Deriving Taylor series without applying Taylor's theorem. First, a neat little 'proof' of the Taylor series of $e^x$.
Start by proving with L'Hospital's rule or similar that
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
and then binomial expand into
$$e^x=\lim_{n\to\infty}1+x+\frac{n-1}n\frac{x^2}2+\dots+\frac{(n-1)(n-2)(n-3)\dots(n-k+1))}{n^{k-1}}\frac{x^k}{k!}+\dots$$
Evaluating the limit, we are left with
$$e^x=1+x+\frac{x^2}2+\dots+\frac{x^k}{k!}+\dots$$
which is our well known Taylor series of $e^x$.
As dxiv mentions, we can exploit the geometric series:
$$\frac1{1-x}=1+x+x^2+\dots$$
$$\ln(1+x)=x-\frac12x^2+\frac13x^3-\dots$$
$$\arctan(x)=x-\frac13x^3+\frac15x^5-\dots$$
which are found by integrating the geometric series and variants of it.
I was wondering if other known Taylor series can be created without applying Taylor's theorem.  Specifically, can we derive the expansions of $\sin$ or $\cos$?

Alan Turing, at a young age, derived the series expansion of $\arctan$ without using (and, purportedly without knowing) calculus whatsoever.
Using the identity
$$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x},$$
he obtained
$$\tan(2 \arctan x) = \frac{2x}{1-x^2},$$
and
$$2 \arctan x = \arctan\left( \frac{2x}{1-x^2}\right).$$
Using the geometric series for $|x| < 1$, 
$$\tag{1}2 \arctan x =\arctan [2x(1 + x^2 + x^4 + x^6 + \ldots)]$$
Assuming $\arctan x = a_0 + a_1x + a_2x^2 + \ldots$ and matching coefficients in the expansions of each side of (1), he obtained 
$$\arctan x = a_1\left(x - \frac{1}{3} x^3 + \frac{1}{5}x^5 - \ldots \right).$$
Some basic trigonometry reveals
$$a_1 = \lim_{x \to 0} \frac{\arctan x}{x} = 1.$$ 
