<s>[INST] An algebra problem related to trigonometry Let $\alpha $ be an acute angle such that $\sin \alpha = \dfrac{3 \sqrt 3}{14}$. 
Prove that 
$$\frac{2\cdot7^{n}}{\sqrt 3}\sin \left(n\alpha + \dfrac{\pi}{3} \right) \in \mathbb{Z} \qquad\forall  n>0. $$
 [/INST] This is equivalent to saying that for all $n>0$
$$
\frac{2\cdot7^{n}}{\sqrt{3}}\sin\left(n\alpha+\frac\pi3\right)\in\mathbb{Z}
$$
Note that $\cos(\alpha)=\frac{13}{14}$, so that
$$
e^{i\alpha}=\frac{13}{14}+i\frac{3\sqrt3}{14}
$$
therefore
$$
\begin{align}
e^{i(n\alpha+\pi/3)}
&=\left(\frac{13}{14}+i\frac{3\sqrt3}{14}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right)\\[6pt]
&=\frac1{7^n}\left(\frac{13}{2}+i\frac{3\sqrt3}{2}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right)
\end{align}
$$
Note that
$$
a_n=\left(\frac{13}{2}+i\frac{3\sqrt3}{2}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right)
$$
and $\frac2{\sqrt{3}}$ times its imaginary part are solutions to
$$
a_n=13a_{n-1}-49a_{n-2}
$$
$\dfrac2{\sqrt{3}}\mathrm{Im}(a_0)=1$ and $\dfrac2{\sqrt{3}}\mathrm{Im}(a_1)=8$. Thus, $\dfrac2{\sqrt{3}}\mathrm{Im}(a_n)\in\mathbb{Z}$. Substituting back, we get the desired result.
</s>