Zero limit implies zero sequence

We will use the next lemma:

Lemma. For $\alpha,\beta$ from $(0,1)$ and a positive integer $n$, let $$ I_n(\alpha,\beta)=\frac{2}{n}\sum_{k=1}^n\sin(\alpha \pi k)\sin(\beta \pi k).$$ Then $$\lim_{n\to\infty}I_n(\alpha,\beta)=\left\{\matrix{0&\hbox{if}&\alpha\ne \beta,\cr 1&\hbox{if}&\alpha=\beta.} \right. $$

Proof. Note that, for $\theta\notin 2\pi\mathbb{Z} $, we have $$ \sum_{k=1}^n\cos(k\theta)=-\frac{1}{2}+\frac{1}{2}\sum_{k=-n}^ne^{ik\theta} =-\frac{1}{2}+\frac{1}{2}\frac{e^{i(n+1)\theta}-e^{-in\theta}}{e^{i\theta}-1} =-\frac{1}{2}+\frac{\sin((n+1/2)\theta)}{2\sin(\theta/2)} $$ So $$\forall\,\theta\in\mathbb{R}\setminus2\pi\mathbb{Z}, \quad\lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n\cos(k\theta)=0\tag{1} $$ Now, if $\alpha\ne\beta$ $$ I_n(\alpha,\beta)=\frac{1}{n}\sum_{k=1}^n\cos(k\pi(\alpha-\beta)) -\frac{1}{n}\sum_{k=1}^n\cos(k\pi(\alpha+\beta)) $$ and by $(1)$, $\lim_{n\to\infty}I_n(\alpha,\beta)=0$ in this case. Similarly, if $\alpha=\beta$, we have $$ I_n(\alpha,\alpha)=\frac{1}{n}\sum_{k=1}^n(1-\cos(2k\pi\alpha)) =1-\frac{1}{n}\sum_{k=1}^n\cos(2k\pi\alpha) $$ so, $\lim_{n\to\infty}I_n(\alpha,\alpha)=1$.$\qquad\square$

Now, let us come the the proposed question. Let $$ x_n=\sum_{p=1}^ka_p\sin(n\pi\alpha_p) $$ by assumption $\lim\limits_{n\to\infty}x_n=0$, So, for $q\in\{1,2,\ldots,k\}$ we have $\lim\limits_{n\to\infty}x_n\sin(n\pi\alpha_q)=0$. Using Stolz–Cesàro theorem we conclude that $$ \lim_{n\to\infty}\frac{2}{n}\sum_{m=1}^nx_m\sin(m\pi\alpha_q)=0 $$ This is equivalent to $$ \lim_{n\to\infty}\sum_{p=1}^ka_pI_n(\alpha_p,\alpha_q)=0 $$ or $a_q=0$ according to the Lemma, and the desired conclusion follows.$\qquad\square$

There is also a one-line solution through diophantine approximation.

The sequence $\left\{\left(\left\{\frac{n\alpha_1}{2}\right\},\ldots,\left\{\frac{n\alpha_k}{2}\right\}\right)\right\}_{n\in\mathbb{N}}$ is ergodic in $[0,1]^k$, hence is some $a_p$ differs from zero, $\{x_n=\sum_{p=1}^{k}a_p \sin(\pi n \alpha_p)\}_{n\in\mathbb{N}}$ has a non-Cauchy subsequence and cannot converge.