Graph $y=1/\sin x$ 
Graph $y=\dfrac{1}{\sin x}$

Now, I looked at the graph on google and got this 
Which I thought that $y=\dfrac{1}{\sin x}$ would be $y=\sin^{-1}x$ But it's apparently not. So if anyone can shed some light on this. It's not just about finding a graph and copying it. I would like a better understanding of this. Also, I know the format for graphing trig functions is $y=a\sin k(x+c)+d$. But I don't understand how to  fit the OP into this.
Edit:
So, the OP is $y=(\sin x)^{-1}$ or the inverse of $\sin x$. Now, all I need help with is how to graph accordingly.
 A: Just to put it all in one place:
if there were any justice, nobody would ever say
$$
\sin^{-1}(x).
$$
They would say
$$
\arcsin(x)
$$
if they meant the inverse of the sine function, and
$$
\frac{1}{\sin(x)}
$$
if they meant the reciprocal of the sine function. 
Because there is no justice, there's a horrible rule that $\sin^{-1}(x)$ means the inverse and not the reciprocal. This makes no sense becuase $\sin^2(x)$ means the square of the sine function. This unnecessarily confuses students who are already often confused about the difference between an inverse and a reciprocal. It's just... it's just the worst. I need to go drink a glass of water, I'm foaming at the mouth a little.
A: This answer supposes the OP actually wants to plot $\frac{1}{\sin(x)}$, not $\arcsin(x)$.
HINT: what is the value of the sine function on $\dots, -2\pi, -\pi, 0,\pi, 2\pi, \dots$ and on $\dots, -3\pi/2,-\pi/2,\pi/2,3\pi/2,\dots$? What's its behavior in between these points (positive/negative, increasing/decreasing)?
This is a craftman's hint to the question, which exploits the fact that we know exactly how $\sin(x)$ behaves.
If we are more clever, we can also exploit another fact of the sine function, the fact that it is periodic. You know that $\sin(x+2\pi)=\sin(x)$ for all $x\in \mathbb{R}$. This allows you to plot the sine function just on $[0,2\pi)$ and then "copy it" appropiately to get the graph on all $\mathbb{R}$.
We can exploit this in this case too, since $\frac{1}{\sin(x+2\pi)}=\frac{1}{\sin(x)}$ whenever $\sin(x)$ doesn't vanish. This means that, to plot $\frac{1}{\sin(x)}$, you might just as well plot it on the points of $[0,2\pi]$ where it is defined, and then copy the graph appropiately. This might make it easier, and puts on paper what you surely observed the moment you looked at the graph, that is, it is the same on $[-2\pi,0]$ and on $[0,2\pi]$.

There is another approach which is more mechanical and uses calculus:
HINT: Find out where the function is defined. At the points where it is not defined, find out the lateral limits. Now what calculus tool lets you find out if a function is increasing/decreasing? Compute it, and you will also find the local maxima/minima.
A: The notation is a bit misleading. Instead of $\dfrac1{\sin(x)}$ (the reciprocal of the sine), what $\sin^{-1}(x)$ means is the arcsine, the inverse function of sine.
