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

Now, I looked at the graph on google and got this y=1\sin(x)

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.
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.

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$\sin^{-1} x$ is one of those poor notations given that, for example, $\sin^2 x = (\sin x)^2$. $\sin^{-1} x$ means the inverse sine of $x$ (i.e. $\arcsin x$), not $1/\sin x$. –  Henry T. Horton Jul 19 '12 at 23:41
Okay. Thanks. But how would I graph this? Using the given equation. But I also thought that $\dfrac{1}{x}=x^{-1}$ Thus, $\dfrac{1}{\sin x}=\sin^{-1}x$ –  Austin Broussard Jul 19 '12 at 23:42
aka $\text{arcsin}.$ –  user2468 Jul 19 '12 at 23:48
Right, so back to the OP; $y=\dfrac{1}{\sin x}=\arcsin x$, right? But, if that's the case. Then those two equations are two different graphs –  Austin Broussard Jul 19 '12 at 23:50
For further notes on the confusing notation, see this thread. –  Ｊ. Ｍ. Jul 20 '12 at 0:11

3 Answers 3

up vote 2 down vote accepted

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.

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So, you're saying plot the points of all the sine functions and then connect the dots in a sense? –  Austin Broussard Jul 19 '12 at 23:57
@Austin: I believe you meant to say "plot the sine of all those points", which is not quite the same. Well, plotting some crucial points is never enough to plot a function (you could "connect the dots" in so many different ways!), but that is the purpose of my second question. If you know the value of the function on some points, and you know the behavior of the function in between, you can then make a reasonable plot. –  Bruno Stonek Jul 19 '12 at 23:59
I mean like, how $\sin (2\pi)=0$ I would then put a dot there? and so on. But that wouldn't seem to work. Your second question about the behavior. Can you elaborate on the question without giving the answer, I need to try and figure this out myself! Thanks. –  Austin Broussard Jul 20 '12 at 0:03
Be careful! You're not plotting $\sin(x)$ but $\frac{1}{\sin(x)}$. If $\sin(2\pi)=0$, then what happens with $\frac{1}{\sin(x)}$ in $2\pi$? –  Bruno Stonek Jul 20 '12 at 0:07
Now, this can be a problem considering this is a calculus assignment when I have not taken the course yet and also, I skipped pre-calculus. So I'm trying to answer this question with only trigonometry skills. Which is probably why I'm completely confused. And would that make it $(-2\pi)$? –  Austin Broussard Jul 20 '12 at 0:07

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.

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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.

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Of course, you then have those people who use $\log^2 x$ to denote the 2-fold iterate of the logarithm, $\log\log\,x$. Oh, notation... –  Ｊ. Ｍ. Jul 20 '12 at 0:19

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