How to approximate $\pi$ using the Maclaurin series for $\sin(x)$ We have that
$$\sin(x)=x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
Now plugging in $x=\pi$,
$$0=\pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + \cdots$$
Is there a way we can use this beautiful result to calculate better and better approximations to $\pi$?
EDIT: This is a problem from a Mathologer video here titled Euler's real identity NOT e to the i pi = -1 at minute 7:06. I think this problem is meant to be doable, so please try it and don't just say that it is impossible and pointless, because if that were the case, Mathologer wouldn't have put it in his video (hopefully?)

See the comment of Steve the Cat Couch under the original Youtube video, which is approved by Mathologer.
Basically, we could solve the root of truncated polynomial 
$$0=x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
by Newton's method. So there are two steps of approximation. One is from the polynomial we use and another is from Newton's method.
