How can the fact that
$$\pi=\displaystyle\lim_{n\to\infty}n\sin\frac{180^\circ}{n}$$
be useful when calculating $\pi$? I mean, $180^\circ =\pi$, isn't that a circular reasoning?
I got the idea that Lagrange interpolation might help, since every argument is divided by the corresponding power of $\pi$, therefore avoiding the circularity when we choose to interpolate
$$0\to 0,\, \frac{\pi}{2}\to 1,\, \pi\to 0,\, \frac{3\pi}{2}\to -1,\, 2\pi\to 0.$$
This interpolation yields
$$\sin x\approx \dfrac{8x^3}{3\pi ^3}-\dfrac{8x^2}{\pi ^2}+\dfrac{16x}{3\pi}.$$
But this is problematic since it's a polynomial and its behavior at $\infty$ is very different from $\sin$ at $\infty$, so that can't be used. Using
$$\sin x=\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n\, x^{2n+1}}{(2n+1)!}$$
or
$$\sin x=x\displaystyle\prod_{n=1}^\infty \left(1-\dfrac{x^2}{n^2\pi^2}\right)$$
doesn't help, since not every $x$ is divided by corresponding power of $\pi$; using such series or products to calculate $\pi$ would be circular. So, how can the formula in the question be used to calculate $\pi$?
Best Answer
One way of using this fact is to stick to a certain subset of $\mathbb N$, namely $n = 2^k$, and evaluate $\sin\frac{\pi}{2^{k+1}}$ in terms of $\sin\frac{\pi}{2^k}$. This essentially is what Viete did to arrive to his formula.
Notice that you don't need to know $\pi$ to compute $\sin \frac{\pi}{4}$.
Another way is to interpret it in purely geometrical way: divide the half circle into $n$ congruent arcs, construct a corresponding sine, and replicate it $n$ times to get a geometric approximation of $\pi$.