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!
 A: First, since we deal with a power series which uniformly convergent in any compact set included in the disc of convergnece, we can switch $\lim$ and $\sum$. We have:
\begin{align}\lim_{h\to\infty}\frac{\sin(x+h)-\sin(x)}{h}&=\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!}\lim_{h\to0}\frac{(x+h)^{2n+1}-x^{2n+1}}{h} \\
&=\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!}(2n+1)x^{2n}=\cos(x).\end{align}
A: Use $\sin(x+h)=\sin(x)\cos(h) + \sin(h) \cos(x)$.
Or more explicit
$$\frac{\sin(x+h)-\sin(x)}{h}=\frac{\sin(x)(\cos(h)-1)}{h}+ \frac{\sin(h)\cos(x)}{h}$$
As 
$$\lim_{h \to 0} \frac{\sin(x) (\cos(h)-1)}{h}=0$$
And 
$$\lim_{h\to 0} \frac{\sin(h)\cos(x)}{h}=\cos(x)$$ 
you get that $\sin'(x)=\cos(x)$.  
For $\cos$ you get the hint $\cos(x+h)=\cos(x)\cos(h)-\sin(x)\sin(h)$$
A: By using your definition, if you can prove that you may exchange summation and differentiation, then you may write
$$ {d \over dx} \sin x = {d \over dx} \sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!} = \sum^{\infty}_{n=0}(-1)^n {d \over dx} \frac{x^{2n+1}}{(2n+1)!} = \\
= \sum^{\infty}_{n=0}(-1)^n \frac{(2n+1) x^{2n}}{(2n+1)!} = \sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{(2n)!} = \cos x$$
and similarly for the second identity (be careful and make sure you get the minus sign though - hint: what happens to the first term?).
To prove you may exchange summation and differentiation, it suffices to prove that the second series (the series of derivatives) converges uniformly (locally uniformly is also good). In this case, you may do this by using some easy estimates on the factorial.
A: The radius of convergence of the power series is infinity, so you can interchange summation and differentiation.
$\frac{d}{dx}\sin x = \frac{d}{dx} \sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!} = \sum^{\infty}_{n=0}(-1)^n \frac{d}{dx} \frac{x^{2n+1}}{(2n+1)!} = \sum^{\infty}_{n=0}(-1)^n  \frac{x^{2n}}{(2n)!} = \cos x$
If you wish to use the definition, you can do:
$\frac{\sin(x+h)-\sin x}{h} = \frac{\sum^{\infty}_{n=0}(-1)^n \frac{(x+h)^{2n+1}}{(2n+1)!}-\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}}{h} = \sum^{\infty}_{n=0}(-1)^n \frac{1}{(2n+1)!}\frac{(x+h)^{(2n+1)}-x^{(2n+1)}}{h}$
Then $\lim_{h \to 0} \frac{(x+h)^{(2n+1)}-x^{(2n+1)}}{h} = (2n+1)x^{2n}$ (using the binomial theorem).
(Note $(x+h)^p-x^p = \sum_{k=1}^p \binom{p}{k}x^{p-k}h^k = p x^{p-1}h + \sum_{k=2}^p \binom{p}{k}x^{p-k}h^k$. Dividing across by $h$ and taking limits as $h \to 0$ shows that $\frac{d}{dx} x^p = p x^{p-1}$.)
A: 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.
