Suppose $(a_n)$ is such that for all $t\in I=[0,1],~\lim~a_n\sin(nt)$ exists and $=0$. You are in effect saying that the sequence of functions $f_n:I\rightarrow \mathbb R, t\mapsto a_n\sin(nt)$ converges simply (direct translation from french, maybe "pointwise" in english) to the $0$ funtion. What you want to show is that this implies uniform convergence of the sequence $(f_n)$ to the zero function (since for $n\geq2>\pi/2,~\|f_n\|_{\infty}=|a_n|$.)
Suppose $(a_n)$ doesn't tend to $0$. Then there is a positive real number $\epsilon>0$ and a strictly increasing sequence of positive integers $\varphi(n)\uparrow +\infty$ such that $|a_{\varphi(n)}|>\epsilon$ for all $n$. Without loss of generality, we may assume that for all $n,~\varphi(n+1)\geq 100\varphi(n)$ and $\varphi(0)>100\pi$ ($100$ isn't special, it is just large enough for our purpose.)
For $x\in \mathbb R$ and $r>0$, we call $[x-r,x+r]$ the segment centered at $x$ and of length $2\times r$. We construct a sequence of nested closed intervals $(I_n)$ centered around maxima of $|f_{\varphi(n)}|$ and of length $2\times\frac{\pi}{3\varphi(n)}$.
The first one is defined to be $I_0\subset [0,1]$ centered around one of the maxima of $|f_{\varphi(0)}|$ of the form $\frac{\pi/2+k\pi}{\varphi(0)}$ and of length $2\times\frac{\pi}{3\varphi(0)}$. Since $\varphi(0)$ is quite large, $f_{\varphi(0)}$ has plenty of time to complete many of full oscillations, so there is room enough to fit $I_0$ into $I$.
Suppose the $n^{th}$ such nested interval $I_n$ has been constructed. Because $\varphi(n+1)\geq 100\varphi(n)$, the $\varphi(n+1)^{th}$ function has time to do several full oscillations inside $I_n$, and we just pick one point in $x_{n+1}\in I_n$ that realizes a maxima for $|f_{\varphi(n+1)}|$, is of the form $\frac{\pi/2+k\pi}{\varphi(n+1)}$, and is close enough to the center of $I_n$ so that the segment centered at this maxima and or length $2\times\frac{\pi}{3\varphi(n+1)}$ is completely contained in $I_n$.
Saying that the maxima of $|f_{\varphi(n)}|$ we choose are of this particular form is not necessary, since by hopothesis the $a_{\varphi(n)}$ are non zero, and so the maxima of $|f_{\varphi(n)}|$ are those of $|\sin(~\cdot\times\varphi(n))|$. I include this extra information only to facilitate the verification of a calculation further down which relies upon the fact that $|\sin(x)|\geq 1/2$ on $[\pi/2-\pi/3,\pi/2+\pi/3]$.
We thus get a nested sequence of closed intervals whose length tend quickly to $0$, and by compactness there exists $t\in I$ with $\bigcap_{n\in\mathbb N}I_n=\lbrace t\rbrace.$ Our choice of $I_n$ (the fact that it is centered around a maxima of $|f_{\varphi(n)}|$ and of length $2\times\frac{\pi}{3\varphi(n)}$ ensures that for all $s\in I_n,~|\sin(\varphi(n)s)|\geq 1/2$. Since for all $n$ we have $t\in I_n$, we have for all $n$ that $|f_{\varphi(n)}(t)|\geq |a_n|/2>\epsilon/2$ which contradicts the assumption that $|f_n(t)|$ must tend to $0$.