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?)

It is easy to find better approximations to $\pi$ by iteration. Let $a_{n+1}=a_n+\sin(a_n)$. If you start with $a_1$ close enough to $\pi$ the sequence converges to $\pi$. For example, $a_1=3$ will work. You can replace $\sin(x)$ with a truncated Taylor series $f(x)$ and the iteration will converge to the root of $f(x)$ closest to $\pi$.
