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?)
 A: I'd rather work with $\dfrac\pi6$ for a faster converging series.
Then you can indeed solve for the roots of
$$\sin\frac\pi6=\frac12=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
by Newton and progressively increase the number of terms and the accuracy with which you compute the roots.
For instance, the given quintic equation yields $\pi\approx3.1415778790776$, and the next approximation is $3.1415927099226$.
I don't think that this is a very efficient method as it makes you evaluate the sine for many values using a (truncated) series during the resolution of the equations, instead of a direct evaluation, for instance by the Machin formula.
A: 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 $\pi$.
A: 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.
A: We can write $\pi$ like this:
$\pi = 4\cdot \frac{\Large\pi}{\Large4}=4\cdot\arctan(1)$
And use Maclaurin Series for $\arctan(x)$:
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+~...~+(-1)^n\cdot\frac{x^{2n+1}}{(2n+1)!}+~...$$
Hope my answer will help you =)
