Proving the product expansion of $\sin\theta$. Where did I go wrong? You can prove the expansion$$\frac {\sin\theta}{\theta}=\left\{1-\left(\frac {\theta}{\pi}\right)^2\right\}\left\{1-\left(\frac {\theta}{2\pi}\right)^2\right\}\ldots$$By taking the expansion$$\sin n\phi=2^{n-1}\sin\phi\cos\phi\left(\sin^2\frac {\pi}n-\sin^2\phi\right)\left(\sin^2\frac {2\pi}{n}-\sin^2\phi\right)\ldots$$substituting $\phi=\theta/n$ and dividing by$$n=2^{n-1}\sin\frac {\pi}n\sin\frac {2\pi}n\ldots\sin\frac {\pi(n-1)}{n}$$

However, when I try, I always get zero as the answer. I started off with$$\sin\theta=2^{n-1}\sin\frac {\theta}n\cos\frac {\theta}n\left(\sin^2\frac {\pi}n-\sin^2\frac {\theta}n\right)\left(\sin^2\frac {2\pi}n-\sin^2\frac {\theta}n\right)\ldots$$And divided it to get$$\frac {\sin\theta}n=\frac {\sin\frac {\theta}n\cos\frac {\theta}n\left(\sin^2\frac {\pi}n-\sin^2\frac {\theta}n\right)\left(\sin^2\frac {2\pi}n-\sin^2\frac {\theta}n\right)\ldots}{\sin\frac {\theta}n\cos\frac {\theta}n\ldots\sin\frac {\pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$\begin{align*}\sin\theta & =\theta\left\{\frac 12-\frac 12\left(\frac {\theta}n\right)^2\right\}\left\{\frac 13-\frac 13\left(\frac {\theta}{2\pi}\right)^2\right\}\ldots\\\\ & =\theta\prod\limits_{k=1}^{\infty}\left\{\frac 1{k+1}-\frac 1{k+1}\left(\frac {\theta}{k\pi}\right)^2\right\}\\ & =0\end{align*}$$

Question: I'm trying to prove$$\frac {\sin\theta}{\theta}=\prod\limits_{k\geq1}\left\{1-\left(\frac {\theta}{k\pi}\right)^2\right\}$$So where did I go wrong?

 A: We have
\begin{align}
\sin\theta&=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\
&=2\sin\frac{\theta}{2}\sin\left(\frac{\pi}{2}+\frac{\theta}{2}\right)..........(1).
\end{align}
Similarly in (1) changing $\theta$ into $\frac{\theta}{2}$ and $\frac{\pi}{2}+\frac{\theta}{2}$ successively, we have
$$\sin\frac{\theta}{2}=2\sin\frac{\theta}{2^2}\sin\left(\frac{\pi}{2}+\frac{\theta}{2^2}\right)=2\sin\frac{\theta}{2^2}\sin\left(\frac{2\pi}{2^2}+\frac{\theta}{2^2}\right),$$
and
\begin{align}
\sin\left(\frac{\pi}{2}+\frac{\theta}{2}\right)&=2\sin\left(\frac{\pi}{2^2}+\frac{\theta}{2^2}\right)\cdot\sin\left(\frac{\pi}{2}+\frac{\pi}{2^2}+\frac{\theta}{2^2}\right)\\
&=2\sin\left(\frac{\pi}{2^2}+\frac{\theta}{2^2}\right)\cdot\sin\left(\frac{3\pi}{2^2}+\frac{\theta}{2^2}\right).
\end{align}
Substituting these values in the right-hand side of (1) we have, after rearranging,
$$\sin\theta=2^3\sin\frac{\theta}{2^2}\sin\frac{\pi+\theta}{2^2}\sin\frac{2\pi+\theta}{2^2}\sin\frac{3\pi+\theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$\sin\theta=2^{p-1}\sin\frac{\theta}{p}\sin\frac{\pi+\theta}{p}\sin\frac{2\pi+\theta}{p}...\sin\frac{(p-1)\pi+\theta}{p} ......(3),$$
where $p$ is a power of 2.
The last factor in (3)
$$=\sin\left[\pi-\frac{\pi-\theta}{p}\right]=\sin\frac{\pi-\theta}{p}.$$
The last but one
$$=\sin\frac{(p-2)\pi+\theta}{p}=\sin\frac{2\pi-\theta}{p},$$
and so on.
Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$\sin\theta=2^{p-1}\sin\frac{\theta}{p}\left\{\sin\frac{\pi+\theta}{p}\sin\frac{\pi-\theta}{p}\right\}\left\{\sin\frac{2\pi+\theta}{p}\sin\frac{2\pi-\theta}{p}\right\}...\ \ ....(4).$$
The last factor is
$$\sin\frac{\frac{p}{2}\pi+\theta}{p}\\=\sin\left(\frac{\pi}{2}+\frac{\theta}{p}\right)=\cos\frac{\theta}{p}.$$
Hence (4) is
$$\sin\theta=2^{p-1}\sin\frac{\theta}{p}\left[\sin^2\frac{\pi}{p}-\sin^2\frac{\theta}{p}\right]\left[\sin^2\frac{2\pi}{p}-\sin^2\frac{\theta}{p}\right]\\...\left[\sin^2\frac{\left(\frac{p}{2}-1\right)\pi}{p}-\sin^2\frac{\theta}{p}\right]\cdot\cos\frac{\theta}{p}.........(5)$$
Divide both sides of (5) by $\sin\frac{\theta}{p}$ and let $\theta\to 0$. Then $\lim_{\theta\to 0}\frac{\sin\theta}{\sin\frac{\theta}{p}}=p,$ and we have
$$p=2^{p-1}\cdot\sin^2\frac{\pi}{p}\cdot\sin^2\frac{2\pi}{p}\cdot\sin^2\frac{3\pi}{p}...\sin^2\frac{\left(\frac{p}{2}-1\right)\pi}{p}............(6)$$
Dividing (5) by (6), we have
$$\sin\theta=p\sin\frac{\theta}{p}\left[1-\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{\pi}{p}}\right]\left[1-\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{2\pi}{p}}\right]\left[1-\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{3\pi}{p}}\right]...\\...\left[1-\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{\left(\frac{p}{2}-1\right)\pi}{p}}\right]\cos\frac{\theta}{p}.......(7)$$
Now let $p\to\infty$.
Since
\begin{align}
\lim_{p\to\infty}\left[p\sin\frac{\theta}{p}\right]&=\theta,\\
\lim_{p\to\infty}\left[\frac{\sin^2\frac{\theta}{p}}{\sin^2\frac{\pi}{p}}\right]&=\frac{\theta^2}{\pi^2},
\end{align}
and so on, we have
$$\sin\theta=\theta\left(1-\frac{\theta^2}{\pi^2}\right)\left(1-\frac{\theta^2}{2^2\pi^2}\right)\left(1-\frac{\theta^2}{3^2\pi^2}\right)...$$
