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.