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

Question

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 $(\tan x)(\cot x)$.

Answer 1

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.

Answer 2

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".

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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, $\cos(x) \sec(x) = 1$.

Answer 3

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.

Answer 4

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.