# Is the graph of $(\cos x)(\sec x)$ discontinuous?

I suspect that the graph of $\cos x\times \sec x$ must be discontinuous because,at $x=90^0$ the function becomes $\frac{0}{0}$ so it must be undefined or just point to nothing on the graph at that point.

But,Desmos and Wolfram gives me the plot of the function as a straight line passing through $(0,1)$ without any discontinuation.

So,which one is correct?Am I wrong in thinking that $\cos x\times\sec x=1$ is not true for all values of $x$?

Wolfram-

Desmos-

Note:-Same goes for $(\sin x)(\text{cosec} x)$ and $(\tan x)(\cot x)$.

• It might be better to format the function as $(\cos x)(\sec x)$ or simply $\cos x \sec x$ rather than using the $\times$ operator, to avoid giving the impression you are forming a Cartesian product of two sets. – hardmath Aug 2 '16 at 16:44
• You could try $y=\dfrac{x}{x}$ in Wolfram Alpha to see that it does not always capture the obvious – Henry Aug 2 '16 at 16:44
• @AakashKumar It's not discontinuous, it is undefined. Those are two different things. – Arthur Aug 2 '16 at 16:46
• @Arthur Sorry , you are correct . – Aakash Kumar Aug 2 '16 at 16:47
• @Arthur: I would want to avoid saying that $f$ is discontinuous at $\pi/2$. However, the sentence "It is not the case that $f$ is continuous at $\pi/2$" is true. – André Nicolas Aug 2 '16 at 17:02

To simplify things, consider the function $f(x)=\frac{x}{x}$. For all $x\not=0$, $f(x)=1$. But, $f(0)$ is undefined. Thus, $f$ is equivalent to $1$ except at $x=0$ where it is undefined. In the same way, $\cos x\times\sec x=\frac{\cos x}{\cos x}$ is $1$ for all $x$ such that $\cos x\not=0$. I.e., for all $x$ except $x=\pm \frac{\pi}{2},\pm\frac{3\pi}{2},\ldots$. On a graph you can indicate this by interrupting the line $y=1$ with hollow circles at these points indicating that the function is undefined there.

• This is the true answer. Just as the more famous (and counterintuitive) example $\frac 1x$, the function doesn't have any discontinuities. It just has points where it's not defined, and is continuous everywhere else. – Arthur Aug 2 '16 at 16:44

It depends what you mean.

If you are considering $\cos(x) \sec(x)$ to be the partial* function whose values are given by mechanically inserting a value for $x$ and computing the indicated sequence of operations, then yes, its graph should have removable discontinuities at every $x = (2n+1) \pi/2$.

(note that there are no discontinuities in $x \in [-1,1]$, but that's moot since you get the same behavior from wolframalpha on an interval that does contain a discontinuity)

*: Throughout the post, I follow the common abuse of notation and say "function" when I mean "partial function".

However, that is often not what one means by such notation. One example is that one might instead intend to continuously extend the resulting function, in the same way we might say $x/x=1$.

Here, we are treating $\cos$ and $\sec$ as mathematical objects in their own right, and performing arithmetic on those objects rather than on their values.

We can define multiplication of two functions $f$ and $g$ to be the function defined in the following way:

• First compute the function $h$ defined by pointwise multiplication.
• Let $S$ be the set of removable discontinuities of $h$
• The product of $f$ and $g$ is the function given by

$$x \mapsto \begin{cases} h(x) & x \notin S \\ \lim_{y \to x} h(y) & x \in S \end{cases}$$

With this definition of product, $\cos(x) \sec(x) = 1$.

How would you prefer for these graphing programs to illustrate the hole? With a hollow dot? With a break in the curve/line? Then how wide should the dot be? How wide should the break be? Any positive width is perhaps inherently misleading.

But more fundamentally, how should these programs know there is a hole there in the first place? If at some stage, it tries to evaluate at $x=\pi/2$, then it would be reasonable for the software to recognize there is no output there. But there is a continuum of $x$-values to display, and the software can only make finitely many calculations before it plots. So it is bound to simply never have looked directly at the $x$-values corresponding to some of the holes. Instead it will have calculated at some $x$-value to the left, and some other tot he right, and just interpolated a small line segment between them.

• WolframAlpha has access to a sophisticated symbolic manipulation package; it's not limited to simply plugging in values and hoping for the best! – user14972 Aug 2 '16 at 17:21
• How do you show that discontinuty in a graph? – tatan Aug 2 '16 at 17:22
• @Hurkyl But it is still limited. Find some function $f$ with zeros all over the place in a pattern that has not been studied and put into some specific package, and then it will certainly fail to find the holes in $\frac{f(x)}{f(x)}$. Furthermore, missing the holes can be precisely the fault of the symbolic manipulation. When $\cos(x)\sec(x)$ is symbolically simplified to $1$, that's how holes get missed. – alex.jordan Aug 2 '16 at 17:47

As others said, it depends what you mean. Here, $\cos x$ and $\sec x$ are both meromorphic functions on $\mathbb C$. Their product as meromorphic functions is the meromorphic function $1$ on $\mathbb C$. So, when you consider them not pointwise, but instead as meromorphic functions, the product has no discontinuities.

• What is $\mathbb C$?I am not too confident in complex analysis...so can you make your answer more informative...it will help me a lot..thanks... – tatan Aug 3 '16 at 5:12
• From your answer and the other answers,I am a little confused....you said that as it is a meromorphic function,it must be $1$...so,graph must be straight line...other answers say that it is not $1$ as the function is discontinuous as $\lim_{x\to\frac{\pi}{2}}\cos x\sec x\neq f(\frac{\pi}{2})$ as it is not defined there...So,which one should I learn myself or teach others?$\cos x\sec x=1$ or $\neq1$ or both are equally correct? – tatan Aug 3 '16 at 13:27