### 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 $x$:

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.

From here, you then probably say to yourself, “We need to divide by $3$ on each side.” So, that’s what we do:

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:

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:

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:

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:

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.

### Summer Research

Back when I was in CÉGEP, I spent my summer months working as a gardener. It was a radical change from my usual routine of doing mathematics and physics to planting flowers and cleaning flower beds in the heat of the summer. It wasn’t bad by any means, but it was quite different from what one would normally expect myself to do. In essence, it was a job of convenience. I didn’t hate it, but I did look forward to doing something else. In fact, I remember telling my coworkers for multiple summers that I was working as a gardener only until I could finally work in my own domain of interest.

Fast-forward to now, and I finally have that job I’ve wanted. I’m working as a summer research assistant for one of the professors at my school, and it has been an interesting experience so far. I wanted to post updates on the blog about the things I’m learning, and how I’m finding the job.

So far, I’ve been playing catch-up for the last few weeks. I’m attempting to learn the basics of general relativity, and it hasn’t been a simple task. The reason is that I’ve only taken my basic physics and mathematics courses, so while I do have the mathematical tools of calculus, differential equations, and linear algebra, I don’t have things like calculus of variations, Lagrangian and Hamiltonian mechanics, and so on. As such, it hasn’t been easy to generalize my mathematical tools to curved spaces.

I’ve been learning by going through a textbook, which has its ups and downs. The book is Sean Carroll’s Spacetime and Geometry, and I like the book and the author, but I feel like restricting oneself to only a textbook makes it difficult to learn the material fully. The book packs in a lot of information, but it’s often at the expense of carefully going through each result. I’m sure for those who have more background to the subject, the book would flow well. The trouble is that I can get stuck on a single line for a long time, simply because no explanation is provided and I don’t know what to do. That’s where learning the subject with someone else would be helpful. The ability to ask a question right when it comes up could make learning the material so much easier. This is exactly what I try to do as a tutor when I help younger students. I want them to feel confident in what they’re doing, but be able to ask a question whenever it comes up.

Despite the difficulties, I do like what I’m doing. Learning is something that I truly enjoy, so I am happy to do the difficult job of learning a subject on my own. The ideas are interesting, and the mathematics that I’m learning is also useful, so I know that whatever I learn will indeed be useful. I think the key is to keep on pressing with the material, and by working with the ideas, I’ll get more familiar and comfortable with the subject. I’m enjoying the start of my summer work, and I can see myself doing this kind of stuff for a long time to come.

### The Friends/Strangers Problem

I recently came across a pretty interesting puzzle that used a different kind of mathematics – called Ramsey theory – in order to tackle the problem.

Here’s the problem:

Suppose you have a group of six people. Show that you will necessarily have at least three mutual friends or three mutual strangers.

This is a sort of pigeonhole problem, which sort of means that we have to make a binary choice for each person. But what I love about this proof so much is that it’s entirely visual and fairly easy to grasp.

First, let’s imagine we have six people, who we can each represent as dots.

From there, we want to show the relationships between people. To do this, we will use lines to mark the type of relationship two people have. We will use orange to represent friends, and blue to represent strangers.

To begin, we need to start drawing lines. The first thing that you need to know is that there will always be at least three lines of any one colour that goes out from one person. The reason that this is true is because if we look at one person, the five others have no choice but to be friends or strangers. As such, every person will have at least three orange or blue lines connecting to other people, so we can simply choose one scenario of having three friends, which means three orange lines. We’ll connect them to other people at random, and so we get the following scenario.

Now that we’ve got three lines down, we know that each line represents a friendship between the people. But notice that, to show what we’ve set out to do, we only need to have one of those three friends also be friends. Put differently, if we can find a friendship such that we get a triangle of a single colour, we will prove our statement. However, it’s trivial to simply draw a triangle of one colour and say that we’re done. Instead, we need to show that we’ll be forced to make a triangle of one colour, no matter how much we don’t want to.

Suppose we consider one of those friends who were touched by our original rays. If we look at Person 4, they are friends with Person 1. Additionally, we see that Person 1 is friends with Person 3 and Person 5. Therefore, if we draw a friendship line between either Person 4 and Person 3, or Person 4 and Person 5, we will create a triangle of one colour, which means three people are friends. Since we’re trying to avoid that at all costs, we will draw two blue lines between these people, making sure that no orange triangles are created. This give us the following:

But what, we now have a problem. Look at the line we will need to eventually draw between Person 3 and Person 5. If we draw an orange line between them, then we will create an orange triangle between Person 1, Person 3, and Person 5. But on the other hand, if we draw a blue line between Person 3 and Person 5, we’ll be creating a blue triangle between Person 3, Person 4, and Person 5. Therefore, we’re stuck!

This means that no matter what, a group of six people will have at least three mutual friends or three mutual strangers. Using this technique, we’ve pigeonholed, or forced, ourselves into this situation.

