Development of the sine function into the power series I would like to check my solution for the development of the real sine function, $f(x) = \sin x$ into the power series. Here is my solution:
First, we have that $\mathcal{D}_f = \mathbb{R}$. Now, we have to check if the given function is infinitely diferentiable and where:
$$f'(x) = \cos x = \sin (x + \frac{\pi}{2}),$$
$$f''(x) = (\sin(x + \frac{\pi}{2}))' = \cos(x + \frac{\pi}{2}) = \sin (x + \frac{2\pi}{2}),$$
$$f'''(x) = (\sin(x + \frac{2\pi}{2}))' = \cos(x + \frac{2\pi}{2}) = \sin(x + \frac{3\pi}{2}),$$
$$\vdots$$
$$f^{(n)}(x) = \sin(x + \frac{n\pi}{2}).$$
So, we, have that given function is infinitely diferentiable on the set $\mathbb{R}$.
Now, notice the point $x_0 = 0$, in which we are going to try to develop the given function.
Because $f$ is infinitely diferentiable, from that follows that $f$ is also continuous with all of its derivatives up to the $n$ - th order, in the arbitrarily neighborhood $U$ of the point $x_0 = 0$, and it has the derivative of the $n + 1$ - st order in $U$. Now, we have that $f$ fulfills the conditions of the theorem from here:
Checking my understanding of the process of developing function into power series
Now, we have that the $f$ can be represented with Maclaurin series:
$$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n(x) =$$
$$= \sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n + R_n(x).$$
Next, I need to examine if $\lim\limits_{n \to \infty} R_n(x) = 0$:
We can notice that $\forall n \in \mathbb{N}$, $\forall x \in \mathbb{R}$, $|f^{(n)}(x)| \le 1$.
Further, we can notice that for $\forall h \in \mathbb{R}$, $f$ is infinitely diferentiable in $[x_0 - h, x_0 + h]$ and that there is a constant $M = 1$, such that $\forall n \in \mathbb{N}$ and $\forall x \in [x_0 - h, x_0 + h]$ it is $|f^{(n)}(x)| \le M$. 
So, We have that conditions of the lema from the question I already linked, are fulfilled. Because of that, we can conclude that $\lim\limits_{n \to \infty} R_n(x) = 0$ for $\forall x \in [x_0 - h, x_0 + h]$, or in this case $\forall x \in \mathbb{R}$.
Conclulsion:
We can conclude that  the function $f(x) = \sin x$, can be developed into the power series 
$$\sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n$$
in the neighborhood of the point $x_0 = 0$.
Edit:
I tried to simplify the power series I got, to try and get the shape suggested in the answer by Chris Custer. I think I did it:
$$\sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n = \frac{0 x^0}{0!} + \frac{1 x^1}{1!} + \frac{0 x^2}{2!} + \frac{(-1) x^3}{3!}+\frac{0 x^4}{4!} + \frac{1 x^5}{5!} + \frac{0 x^6}{6!} + \frac{(-1) x^7}{7!} + \cdots =$$
$$= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} + \frac{x^{13}}{13!} - \frac{x^{15}}{15!} + \cdots = $$
$$= (-1)^0 x^{2\cdot 0 + 1} + (-1)^1 \frac{x^{2\cdot 1 + 1}}{(2\cdot 1 + 1)!} + (-1)^2 \frac{x^{2\cdot 2 + 1}}{(2\cdot 2 + 1)!} + (-1)^3 \frac{x^{2\cdot 3 + 1}}{(2\cdot 3 + 1)!} + (-1)^4 \frac{x^{2\cdot 4 + 1}}{(2\cdot 4 + 1)!} + (-1)^5 \frac{x^{2\cdot 5 + 1}}{(2\cdot 5 + 1)!} + \cdots = $$
$$= \sum\limits_{n = 0}^{\infty} (-1)^n \frac{x^{2n + 1}}{(2n + 1)!}$$
Please, could you tell me if my solution is correct, and if it isn't could you tell me where I made a mistake?
 A: Using Taylor's theorem:   $f(x)=\sum_{n=0}^{\infty}\dfrac {f^{(n)}(0)}{n!}x^n$, as you appear to have done, you should get $\sin x=\sum_{n=0}^\infty(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$.
Note that $\sin n\frac{\pi}2$ takes values $0,1,0,-1$ in succession. 
A: Take this as a supplement, relying on the (omnipotent) power series of the exponential function
$$\exp( z ) \:=\:\sum_{k=0}^\infty \frac{z^k}{k!}\:=\:1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + \ldots$$
in the complex plane, which is given instead of developed here.
Papa Rudin's Prologue(*) is dedicated to it and defines the exponential function by the above series (as do many others), being valid for all $\,z\in\mathbb C\:\!$, in other words, the radius of convergence is $\infty\,$.
Pursuing this definition line one obtains $\,\cos\,$ and $\,\sin\,$ as the real and imaginary part of $\,\exp(ix)$ for $x\in\mathbb R$:
$$\exp(ix)\:=\:\sum_{k=0}^\infty \frac{(ix)^k}{k!}
\;=\;\underbrace{\sum\nolimits_{k=0}^\infty\: (-1)^k\frac{x^{2k}}{(2k)!}}_{\cos x} \:+\:i\:\underbrace{\sum\nolimits_{n=0}^\infty\:(-1)^n\frac{x^{2n+1}}{(2n+1)!}}_{\sin x}$$
This corresponds to the result you have accomplished above.
Furthermore, the "neighbourhood of the point $_0=0$" you mention is the largest possible, namely all of $\mathbb R$.
* Walter Rudin: "Real and Complex analysis" 
