Finding the derivatives of sin(x) and cos(x) We all know that the following (hopefully):
$$\sin(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}\ , \ x\in \mathbb{R}$$
$$\cos(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{(2n)!}\ , \ x\in \mathbb{R}$$
But how do we find the derivates of $\sin(x)$ and $\cos(x)$ by using the definition of a derivative and the those definition above?
Like I should start by doing:
$$\lim_{h\to\infty}\frac{\sin(x+h)-\sin(x)}{h}$$
But after that no clue at all.  Help appreciated!

Well, in the case that you really need to show it, let's examine a general term in $\sin(x+h)-\sin(x)$.
\begin{align*}
\frac{(-1)^n}{(2n)!}\cdot((x+h)^{2n}-x^{2n})&=\frac{(-1)^n}{(2n)!}\cdot\left(x^{2n}+2nhx^{2n-1}+\ldots+h^2n-x^{2n}\right)\\
&=\frac{(-1)^n}{(2n)!}\cdot h\left(2nx^{2n-1}+[\text{a bunch of stuff depending on h}]\right)
\end{align*}
Then if we divide by $h$ then let $h\to 0$,
\begin{align*}
\lim_{h\to0}\frac{(-1)^n}{(2n)!}&\cdot \left(2nx^{2n-1}+[\text{a bunch of stuff depending on h}]\right)=(-1)^n\frac{2nx^{2n-1}}{(2n)!}\\
&=(-1)^n\frac{x^{2n-1}}{(2n-1)!}
\end{align*}
So there's the derivative of a general term. I think the other answers have finished the rest of the proof.
