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March 13, 2021

Proving Identities

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Textbook Reference: Section 7.1 Solving Trigonometric Equations with Identities

Take a moment and think about how well you've gotten to know the unit circle and the circular functions so far this semester.

Seriously, take a moment. I'll tell you why.

When learning math, it can sometimes feel like you're treading water as you try to keep up with everything — all the definitions, formulas, and techniques can be a bit overwhelming at times. So it helps to take stock of what you've done thus far:

  • You've learned how the unit circle gives us a more "active" way of looking at angles, and how those angles can be any real number — not just limited to between \(0^\circ\) and \(360^\circ\), but any real number, positive or negative.
  • You've learned how radians give a more natural way of measuring angles, because they're not based on an arbitrary number like \(360\).
  • You've learned how the circular functions give us useful measurements about these angles
    • The sine measures the height (the \(y\)-coordinate), and the cosine measures the overness (the \(x\)-coordinate).
    • The tangent measures the slope.
    • The other three functions — cosecant, secant, and cotangent — are reciprocals of the main three.
    • The inverse circular functions let you find an unknown angle given the value of one of the functions (though it only gives you one of multiple possible angles).
  • You've learned how to solve equations with circular functions, including seeing how the equation-solving techniques you've learned in previous classes fit in with these new functions.
  • You've learned some applications of these functions:
    • The sinusoidal functions (sine and cosine) functions can model all sorts of situations where things fluctuate back and forth.
    • The functions can be used in trigonometry — measuring triangles — using simple tools like the Pythagorean Theorem and SOH-CAH-TOA for right triangles, as well as more powerful tools like the Law of Sines and Law of Cosines for oblique triangles.
  • You've learned important identities, capturing fundmental properties of the circular functions and relationships between them.

To finish up our main study of the circular functions, we're going to dive deep into these identities. In particular, you're going to be proving them. And there's a very good reason:

In some sense, this is going to be your first taste of real mathematics.

In previous classes, you may have been used to learning a number of techniques, where all you have to do is pick the right technique for the problem and chug through it to get an answer.

Proving identities is a bit different.

With proving identities, you're going to find that it may not be immediately obvious (at least at first) how to proceed with a problem. You might have to try a few different things. You might have some false starts before you find something that works. It's less of a science and more of an art.

It's also worth keeping in mind that one of the goals of this course is to prepare you for calculus (MAT 131, 132, 233, and 234 at Oglethorpe). When you take calculus, and you learn about "integrals," you'll find yourself in this "less-of-a-science-and-more-of-an-art" situation again.

And if you choose to go further in mathematics — and I definitely hope that some of you do — you'll find that more and more of mathematics is about less about following a set of steps and more about playing with things until they work.

So let's get started.

Simplifying expressions

One of the benefits of having so many identities is that we can use them to simplify expressions. You might have an expression that looks incredibly complicated, but if you do a little bit of work, you can replace it with a much simpler expression that works just as well.

For example, consider the following equation: \[\cos(t)+\tan(t)\sin(t)=2\] None of our usual techniques work for this equation, so we might seem out of luck. But let's play with this expression a bit.

First, remember when I said earlier that everything really just comes down to sines and cosines? Let's make use of that, and rewrite our tangent as a ratio of sine to cosine: \[\cos(t)+\left(\dfrac{\sin(t)}{\cos(t)}\right)\sin(t)=2\] We can combine everything in the second term to a single fraction: \[\cos(t)+\dfrac{\sin^2(t)}{\cos(t)}=2\] If we rewrite the first term as \(\dfrac{\cos^2(t)}{\cos(t)}\), we can combine both terms into a single fraction: \[\dfrac{\cos^2(t)+\sin^2(t)}{\cos(t)}=2\] Hey, that numerator looks familiar! One of our Pythagorean identities tells us that numerator is just equal to \(1\): \[\dfrac{1}{\cos(t)}=2\] And now we can recognize one of our reciprocal identities — this is just the definition of the secant function! \[\sec(t)=2\] This is now much easier to solve than our previous equation. (It's equivalent to saying that \(\cos(t)=\tfrac12\), so the solutions are \(t=\dfrac\pi3+2\pi n\) and \(t=\dfrac{5\pi}3+2\pi n\).)

Awesome! We turned our difficult problem into a much easier problem. There's just one problem:

How the heck did we know to do those particular steps?

The best answer I have for you is:

It looked like it might work.

I know that answer's probably a bit frustrating, but hear me out. These are the kinds of things that you start to get a feel for after working with identities — you start noticing the relationships between different functions, and you start getting an idea of what might work. For example, I knew to try this in particular because I saw a tangent multiplied with a cosine, and I had a hunch that things might cancel when I combined the fractions.

