What happened to the domain here?

Question

If we start with:

$$\dfrac{1-\cos(2\theta)}{\sin^2(2\theta)}\;,$$

we can simplify that to , $$\dfrac{1-(1-2\sin^2(\theta)}{4\sin^2(\theta)\cos^2(\theta)}\;,$$

which simplifies further to

$$\dfrac{\sin^2(\theta)}{2\sin^2(\theta)\cos^2(\theta)}\;.$$

Now, I can divide both sides by $\sin^2(\theta)$, (given that it doesnt equal $0$), and arrive at the answer $\dfrac{1}{2}$$\sec^2(\theta)$.
Now, surely the domain resitriction I must apply is all the values of $\theta$ for which $\sin^2(\theta)$ equals $0$, such as $\pi$.
Now this is the bit that makes little sense, if I put $\pi$ into the first equation, I get an undefined answer, but in $\dfrac{1}{2}$$\sec^2(\theta)$, I get $0.5$, which is what desmos plots.

So my question is, what is happening here?

Answer 1

What you have calculated, put in more precise words, is this:

For any real number $\theta$ such that $\sin^2(2\theta)\neq 0$, we have $$ \frac{1-\cos(2\theta)}{\sin^2(2\theta)}=\frac{1}{2}\sec^2(\theta). \tag{1} $$

The fact that $(1)$ does not hold when $\theta=\pi$ does not falsify the above result you have shown, because $\sin^2(2\theta)=0$ in that case.

Answer 2

I want to address what Desmos is doing, as the other answers have explained what's going on algebraically.

Desmos is actually pretty smart when it comes to functions with holes in their domain: it can visually distinguish between "points" and "holes" on the graph of a function. To see this, try clicking and holding on a point on the graph of your function. You should see something like this:

Now drag the point to the left, and see what happens when you reach $x = 0$:

Do you see how the point style changes from a solid circle to an open one, and the point label changes to indicate that the function is undefined there?

Now let's see what happens when you click and drag near $x = \pi$:

Why isn't Desmos displaying this "hole" as an open circle? It's because we can't drag the point exactly onto the point where the $x$-coordinate is $\pi$. Desmos is describing the point using a decimal approximation, and the best we can do is get close to $\pi$.

However, you can confirm that Desmos does actually know that the function is undefined there by explicitly asking (in the Algebra pane) for an evaluation of $f(\pi)$:

Once you know that there are supposed to be holes in the graph of the function whenever $x$ is a multiple of $\pi$, you can manually add them to the graph by plotting a sequence of points and setting the point style to be an open circle:

amends the graph as follows (only a portion is shown below):

Answer 3

This is a conceptual question.

Generally, suppose $f$ and $g$ are functions from some set $D$ to the real line $\mathbb{R}$. Let $E$ be the subset of $D$:$E=\{x\in D |g(x)=0\}$. When you try to simplify $\frac{f(x)}{g(x)}$, you need to calculate it on the set $D\setminus E$ (where $g(x) \neq 0$).

Here's a simple example, let $f(x) = x^3$, $g(x) = x^2$, and $x \in \mathbb{R}$. Then, $\frac{f(x)}{g(x)} = x$ (for $x \neq 0$), the right hand side $x$ is defined on $\mathbb{R}$, but $\frac{f(x)}{g(x)}$ is defined on $\mathbb{R} \setminus \{0\}$. There is no essential difference between this example and the question you asked.