No it's not correct. The Laurent series of $\sin(z)$ around $z=1$ is given by $$\sin(z)=\sum_{k=0}^\infty \frac{f^{(k)}(1)}{k!}(z-1)^k,$$
where $$f^{(k)}(1)=\begin{cases}\sin(1)&k\in 4\mathbb N\\ \cos(1)&k\in 4\mathbb N+1\\ -\sin(1)&k\in 4\mathbb N+2\\ -\cos(1)&k\in 4\mathbb N+3.\end{cases}$$
Power (or Taylor or Maclaurin) series are special cases of Laurent series.
By their uniqueness properties (which are often underplayed), any process that produces any such series as an output produces the same thing (where we mean that the center point is the same, and inner radius of an annulus is the same, etc.)
So, yes, a holomorphic function on a disk has Laurent series which is actually a power series: there are no negative-degree terms.
(A function holomorphic on an annulus need not extend to a function with an isolated singularity at the center, and certainly need not extend to a meromorphic function on that punctured disk, to have a Laurent expansion on the annulus.)
So, yes, $\sin z = z+ z^3/3!+...$ is both the power series expansion at $0$ and the Laurent expansion at $0$... with no negative terms.
A-priori, sure, we know that ${\sin z\over z}$ is meromorphic on the complex plane punctured at $0$, so has a Laurent expansion. But/and the singularity at $0$ is removable, so it has a power series expansion (special type of Laurent). Further, we can obtain that power series (or Laurent series, in more complicated cases) by dividing the power series for $\sin z$ by $z$. Bingo. (Since the power series for $\sin z$ has no zero-degree term, the quotient of that power series by $z$ is still a power series... not a Laurent series.)
If you wanted the Laurent expansion of ${\cos z\over z}$ at $0$, divide the power series $\cos z = 1 + z^2/2!+\ldots$ by $z$ to have
$$
{\cos z\over z} \;=\; {1\over z} + {z\over 2!} + \ldots
$$
Here, there is a negative-index term.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{{\pi \over \sin\pars{\pi z}} = \lim_{m \to \infty}\sum_{n = -m}^{m}{\pars{-1}^{n} \over z - n}}:\ {\Large ?}}$.
With $\ds{m \in \mathbb{N}_{\ \geq\ 1}}$: \begin{align} &\bbox[5px,#ffd]{\pi \over \sin\pars{\pi z}} = {1 \over z} + {1 \over z}\bracks{{\pi z \over \sin\pars{\pi z}} - 1} \\[5mm] = &\ {1 \over z} + {1 \over z}\bracks{% \lim_{m \to \ \infty}\sum_{{\Large n\ =\ -m} \atop {\Large n\ \not=\ 0}}^{m} a_{n}\pars{{1 \over z - n} + {1 \over n}}} \end{align} where $\ds{a_{n} \equiv \mrm{Res}\pars{{\pi z \over \sin\pars{\pi z}} - 1, z = n} = \lim_{z \to n} \braces{\pars{z - n}\bracks{{\pi z \over \sin\pars{\pi z}} - 1}} = \pars{-1}^{n}\, n}$.
See Mittag-Leffler Expansion. Then, \begin{align} &\bbox[5px,#ffd]{\pi \over \sin\pars{\pi z}} = {1 \over z} + {1 \over z}\bracks{% \lim_{m \to \ \infty}\sum_{{\Large n\ =\ -m} \atop {\Large n\ \not=\ 0}}^{m} {\pars{-1}^{n}\,z \over z - n}} = {1 \over z} + \lim_{m \to \ \infty}\sum_{{\Large n\ =\ -m} \atop {\Large n\ \not=\ 0}}^{m} {\pars{-1}^{n} \over z - n} \\[5mm] = &\ \bbx{\large\lim_{m \to \ \infty}\sum_{n\ =\ -m}^{m} {\pars{-1}^{n} \over z - n}} \\ & \end{align}