Period of a product of $\sin$ and $\cos$ 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?
 A: Here's a more self-contained proof that $f(t) = \sin(t) \cos(\pi t)$ is not periodic, using only the fact that $\pi$ is irrational.  If    $f(t)$ had period $p$, then we'd also have $f(p) = f(0) = 0$.  Now this implies
either $\sin(p) = 0$, i.e. $p = n \pi$ for some nonzero integer $n$, or $\cos(\pi p) = 0$, i.e.
$p = n+1/2$ for some integer $n$.  But 
$$f(1/2 + n \pi) - f(1/2) = (-1)^{n+1} \sin(1/2) \sin(n \pi^2) \ne 0$$ since neither $1/2$ nor $n \pi^2$ is  an integer multiple of $\pi$ (if it were $m \pi$, then $\pi = m/n$ would be rational).
Similarly, 
$$f(\pi + (n+1/2)) - f(\pi)= (-1)^n \sin(\pi^2) \sin(n+1/2) \ne 0$$ since neither $\pi^2$ nor $n+1/2$ is an integer multiple of $\pi$.
A: 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.
A: We show that $\sin(t)\cos(\pi t)$ is not periodic. Suppose to the contrary that it is periodic. Let $f(t)=|\sin(t)\cos(\pi t)|$. Then $f(t)$ is periodic. Let $p$ be a period of $f(t)$.  
Let $m$ be the maximum value of $f(t)$ in the interval $[0,p]$.  If $f(t)$ is periodic, then $m$ is the maximum value of $f(t)$ as $t$ ranges over all the reals.  We will show that this is not the case, by showing that there is a $t$ such that $f(t)>m$.
Note first that $m\ne 1$. For if  $f(t)$ ever takes on the value  $1$, then   $|\sin(t)|$ and $|\cos(\pi t)|$ must be  simultaneously equal to $1$. So $t$ is  an odd multiple of $\pi/2$, say $t=q \pi/2$. Also, $\pi t$ is a multiple of $\pi$, so $t$ is an integer. It follows that  $\pi=2t/q$. This is impossible, since $\pi$ is irrational. 
We now show that there is a $t$ such that $f(t)>m$.  This is easy, but uses some machinery.
The sequence $(\sin(n))$ is dense in the interval $[-1,1]$.  Thus there is an integer $t$ such that $\sin(t)>m$.  Since $|\cos(\pi n)|=1$, it follows that $f(t) >m$.  
Comment: A quick search shows that there are many proofs of the fact that the sequence $(\sin(n))$ is dense in $[-1,1]$. Indeed the problem has been posed and solved on MSE. The most intuitive argument shows that the points $(\cos(n), \sin(n))$ are dense on the unit circle. The result for $(\sin(n))$ then follows by projecting on the $y$-axis. In general, if $\theta$ is not a rational multiple of $\pi$, then the points $(\cos(n\theta), \sin(n\theta))$ are dense on the unit circle.     