The techniques you'll learn over the next couple of weeks will help you develop that intuition.

Besides, it's worth keeping in mind that whenever you see these nice, clean proofs in textbooks, what you usually don't see is all the false starts and failed attempts before the "a-ha" moment that led to the correct solution. Those failures are an important part of doing math — it's just as important to know what didn't work as it is to know what did.

How to prove identities

Suppose we have an identity like this: \[\dfrac{\cos(x)}{1-\sin(x)}=\dfrac{1+\sin(x)}{\cos(x)}\] How can we prove that it's true?

One way to do it would be to play with one side — say, the left side — and attempt to turn it into the other one through a series of manipulations — what I like to call "algebraic massage."

We can start by multiplying the left side by \(\dfrac{1+\sin(x)}{1+\sin(x)}\): \[\dfrac{\cos(x)}{1-\sin(x))}\cdot\dfrac{1+\sin(x)}{1+\sin(x)}=\dfrac{\cos(x)(1+\sin(x))}{(1-\sin(x))(1+\sin(x))}\] Now let's multiply out the denominator: \[=\dfrac{\cos(x)(1+\sin(x))}{1-\sin^2(x)}\] Hey look, we can use a Pythagorean identity! Since \(\sin^2(x)+\cos^2(x)=1\), we know that \(1-\sin^2(x)=\cos^2(x)\): \[=\dfrac{\cos(x)(1+\sin(x))}{\cos^2(x)}\] Seeing a factor of \(\cos(x)\) in both the numerator and the denominator, we can cancel them out (since \(\dfrac{\cos(x)}{\cos(x)}=1\)): \[=\dfrac{1+\sin(x)}{\cos(x)}\] And look — we have the other side of what we wanted to prove. We did it!

What about cross-multiplying?

You might have seen an easier way to establish the identity — couldn't we just cross-multiply? \[\cos(x)\cos(x)=(1+\sin(x))(1-\sin(x))\] Then you could multiply everything out: \[\cos^2(x)=1-\sin^2(x)\] Finally just add \(\sin^2(x)\) to both sides: \[\sin^2(x)+\cos^2(x)=1\] This is one of our known Pythagorean identities! Since we've simplified our original identity to something we already know is true, our original identity must have been true.


Well... you have to be REALLY careful with this line of thought.

Doing the same thing to both sides is usually something you're only allowed to do if you already know both sides are equal. Otherwise you can run into "proofs" like this: \[ \begin{align*} \sin(x) &= -\sin(x)\\ \sin^2(x) &= \sin^2(x) & \text{(Square both sides)} \end{align*} \] Just because \(\sin^2(x)=\sin^2(x)\) is true does NOT mean that \(\sin(x)=-\sin(x)\) is true. (If \(x=\dfrac\pi2\), then this would be claiming that \(1=-1\).) So what happened?

The problem is that squaring both sides isn't a reversible step. If we square both sides, there's no guarantee that the original two expressions were equal — they might have been negatives of each other. More technically, squaring isn't a one-to-one function.

So, most textbooks will tell you that you can just never ever work with both sides of the equation at the same time, as a rule.

That's where I have a different opinion.

As far as I'm concerned, you ARE allowed to work on both sides of an equation at the same time, as long as you ensure that all of your steps are reversible.

If you want to know more about why I'm going against most of the textbooks, I wrote a blog post about exactly this issue a few years ago, which goes into more detail.

Preview Activity 14

Answer these questions and submit your answers as a document on Moodle. (Please submit as .docx or .pdf if possible.)

  1. Simplify the expression \(\dfrac{\sin(x)\sec(x)}{\tan(x)}\) as much as possible.
  2. Show that \(\tan(\phi)+\cot(\phi)=\sec(\phi)\csc(\phi)\). (Start by writing the left in terms of sines and cosines, and then make sure both fractions have a common denominator.)
  3. Use Desmos to graph the equations \(y=\dfrac{\cos(x)}{1+\sin(x)}\) and \(y=\dfrac{1-\sin(x)}{\cos(x)}\).
    1. What do you notice about the graphs?
    2. Suppose one of your friends claims to have proven that \(\tan(x)(1+\sin(x))=\dfrac{\sin(x)\cos(x)}{1+\sin(x)}\). Are they likely to be correct? Why or why not?
  4. Based on what you've seen so far in the reading and the exercises, what techniques do you notice that can be used to simplify expressions with circular functions and prove identities?
  5. Answer AT LEAST one of the following questions:
    1. What was something you found interesting about this activity?
    2. What was an "a-ha" moment you had while doing this activity?
    3. What was the muddiest point of this activity for you?
    4. What question(s) do you have about anything you've done?

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