What's a purely algebraic way to prove that $\pi-\frac{\pi^3}{3!}+\frac{\pi^5}{5!}-\dots=0$? I'm sure that the first step is to write $\pi=4-\frac43+\frac45-\dots$, but I haven't been bold enough to complete the algebra, nor do I think it is feasible. :(
I guess in general, what's the algebraic way to prove that the Taylor series of $\sin x$ converges to the geometric definition of $\sin x$?
Not sure about an "algebraic proof." A direct power series substitution using the power series for $\arctan$ would need to give you:
$$\sin(4\arctan(x))=\frac{4x(1-x^2)}{(1+x^2)^2}=\frac{8x}{(1+x^2)^2}-\frac{4x}{1+x^2}=4\sum_{n=0}^{\infty}(-1)^n(2n+1)x^{2n+1}$$
That's gonna be messy but not necessarily impossible to show. Note the series does not converge when $x=1$, but arguments can be made as $x\to 1-$.
The key to the second part is to prove that $\exp(ix)=\cos x + i\sin x$, where $\exp(z)=\sum_k z^k/k!$. One way to prove this is to show that $f(x)=1+ix$ is "close enough" to $\mathrm{cis}(x)=\cos x + i\sin x$ when $x$ is near zero so that $\lim f(x/n)^n = \lim \mathrm{cis}(x/n)^n$. But you can prove that $\mathrm{cis}(x/n)^n=\mathrm{cis}(x)$ by induction, and to show that $\lim_{n\to\infty} f(x/n)^n = \exp(ix)$.
The "close enough" part of the proof - that $\mathrm{cis}(x)=1+xi + o(x)$ - can be made geometric.
Here's an answer to another question that covers how close "close enough" needs to be to get this result.
Outline of the geometric part of the proof
First show that $\sin x<x$ for $x$ positive, since $\sin x$ is the shortest distance from a point $(\cos x,\sin x)$ to the $x$-axis, and $x$ is the distance from the same point along the circle.
Then show that $1-\cos x =\frac{\sin ^2 x}{1+\cos x}<\sin^2 x<x^2$.
The last step is harder - showing that $x-\sin x$ is "small enough." It requires a theorem about convex sets - the shortest path between two points on the surface of a convex set which never enters the interior must lie entirely on that convex set. This is sort of intuitive.
From that geometric result, you can show that $\tan x > x$, and hence that $x-\sin x < \tan x - \sin x < \sin x\frac{1-\cos x}{\cos x} = o(x)$ when $x$ near zero.
More on algebraic approach
The power series approach is going to be brutal. Let's just try the case $\sin 2\arctan x = \frac{2x}{1+x^2}=2\sum_{n} (-1)^n x^{2n+1}$.
Now, for $k$ odd, the contribution of $\arctan^k x$ to the coefficient $x^n$ in $\sin 2\arctan x$ is in terms of the set $P_{n,k}=\{(n_1,\dots,n_k)\mid n_i\text{ odd, } \sum n_i=n\}$. The contribution of $(n_1,\dots,n_k)$ is $$\frac{(-1)^{(k-1)/2}4^k}{k!}\prod_i \frac{(-1)^{(n_i-1)/2}}{n_i} = \frac{(-1)^{(n-1)/2}4^k}{k!\prod n_i}$$
Basically, you need that:
$$\sum_{n_1+\cdots + n_k=n} \frac{4^{k-1}}{k!n_1n_2\cdots n_k} = n$$
where the $n$ is odd and the sum is over partitions under the condition that each $n_i$ is odd.
It's not at all clear how summing these is even going to be an integer.
It's easy to verify for small $n$. $n=1$ is trivial. For $n=3$, the partitions are $(3)$ and $(1,1,1)$ which gives $\frac{1}{3}+\frac{4^2}{6}=3$. For $n=5$, the partitions are $(5),(3,1,1),(1,3,1),(1,1,3),(1,1,1,1,1)$ and you get:
$$\frac{1}{5} + 3\cdot \frac{4^2}{3!\cdot 3} + \frac{4^4}{5!}=5$$
An interesting fact is that this equality seems to be true for $n$ even, as well (still with the condition that the $n_i$ are odd.) I wonder if there is a probability explanation for this - some process for choosing partitions of $n$ into odd integers where the probability of getting $(n_1,\dots,n_k)$ is $\frac{4^{k}}{n k!n_1\dots n_k}$.
Hint:
Along the circle of equation $x^2+y^2=1$, the arc length is (in the first quadrant)
$$\theta=\int_0^y\frac{dy}{\sqrt{1-y^2}}=\arcsin(y)$$ so that $y$ is the geometric definition of the sine.
This relation can also be written as the differential equation
$$y'^2(\theta)+y^2(\theta)=1,$$
which by derivation gives
$$2y''(\theta)y'(\theta)+2y'(\theta)y(\theta)=0,$$
or $$y''(\theta)=-y(\theta).$$
With $y(0)=0$, this leads to the well-known Taylor development.
