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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.
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.

Austin Broussard
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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 $\text{arcsin}.$ –  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
  • It's true that $1/x=x^{-1}$, but that implies that $\frac{1}{\sin{x}}=(\sin{x})^{-1}$, which is not the same as $\sin^{-1}(x)$, which is just a notation for $\arcsin$ (or viceversa), which is the inverse function of the sine function. The quid here is that "the inverse function $f^{-1}$ to a function $f$ is not computed in some value $x$ as the multiplicative inverse to the value $f(x)$, that is: it is not true that $f^{-1}(x)=\frac{1}{f(x)}$. In particular, it is not true for the sine function. – Bruno Stonek Jul 19 '12 at 23:52
  • So, $y=\dfrac{1}{\sin x}=\arcsin x$? – Austin Broussard Jul 19 '12 at 23:53
  • No, $\arcsin(x)$ is by definition some $t$ such that $\sin(t) = x$. This is not at all the same as $\dfrac{1}{\sin(x)}$, which is $y$ such that $y \sin(x) = 1$. It's just that the notation $f^{-1}$, where $f$ is the name of a function, is interpreted as the inverse function, whereas $x^{-1}$, where $x$ is a number, is the reciprocal. – Robert Israel Jul 19 '12 at 23:54
  • Oh, okay. So $y=\dfrac{1}{\sin x}=(\sin x)^{-1}$. But how would I graph that is my next problem. – Austin Broussard Jul 19 '12 at 23:56
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    For further notes on the confusing notation, see this thread. – J. M. ain't a mathematician Jul 20 '12 at 00:11

3 Answers3

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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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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.

Bruno Stonek
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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 00:03
  • 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 00:07
  • 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 00:07
  • @Austin: Then stick to the first hint which uses no calculus, only knowledge of the sine function. The second hint is just another way to solve the same problem. Now, I ask you this (very related) question: is the function $\frac{1}{x}$ defined in 0? – Bruno Stonek Jul 20 '12 at 00:09
  • What do you mean by is the function $\frac{1}{x}$ defined in $0$ – Austin Broussard Jul 20 '12 at 00:15
  • @Austin, do you know how to plot the function $\frac{1}{x}$? Do you know what the domain of definition of a function is? – Bruno Stonek Jul 20 '12 at 00:17
  • Yes, I know what $\frac{1}{x}$ looks like, and I know what domain is. For instance the domain for $y=\sqrt[3]{x}$ is $[x \in \mathbb R]$. Because in the graph for $\sqrt[3]{x}$ it will eventually cover the $x$ until infinity. – Austin Broussard Jul 20 '12 at 00:24
  • @Austin: Yes, that is right. The question is then, is 0 in the domain of $1/x$? Or to put it differently, can you divide 1 by 0? – Bruno Stonek Jul 20 '12 at 00:26
  • No, you can't divide by $0$. The graph will reach infinity just to the right of $x=0$. If I'm not mistaken. So no, you can not divide by zero because you will get undefined. – Austin Broussard Jul 20 '12 at 00:28
  • @Austin: that's right. Now, I ask again: is $\frac{1}{\sin(2\pi)}$ defined? – Bruno Stonek Jul 20 '12 at 00:29
  • No. Let's see, considering $2\pi =360$ which is also $0$. That would be a no. Are you trying to say that $\frac{1}{\sin x}$ is equivalent to $\frac{1}{x}$? But only when $x=0$? – Austin Broussard Jul 20 '12 at 00:30
  • @Austin: I was just leading you to see that $\frac{1}{\sin(2\pi)}$ is not defined, because $\sin(2\pi)=0$ and $1/0$ is not defined. Now back to the question at hand: you know that in $2\pi$ your function is not defined. What about $\dots, -2\pi,0,4\pi,\dots$? – Bruno Stonek Jul 20 '12 at 00:32
  • Oh, alright. And let's see. $$\sin(-2\pi)=0$$ $$\sin(0)=0$$ $$\sin(4\pi)=0$$ – Austin Broussard Jul 20 '12 at 00:34
  • @Austin: Right. So in those points, $\frac{1}{\sin(x)}$ is not defined. Now try to continue through by yourself. – Bruno Stonek Jul 20 '12 at 00:35
  • $$\sin(\pi )=0$$ $$\sin\left(-\frac{3\pi }{2}\right)=1$$ $$\sin\left(\frac{3\pi }{2}\right)=-1$$ $$\sin\left(-\frac{\pi }{2}\right)=-1$$ $$\sin\left(\frac{\pi }{2}\right)=1$$ – Austin Broussard Jul 20 '12 at 00:40
  • @Austin: Very true. I updated my answer to reflect the fact (which you surely already observed) that the sine function is periodic, and how you can exploit that. – Bruno Stonek Jul 20 '12 at 00:43
  • So the period of this graph is every $\pi$? – Austin Broussard Jul 20 '12 at 00:51
  • I got it now, and I understand everything. Thank you so much for taking your time to help me! – Austin Broussard Jul 20 '12 at 01:05
  • @Austin: you're welcome, I'm glad you understood :) – Bruno Stonek Jul 20 '12 at 01:11
  • Me too. And sorry for the questions. It's just hard to relate with not taking pre-calc and jumping straight to calculus. I also just like to KNOW how to do things. Not just the answers! – Austin Broussard Jul 20 '12 at 01:17
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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.

robjohn
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