From the formula, $$\sin\vartheta = 2\cos\frac{\vartheta}2\sin\frac{\vartheta}2\tag1$$ by dividing both sides by $\cos\vartheta$ we could derive that $$\tan \vartheta = \frac 2{\cos \vartheta}\cos^2\frac{\vartheta}2\tan\frac{\vartheta}2$$ and hence by iteration through $\tan\vartheta\mapsto\tan\frac{\vartheta}2$ and simplifying, we can show that $(1)$ is a special case $n=1$ of the broader theorem, $$\sin\vartheta=2^n\sin\frac{\vartheta}{2^n}\prod_{k=1}^n\cos\frac{\vartheta}{2^k}\tag2$$ for a natural number $n\geqslant 1$. When we iterate through $\frac{\sin\vartheta}{\vartheta}\mapsto\cfrac{\sin\frac{\vartheta}{2^n}}{\frac{\vartheta}{2^n}}$, we obtain \begin{align}\frac{\sin\vartheta}{\vartheta}&=\bigg(\prod_{k=1}^n\cos\frac{\vartheta}{2^k}\bigg)\bigg(\prod_{k=1}^n\cos\frac{\vartheta}{2^{k+n}}\bigg)\bigg(\prod_{k=1}^n\cos\frac{\vartheta}{2^{k+2n}}\bigg)\cdots \\ &=\prod_{k=1}^\infty\cos\frac{\vartheta}{2^k}\tag3\end{align} and this is where I am stuck. By $(3)$, wouldn't $(2)$ imply that $$\eta\sin\frac{\vartheta}{\eta}\stackrel{\eta\to\infty}{\longrightarrow}1$$ for some $\eta$ (in this case, $\eta=2^n$), which could only mean $\vartheta=1$? This is strange to me because $(2)$ is meant to work for all $\vartheta$ and not just $\vartheta=1$.
Thanks in advance, and apologies for the nonrigorous approach.
Set $\xi = \frac{1}{\eta }$. So that your limit becomes
$$ \mathop {\lim }\limits_{\eta \to \infty } \left( {\eta \sin \frac{\vartheta } {\eta }} \right) = \mathop {\lim }\limits_{\xi \to 0} \frac{{\sin \left( {\vartheta \xi } \right)}} {\xi } = \vartheta \mathop {\lim }\limits_{\xi \to 0} \frac{{\sin \left( {\vartheta \xi } \right)}} {{\vartheta \xi }} = \vartheta $$
Hope this helps