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!

Solution 1:

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}

Solution 2:

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.

Solution 3:

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)$$