$ \lim_{n\to\infty}\prod_{i=1}^k\sin(nx_i)\ =\ 0 $ Assume that real numbers $x_1,\dots,x_k$ satisfy 
$$
\lim_{n\to\infty}\prod_{i=1}^k\sin(nx_i)\ =\ 0
$$
It implies that one of them is a multiplicity of $\pi$, or not necessarily?
I could not find the answer myself...
 A: To avoid lots of technical writing, I will show it for $k=2$. If you understand it for $k=2$, the general case is easy to see.
We assume that neither $x_1$ nor $x_2$ is a multiplier of $\pi$, i.e. $x_1,x_2 \not \in \pi \mathbb{Z}$. Write $x := x_1$ then $x \in \pi(\mathbb{Q} \setminus \mathbb{Z})$ or $x \in \pi(\mathbb{R} \setminus \mathbb{Q})$
First case: $x \in \pi(\mathbb{Q} \setminus \mathbb{Z}) \implies x=\frac{p}{q}\pi$ for a $p \in \mathbb{Z}\setminus \{0\}$ and a $q \in \mathbb{N} \setminus \{1\}$. Now for every $k \in \mathbb{N}$ we have $$\sin(kqx)=\sin(kp\pi)=0$$. Because of 
$$\sin((2qk+1)\frac{p}{q}\pi) = \sin(2pk\pi + \frac{p}{q}\pi) = \sin(\frac{p}{q}\pi) \not = 0$$ 
the sequence $(\sin(nx))_{n \in \mathbb{N}}$
is $2q$ periodic and not convergent.
Second case: $x \in \pi(\mathbb{R} \setminus \mathbb{Q})$: In this case we know (I will not prove it) that the set $\{ \sin(nx \mod 2\pi) \ | \ n \in \mathbb{N} \}$ is dense in $[0,2\pi]$. We can conclude that $\sin(nx)$ is not periodic, not convergent and never 0 for $x \in \pi(\mathbb{R} \setminus \mathbb{Q})$.
Now the actual prove:
Assume that neither $x_1$ nor $x_2$ is a multiplier of $\pi$, i.e. $x_1,x_2 \not \in \pi \mathbb{Z}$ and assume that $\lim_{n \to \infty} \sin(nx_1)\sin(nx_2) = 0$. If $x_1$ or $x_2$ is an irrational multiplier of $\pi$ then according to the "second case" we can ignore it, since it doesn't converge and is never 0. So we can assume that $x_1$ and $x_2$ is a rational (but not integer) multiplier of $\pi$: $x_1=\frac{p_1}{q_1}\pi, \ x_2=\frac{p_2}{q_2}\pi$. Let $\epsilon := |\sin(\frac{p_1}{q_1}\pi)\sin(\frac{p_2}{q_2}\pi)|$. Then there is a $N \in \mathbb{N}$ such that $|\sin(nx_1)\sin(nx_2)| < \epsilon \ \forall n \geq N$. For $k$ big enough we have $2q_1q_2k+1 \geq N$ and $$|\sin((2q_1q_2k+1)x_1)\sin((2q_1q_2k+1)x_2)| = |\sin(2p_1q_2k\pi+\frac{p_1}{q_1}\pi)\sin(2p_2q_1k\pi+\frac{p_2}{q_2}\pi)|$$ $$=|\sin(\frac{p_1}{q_1}\pi)\sin(\frac{p_2}{q_2}\pi)| < \epsilon = |\sin(\frac{p_1}{q_1}\pi)\sin(\frac{p_2}{q_2}\pi)|$$
which is a contradiction. Hence our assumtion is false and $x_1$ or $x_2$ is a multiplier of $\pi$
