TITLE: Period of a product of $\sin$ and $\cos$
QUESTION [5 upvotes]: I want to find the period of $\sin(t) \cos(\pi t)$.
I started off by transforming that into
$\frac{1}{2}\left [ \sin((\pi +1)t) - \sin((\pi - 1)t\right ]$, but then I get stuck.  How do I find the least common multiple of $\pi + 1$ and $\pi - 1$? Is that what I need to do to find the period of the whole thing?

REPLY [3 votes]: Suppose there is a common multiple $p$ of $\pi+1$ and $\pi-1$.  Then
$$
\begin{align}
p & = n(\pi+1) \\
p & = m(\pi-1)
\end{align}
$$
(Later note: The context of the problem should make it clear that this means $n$ and $m$ are positive integers.)
It follows via a bit of algebra that
$$
\pi=\frac{n+m}{n-m}.
$$
Therefore $\pi$ is rational.  But in this article it is proved that $\pi$ is irrational.  At least two of the proofs given there can be understood by someone who knows nothing beyond first-year calculus.
The function is therefore not periodic, but it is almost periodic.