What am I doing wrong while finding the Laurent series of $\frac{1}{\sin(z)}$? I'm trying to compute the Laurent series of $\frac{1}{\sin(z)}$ at $z_0=0$.
From what I've seen on the internet this is given as 
$f(z)=\frac{1}{z}+\frac{z}{3!}+\frac{7z^3}{360}+...$
My attempt:
$f(z)=\frac{1}{\sin(z)}$ can be rewritten as $f(z)=\frac{1}{z}\frac{z}{\sin(z)}$
We can express $\sin(z)$ as $\sin(z)=z-\frac{z^2}{3!}+\frac{z^4}{5!}+...$
$\therefore \frac{z}{\sin(z)}=(\frac{\sin(z)}{z})^{-1}=(1-\frac{z}{3!}+\frac{z^3}{5!}+...)^{-1}$
To get this to equal what I know it should, then we should have $(\frac{\sin(z)}{z})^{-1}=(1+\frac{z}{3!}+\frac{7z^3}{360}+...)$ but how can this be what do I do with my negative sign and how do I obtain the $\frac{7}{360}$ coefficient?

You are wrong about $\frac{\sin z}z$. Its Taylor series is$$1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots$$Since it is an even function, the Taylor series of $\frac z{\sin z}$ is of the type $a_0+a_1z^2+a_2z^4+\cdots$. So, you have$$\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots\right)(a_0+a_1z^2+a_2z^4+\cdots)=1,$$which means that$$a_0+\left(a_1-\frac{a_0}{3!}+\right)z^2+\left(a_2-\frac{a_1}{3!}+\frac{a_0}{5!}\right)z^4+\cdots=1.$$This, in turn, means that$$a_0=1\text{, }\ a_1-\frac{a_0}{3!}=0\text{, }\ a_2-\frac{a_1}{3!}+\frac{a_0}{5!}=0$$and so on. And from this you can compute $a_0,a_1,a_2,\ldots$