What I like the most about this proof is that it’s visual and easy to understand. It’s what happens when you apply logic at each step, until you arrive at an undeniable conclusion. Furthermore, it’s a glimpse into a type of mathematics that wasn’t shown to me in school, so I thought it would be interesting to share. As I mentioned, I first saw this from a video by Raj Hansen who explains this problem in the introductory video to his Ramsey theory series. You should definitely check it out if you find this proof interesting. Personally, I’m excited to see what other kinds of applications we can find for Ramsey theory.

### Introduction to Vectors

If you were talking to a friend about the drive you made this morning, you might tell them something along the lines of, “I drove from my place to hear in about an hour, going east for the entire time.” Your friend, who knows where you live and has a rough idea of the route to get to where you both are, understands implicitly that you don’t really mean that you went and drove in one direction only for the entire hour. After all, you first had to exit the property of wherever you were parked, then you probably had to navigate through the neighbourhood of where you live, before finally getting onto a highway or a direct route to your destination. It’s only at that point that you actually began going east. Before then, you may have driven in all directions, for varying amounts of time.

In physics, we call this route a trajectory. It’s how you move from one place and get to another. In order to describe the specific trajectory of your drive, we would have to do a lot of specifying, since your trajectory was complicated. But we’ll start off much more simply. Bearing in mind the impossibility of it all, imagine you just had a single block moving in a straight line, like so:

In this case, it’s quite a bit easier to describe how the route of this block will change over time. You don’t even need to be that great with the mathematics of it all for common sense to tell you that the block will move a distance d from wherever it is now by multiplying the speed at which it is moving and the time it travelled. If the block was moving at $5m/s$ and went for $3.5 s$, we would expect for it to move a distance of $d = (5m/s)(3.5s) = 16.5 m$.

This is a simple, but profound idea. We can figure out how far something has travelled by knowing its speed and multiplying it by the amount of time it has travelled for. Simple, right?

Unfortunately, that’s not quite the whole story. As you will see, a lot of learning mathematics is about understanding the special case, before moving on and getting more general. It’s the same story here. In the above example, it was easy to figure out where the block will be after a certain amount of time. Just multiply speed with time! But what if I asked you to look at this:

It isn’t so easy to figure out how far the cannonball will travel in a certain amount of time, even if you know how fast it is to start with. Common sense will once again tell you that, no matter how hard the cannon fires upwards, the ball will eventually fall back down to the ground. As such, the speed the cannonball is moving will change over time. Therefore, we will need a more sophisticated tool to analyze situations like these.

That tool is the vector.

What is a vector? Well, it’s a question that means something slightly different depending on if you’re a physicist, a mathematician, or a computer scientists. For our uses, the most practical way of thinking about a vector is as an arrow. The straight arrows you saw in the above pictures can be considered vectors.

The two main parts of a vector are its magnitude and direction. Direction means exactly what you think it means. In the above examples, the vectors I drew were pointing in the general right direction. So when we were talking about the block and I said it was moving $5m/s$, it would have been more accurate to say, “It’s moving $5m/s$ to the right.

The magnitude of a vector simply means the absolute value of the quantity we’re talking about (we always write magnitudes to be positive). In the first example, the speed of the block was $5m/s$, so that was its magnitude. However, these quantities don’t have to be fixed. Consider again the cannon scenario above. As soon as the cannonball leaves the chamber of the cannon, its speed is changing all the time (and, for that matter, so is its direction). As such, it’s a good idea to remember that while we will start with vectors that don’t change with time, there will come a point in which it is useful to use time.

But how is this helping us with the mathematics?

Fine, I told you about vectors, but how do we describe these situations with vectors? Isn’t it awfully vague to talk about something like a direction?

You’re absolutely right. To help, we need to use a tool that began with Descartes and is invaluable useful today. It’s a way to visualize what is happening in a system, so we don’t have to resort to only writing down equations. I’m talking about, of course, the Cartesian plane, where you’ve undoubtedly seen how to graph things like $y = 6x$ and the like. But now, you’ll find that it’s a fantastic tool for studying vectors.

Let’s go back to the example with the block, but this time, I’ll overlay a grid near the block so we can describe the situation more accurately:

From now on, you’re going to start thinking about space as a sort of grid system. Here, there are two directions: the $x$-direction, and the $y$-direction. Of course, you can go in a combination of these two directions, as the cannonball was doing before. But now that we have this grid system, we easily describe both the magnitude and the direction of any vector. Notwithstanding my bad drawing, the vector above has a magnitude of $5m/s$, and its direction is along the positive $x$-axis. In general, we measure the direction (or angle) of the vector by starting on the positive $x$-axis and rotating counterclockwise until we get to the direction that the vector is pointing.

This is all well and good, but I know that many of you are observant, and might ask, “If we have a vector going diagonally, can we describe it by talking about how much it moves in the $x$ direction and how much it moves in the $y$ direction?”

This highlights an important part of vectors: we can decompose them into their constituent parts. While it might be difficult to think about a certain motion in three dimensions, it’s much easier to think of the motion with a component in each of the $x$, $y$, and $z$ directions. Then, we can calculate things along each of those directions, before finally packaging everything up in the end to give a vector description of the motion.