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 $\cos$?

The series for $\sin$ and $\cos$ can be derived from the expansion of $e^x$.
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040}+\cdots$$
Sub in $ix$ to get:
$$e^{ix}=1+ix+\frac{(ix)^2}{2}+\frac{(ix)^3}{6}+\frac{(ix)^4}{24}+\frac{(ix)^5}{120}+\frac{(ix)^6}{720}+\frac{(ix)^7}{5040}+\cdots$$
$$\cos x+i\sin x=1+ix-\frac{x^2}{2}-\frac{ix^3}{6}+\frac{x^4}{24}+\frac{ix^5}{120}-\frac{x^6}{720}-\frac{ix^7}{5040}+\cdots$$
Compare real and imaginary parts:
$$\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\cdots$$
$$\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots$$
EDIT:
Consider the function $f(x)=(\cos x+i\sin x)e^{-ix}$
$$f'(x)=(-\sin x+i\cos x)e^{-ix}-i(\cos x+i\sin x)e^{ix}$$
$$f'(x)=-e^{ix}\sin x+ie^{ix}\cos x-ie^{-x}\cos x+e^{ix}\sin x$$
$$f'(x)=0$$
Hence $f(x)=c$ and $f(0)=(\cos0+i\sin0)e^0=1$ so $f(x)=1$
Therefore $e^{ix}=\cos x+i\sin x$
SECOND EDIT:
Another way springs to mind as well:
$$f(x)=\cos x+i\sin x$$
$$f'(x)=-sin(x)+i\cos x$$
$$f'(x)=i(\cos x+i\sin x)$$
$$f'(x)=i\cdot f(x)$$
$$\frac{f'(x)}{f(x)}=i$$
$$\ln(f(x))=ix+c$$
$$f(x)=e^{ix+c}$$
$$f(0)=\cos 0+i\sin 0=1\implies c=0$$
$$\therefore f(x)=e^{ix}$$
