# Why Can’t I Divide By $\sin$?

One of the most common misconceptions I see while working with students in secondary school is the notion of an inverse. The idea isn’t too complicated, but the reason that I see students making mistakes with it is because they are in the process of learning about functions and it becomes a cognitive burden to think about these abstract processes such as inverses and other transformations. However, I firmly believe that giving students the right idea of how these different concepts fit together will help them navigate their classes with ease.

## Example: Trigonometric Functions

I want to start right off with an example that is indicative of why it’s important that you understand inverses. Imagine you had the following problem, and you were looking to solve for $x$:

\[10+3\sin(x) = 11\]There’s nothing particularly nasty about this equation. Like all the other ones you see, you need to isolate for $x$, so you start by subtracting $10$ from both sides.

\[3\sin(x)=1\]From here, you then probably say to yourself, “We need to divide by $3$ on each side.” So, that’s what we do:

\[\sin(x)=\frac{1}{3}\]Now, here’s where things can get tricky if you’re not being careful. Depending on what you’ve been taught, you will have several reactions to this. Unfortunately, the one that usually happens is, “Let’s divide both sides by $\sin$”, giving us:

\[x=\frac{1}{3\sin}\]Let’s state it right now: this is *incorrect*. In fact, you can prove it to yourself by trying to enter this value of $x$ into your calculator. It won’t work (unless, you wrote the $3$ after the $sin$, which then *would* give you an answer, albeit an incorrect one).

I wouldn’t mention this if I haven’t seen it enough before, but I think it captures a misunderstanding of something that’s quite a bit more important than simply saying, “You have to use $\sin^{-1}$ to get the answer.” What I want to give you is a *reason* to understand why what I wrote above is wrong, and this is the concept of an inverse.

## What exactly is an inverse?

Like virtually all topics in mathematics, there are many ways to think about inverses. For the purpose of solving equations, I want to present you this simple first thought:

*An inverse “undoes” whatever you’ve done to an expression. It’s similar to an “undo” button that you might use on a computer.*

This isn’t very precise, so let me give you a more mathematical definition:

*Given a function $f(x)$, its inverse, denoted $f^{-1}(x)$, is defined by the following: \(f^{-1}(f(x))=x\).*

This might still seem a little unclear, but with a few examples, you will understand that this isn’t a groundbreaking concept.

### Example: Find the inverse of $x^2$

To find the inverse, we need to first identify what kind of function is “acting” on $x$. In this case, the function that is acting on $x$ take the form of $f(t)=t^2$. Note that I’ve used the variable $t$ in order to make it distinct from $x$. Then, once we substitute $t=x$ into our equation, we get $f(x)=(x)^2$. I also added parentheses around the $x$ in this equation in order to show you that the thing I’m doing to $x$ is *squaring* it.

So far, so good. We now have to the inverse, $f^{-1}(x)$. To do this, ask yourself the question: how do I get rid of the “squaring” function that is acting on the $x$ at the moment? Remember, our goal is to make a new function that when you substitute $(x)^2$ into it, the result you get is $x$. Try it out for yourself.

The operation we need to do is take the *square root*. As such, we define our new inverse function to be $f^{-1}(t)=\sqrt{t}$. What this means is that I have to take the square root of whatever I put into $t$. For our purposes, we are going to feed it our function $f(x)=(x)^2$, giving us:

In other words, if we had the equation $x^2=4$, we know that solving for $x$ means taking the square root on both sides of the equation, giving us $x=\pm 2$. One way to look at this is to say that you’re taking the *inverse* of the square function, which returns the variable by itself (in this case, $x$).

## Revisiting our example

Let’s go back to our trigonometric example from above. To remind you, we were trying to solve:

\[\sin(x)=\frac{1}{3}\]At this point, you should be looking at this and thinking, “Okay, there’s a function that’s acting on $x$ on the left hand side of the equation. In order to solve for $x$, all I need to do is take the *inverse* of that function.”

Indeed, this is *precisely* the purpose of the inverse sine function, denoted $\sin^{-1}(x)$! It’s purpose is to “undo” the work that the sine function did, and return the angle you originally fed the function (in this case, $x$). Explicitly, this is how the manipulation goes:

One thing that I want to very clearly express: the $-1$ superscript on the function is ** not** an exponent. Instead, it’s just a symbol we use to declare that it’s the inverse function, just like our symbol of a generic inverse function for a regular function $f(x)$ is $f^{-1}(x)$.

## Solving equations is just repeatedly applying inverse functions

Once you understand the idea of an inverse function, you start to see that they are *everywhere*. Indeed, when we solve basically any equation, we are implicitly asking, “How do I *undo* what the equation has done?” If you look back to the example we first started with, $10+3\sin(x) = 11$, we first applied the inverse of $f(t)=10+t$ on both sides, namely $f^{-1}(t)=t-10$. This corresponded to subtracting $10$ from both sides of the equation. Similarly, we applied another inverse function to divide both sides by $3$.

Remember, when you’re trying to solve for a certain quantity, you want to do the inverse of what the equation has done. By looking at solving equations by repeatedly applying inverses, you won’t make the mistake of dividing by $sin$ ever again.

## Final note

I just wanted to include a final remark about inverses here. I didn’t explicitly say it above, but when you’re working with algebra (but not special functions like trigonometric ones), you usually have two different kinds of inverses: additive inverses and multiplicative inverses. They aren’t complicated at all, but they are slightly different.

An additive inverse means that, if you have a certain term that we’ll call $a$, then an additive inverse satisfies the following:

\[a+(-a) = 0\]That’s simple enough, and it usually means just slapping on a negative sign to your term.

Next, we have the multiplicative inverse. If you have a term that we’ll call $x$, then the multiplicative inverse satisfies the following:

\[x* \left(\frac{1}{x}\right)=1\]Once again, nothing too complicated. Just be aware that both exist, and that they are both different “kinds” of inverses. You use a specific one depending on what you’re trying to solve for.