, is the function $f=(\sin{x})(\sin{\pi x})$ is periodic?
My first inclination would be two assume that if the periods of the individual sine expressions, $p_1 \text{and}\space p_2$ have the quality that $p_1 \times a = p_2 \times b$ where $a \space\text{and}\space b$ are integers, then the entire function will eventually repeat after a period of $p_1 \times a$. 
If that is true, than I think $f$ might not be periodic due to the fact that two Pi is irrational.
Does anyone know the answer and/or weather my thinking is correct? I've never seen a function like this before, so I'm really curious.
 A: Using the product-to-sum formula,
$$ \sin x \sin\pi x = \frac{1}{2} \left( \cos ((1-\pi)x) - \cos((1 + \pi)x) \right)
$$
but $\frac{1-\pi}{1+\pi} \not\in \mathbb{Q}$ and $\sin x \sin \pi$ is continuous, so this function is not periodic.
A: If $f(x)=\sin(x)\sin(\pi x)$ were periodic, then in particular its set of zeroes would be periodic. But the zero set is $\mathbb Z\cup\pi\mathbb Z$, and $a\mathbb Z\cup b\mathbb Z$ (for nonzero $a$, $b$) is not periodic unless $b/a$ is rational.
(Suppose $a\mathbb Z\cup b\mathbb Z$ is periodic with period $P>0$. Then $0+P\in a\mathbb Z\cup b\mathbb Z$; assume without loss of generality that $P=ka$ for some $k\in Z$. Now $b+P = ka+b$ must be in $a\mathbb Z\cup b\mathbb Z$ too, but if $ka+b=bn$ then $\frac ba=\frac{k}{n-1}$, and if $ka+b=am$, then $\frac ba = m-k$; in both cases $\frac ba$ is rational.)
A: Assume
$$\sin(x+T)\sin(\pi(x+T))=\sin(x)\sin(\pi x)$$ for all $x$.
Then with $x=0$,
$$\sin(T)\sin(\pi T)=0$$ so that $T=k\pi$ or $T=k$.
But you can find examples of
$$\sin(x+k\pi)\sin(\pi x+k\pi^2)\ne\sin(x)\sin(\pi x),$$ i.e. 
$$\sin(\pi x+k\pi^2)\ne\sin(\pi x),$$
and
$$\sin(x+k)\sin(\pi x+k\pi)\ne\sin(x)\sin(\pi x),$$i.e.
$$\sin(x+k)\ne\sin(x).$$

For solutions to be possible, you would indeed need
$$k\pi^2=2k'\pi$$ or $$k=2k'\pi,$$ requiring $\pi$ to be rational.
