Understanding ProofWiki's proof of differentiation of sine power series At ProofWiki's page on the derivative of the sine power series
I've read that being $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ by using the Power Series Differentiable on Interval of Convergence result we get $\sin'(x)=\sum_{n=0}^{\infty}(2n+1) \frac{(-1)^n}{(2n+1)!}x^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=\cos(x)$ but the page Power Series Differentiable on Interval of Convergence result says that if $f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$ then inside the radius of convergence of the power series we have $f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}$ so shouldn't it be $\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$?
Now I understand that this last result has a missing term due to the index of the series being $1$ and not $0$ but being the $1$ mandatory because of the above theorem and since trying to shift index doesn't help (we get $\sin'(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}$) how can the result that ProofWiki shows be rigorously justified? 
(I mean, I think we can't just say "oh, there's a missing term!" after having applied the theorem and magically shift the index of the series to obtain the desired result)
 A: Let's look at a really simple example:
$$\sum_{n=0}^\infty x^n = 1 + x + x^2 + \cdots$$
If we differentiate this without adjusting the index to $n = 1$, we get the following:
$$\sum_{n=0}^\infty nx^{n-1} = \frac{0}{x} + 1 + 2x + \cdots$$
Even though in this case the term $\dfrac{0}{x}$ disappears, so our sum isn't technically incorrect, you have to be careful in general, and when differentiating we should drop the $x^{0}$ term entirely:
$$\sum_{n=1}^\infty nx^{n-1} =  1 + 2x + \cdots$$
But in the case of
$$\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1} = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 + \cdots$$
we don't have a $x^0$ term. So we don't have to drop anything.
A: You are not accurately applying the theorem.
On the one hand we have a series (namely, the sine power series)
that is conventionally written in the form
$$\sum_{n=0}^{\infty}b_n(x-a)^{2n+1}.$$
On the other hand, the theorem about differentiability uses the notation
$$\sum_{n=0}^{\infty}c_n(x-a)^n.$$
You can't just equate $b_n$ with $c_n$ and assume the summation index in the theorem applies directly to the sine power series, because in most cases
$2n+1 \neq n.$
What really matters when looking at a power series is not how we use a summation index to generate terms, but exactly what is the power of $x$ in 
each term.
The theorem says that the $x^0$ (constant) term of $f$ contributes $0$ to the derivative,
the $x^1$ term $c_1x$ contributes $(1)c_1x^{1-1} = c_1$ to the derivative,
the $x^2$ term $c_2x^2$ contributes $(2)c_2x^{2-1} = 2c_2x$ 
to the derivative, and so forth.
In the sine power series, the $x^0$ term is zero.
To construct the $x^1$ term, set $n = 0,$ so that
$$\frac{(-1)^n}{(2n+1)!}x^{2n+1}
 = \frac{(-1)^0}{(2(0)+1)!}x^{2(0)+1}
 = \frac{1}{1!}x^1 = x.$$
The theorem says that due to this $x^1$ term of $f,$
the power series of $f'$ has an $x^{1-1}$ (constant) term
equal to $1.$
The first non-zero term of the correct series is not the $x^2$ term,
although your formula says that is the first term.

To put it another way, in order to apply the theorem correctly, we
set $c_0 = 0,$ $c_1 = \frac{1}{1!},$
 $c_2 = 0,$ $c_3 = -\frac{1}{3!},$
 $c_4 = 0,$ $c_5 = \frac{1}{5!},$ $c_6=0,$ and so forth.
That is,
$$\sin(x) =
0x^0 + \frac{1}{1!}x^1 + 0x^2 - \frac{1}{3!}x^3 + 0x^4 + \frac{1}{5!}x^5
 + 0x^6 +\cdots,$$
and the theorem says that
\begin{align}
\frac{d}{dx}\sin(x) 
& = 1c_1x^0 + 2c_2x^1 + 3c_3x^2 + 4c_4x^3 + 5c_5x^4 + 6c_6x^5 + \cdots\\
&= 1\left(\frac{1}{1!}\right)x^0 + 2(0)x^2 
- (3)\left(\frac{1}{3!}\right)x^3 + 4(0)x^4 + 5\left(\frac{1}{5!}\right)x^5
 + 6(0)x^6 -+\cdots \\
&= \frac{1}{0!}x^0 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 -+\cdots \\
&= \cos(x).
\end{align}
