How to obtain the series of the common elementary functions without using derivatives? First, I'm a freshman student of physics, not of mathematics, so please excuse my ignorance of mathematics :)
Well, I'm reading the book "Huygens and Barrow, Newton and Hooke" by Vladimir Arnold, and one excerpt (at the page 43) called my attention greatly: "he used Taylor's formula for calculating derivatives rather than using the derivatives for the expansion of functions" ("he" refers to Sir Isaac Newton).
My main question is: How to obtain the series of the common elementary functions (trigonometric, exponential...) without using derivatives? And, for example, could calculus be developed in a way that we would get $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ and also the one for cosine and then use this series to find the derivative of sine by seeing that it reduces to the series of the cosine?
P.S. I don't know English very well, I just hope the question asked is that I wanted to ask! Thanks in advance to everyone who answer.

I can't verify that Bromwich didn't use calculus to derive the equation below. But I remember doing something like the derivation below in high school back in 1966. So I will put it forth that there is a series that can be arrived at with recourse to calculus for which the method shown below does work.
Here is one example. Bromwich states that, for odd n
$$
\sin(na) = nx - \frac{n(n^2-1^2)}{3!}x^3 + \frac{n(n^2-1^2)(n^2-3^2)}{5!}x^5 - \cdots
$$
where $x = \sin a$. Now we choose a such that $\theta = na$ as $n\to\infty$. So, as $n\to\infty$, $nx \to n \sin a\to na \to \theta$.
Hence, as $n \to \infty$,
\begin{align}
\sin(\theta)
   &= nx -
      \frac{n(n^2-1^2)}{3!}x^3 +
      \frac{n(n^2-1^2)(n^2-3^2)}{5!}x^5 - \cdots\cr
   &= nx -
      \frac{nx((nx)^2-x^2)}{3!} + 
      \frac{nx((nx)^2-x^2)((nx)^2-(3x)^2)}{5!} - \cdots\cr
   &\to \theta -
      \frac{\theta^3}{3!} + 
      \frac{\theta^5}{5!} - \cdots\cr
\end{align}
Addendum
I'm sure that my logic below is not up to snuff by modern standards. None-the-less, it is still compelling.
Let's assume that n is an odd number. Then
$(\sin \theta + i \cos \theta)^{n} = \sum_{k=0}^{n} \binom{n}k (\sin \theta)^k (i \cos \theta)^{n-k}$
$\sin(n \theta) + i \cos (n \theta) = \sum_{k=0}^{n} \binom{n}k (\sin \theta)^k (i \cos \theta)^{n-k}$
$\displaystyle \sin(n \theta) = \sum_{k=0}^{(n-1)/2} \binom{n}{2k+1} (\sin \theta)^{2k+1} (-1)^k (\cos \theta)^{n-2k-1}$
Let $\theta \to 0$ in such a way that $n \theta \to x$ at the same time.
This can be done by letting $\theta = \dfrac{x}{n}$ and then letting $n \to \infty$.
Now we examine $\binom{n}{2k+1} (\sin \theta)^{2k+1}$ as $n \to \infty$.
\begin{align*}
   \binom{n}{2k+1} (\sin \theta)^{2k+1}
      &= \dfrac{(n)(n-1)(n-2)\cdots (n-2k)}{(2k+1)(2k)(2k-1)
      \cdots (1)}
      (\sin \theta)^{2k+1}\\
   &= \dfrac{(n) \sin \theta}{2k+1}\;
      \dfrac{(n-1) \sin \theta}{2k}\;
      \dfrac{(n-2) \sin \theta}{2k-1}\;\cdots
      \dfrac{(n - 2k) \sin \theta}{1}\\
   &\to \dfrac{(n) \theta}{2k+1}\;
      \dfrac{(n-1) \theta}{2k}\;
      \dfrac{(n-2) \theta}{2k-1}\;\cdots
      \dfrac{(n - 2k) \theta}{1}\\
   &\to \dfrac{x}{2k+1}\;
      \dfrac{x-\theta}{2k}\;
      \dfrac{x-2\theta}{2k-1}\;\cdots
      \dfrac{x - 2k \theta}{1}\\
   &\to \dfrac{x^{2k+1}}{(2k+1)!}
\end{align*}
Since $\cos \theta \to 1$ as $\theta \to 0$, letting $n \to \infty$, we get
$$\displaystyle \sin(x)
  = \sum_{k=0}^{\infty} (-1)^k \dfrac{x^{2k+1}}{(2k+1)!}$$
