A nice proof for the Law of Cosines

We’ve been working on the Law of Cosines in my precalculus classes. And I am having them prove it by scaffolding up from specific triangles to more general triangles. And then with the most general triangles, students consider acute, obtuse, and right triangles.

Kids tend to struggle a bit on the first triangle, but as soon as they realize they need to draw an altitude, they see all that opens up with right triangles and are good. After that, for the rest of the concrete ones, they tend to breeze through. The place where they first stumble again is when they get to the fifth triangle, the one with the angle \beta. They get to L^2=(5\sin\beta)^2+(4-5\cos\beta)^2 but then don’t go any further. But since I know I want them to get to the law of cosines, I tell them to expand and look for something nice. Sometimes I’ll give them the answer (L^2=41-40\cos\beta) and then say: work your work until it looks like this, with one trig function in it. From that point on, kids are in the zone.

For years, I used to teach this by giving kids waaaay too much information.


And I kinda told them what to do… Meh. I was jumping way ahead to get to the formula. We weren’t savoring the thinking to get to the formula. Now we are.

That being said, I ran across something quite beautiful. A stunning proof of the Law of Cosines (at least for acute triangles) on the site trigonography.


I love it because it looks like a proof for the Pythagorean theorem.  Which is nice because  the Law of Cosines is essentially a more generalized version of the Pythagoren theorem.

The area of the bottom square (the green one) is clearly the area of the top two squares (the red and blue ones) minus two green areas. Ummm…. c^2=a^2+b^2-2ab\cos(C) anyone? You see the c^2 and the a^2 and b^2, but what you also see is that you’re taking out some area (the green bits). [1]

When introducing it to my class, I showed them this image:


and said it was a proof of c^2=a^2+b^2-2ab\cos(C).

And I just said to observe. To make statements based on what they see visually…Anything and everything. And if students could, see if they could make connections to the equation (but without writing anything down). After a short while of observations, I opened this geogebra applet and played around. I showed them what happened when we made angle C a right angle.


They saw the green rectangles disappear, and how this would be a proof of the Pythagorean theorem if the blue areas and the pink areas were equal to each other. And then I squiggled and smushed the triangle about and eventually kids conjectured that the blue rectangle areas were equal, the pink rectangle areas were equal, and the green rectangle areas were equal. I told them that was true, and they were going to prove that. But before doing that, I asked them: if this was true, do you see a connection between this diagram and the law of cosines?

And kids eventually got there. They saw this argument, essentially…


…and they realized that the green rectangles were probably the thing that was being subtracted out in the law of cosines!

At this point, I gave my kids a blank paper copy of the diagram, and groups work on proving that the blue rectangle areas were equal, the pink rectangle areas were equal, and the green rectangle areas were equal. They had seen all these right triangles before,  when they were looking at the diagram and making observations, so this went pretty quickly for most of them.

I love this proof of the law of cosines! Of course when I went online, I saw so many other beautiful proofs (look here, and the links at the bottom, for some). Troll the internet and be amazed! They are so elegant! This “scaling up” one might be my favorite. And here’s one that David Butler sent me (that is on the site I linked to above). And I remember proving the Pythagorean Theorem in geometry using the crossed chord theorem, and now the same argument here can be used for the law of cosines.


[1] To be clear, this diagram only works for acute triangles. I haven’t yet modified the argument to work for obtuse triangles.


Some very cool things about the Law of Sines

So yeah, I’m teaching the law of sines and cosines, and I’m finding some awesome things. What’s totally ridiculous is that after I introduced some of it to my class, I looked back at my stuff from last year and apparently I had done the same thing. Like… I don’t remember it at all.

In any case, when I teach the Law of Sines, I tend to have kids derive it by finding the area of a triangle in three different ways.


We set these three different ways to get the area of a triangle equal to each other, and divide by \frac{1}{2}abc to get the Law of Sines:



Now, to be clear, there are some subtleties that have to be addressed here. Like for example, this argument clearly works for acute triangles, but what about an obtuse triangle or right triangle, like:


It turns out with just a tiiiiny little bit of extra work, we can show that the Law of Sines holds. (Here’s a fun little applet you can play with for this… one important thing that can help you for the obtuse triangle proof is that \sin(\theta)=\sin(180-\theta).)

So… yeah. That’s a pretty traditional way to teach the Law of Sines. But did you know that the ratio that pops up with the Law of Sines has a geometric interpretation?

Like, look at this triangle. And look at the ratio of (side length)/(the sine of the angle opposite the side).


That 10.36 has a geometric meaning. Ready for it? READY? I don’t think you are, but I’m going to show it to you anyway…


Dang! HOLY MOTHERFATHER! Yuppers. That triangle has one circle that can perfectly circumscribe it. And twice the radius of that circumscribed circle is that ratio!!! Don’t believe me? Ok, I know you do, but play with this applet I made to see it happen! Maybe try to create a right triangle and see if that reminds you of something you learned in geometry?

Now this year I told my kids that \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R. And I sent ’em up to the whiteboards and asked them to prove it. I gave some hints. Like, for example, the inscribed angle theorem. But eventually kids got it!


This is actually another proof of the Law of Sines! (To be clear, you will also have to make an argument for an obtuse triangle, which requires a tiny bit of modification. You have to see a central angle and one of the angles of the triangle are the same because they both subtend the same arc. And a right triangle.)

So I had a follow up after this… I asked kids to prove that the area of any triangle is: \frac{abc}{4R}, where R is the radius of the circumscribed circle. I asked them to prove it algebraically, and they did:

\text{Area}=\frac{1}{2}ab\sin(C). But we know that \frac{c}{\sin(C}=2R. So let’s manipulate the rea equation to get an R in it.


Now we have \text{Area}=\frac{1}{2}abc\frac{1}{2R}=\frac{abc}{4R}.

I asked kids to show this algebraically. They did it in various ways (all correct), similar to the argument above. However I had a student present me with a stunning geometric argument that proved this area formula. I honestly don’t know if I would have been able to come up with it. It was so stunning I had to take a photo of it. I leave this as an exercise for the reader. MWAHAHAHA.

(All of this Law of Sine stuff was inspired by this webpage.)

A beautiful combinatorics argument

Today a teacher in my department was struggling to understand algebraically and conceptually why:


He was on the way to making a neat Pascal’s Triangle argument. Look at that 70. That’s \binom{8}{4}:


He started working backwards, and saw that 70=35+35.

But each of those 35s came from 15 and 20. So 70=(15+20)+(20+15).

And then going backwards more, we see the 15 comes from 5 and 10, and the 20 comes from 10 and 10. So 70=((5+10)+(10+10))+((10+10)+(10+5)).

And then going backwards once more, we see the 5 comes from 1 and 4, and the 10 comes from 4 and 6. So 70=(((1+4)+(4+6))+((4+6)+(4+6)))+(((6+4)+(6+4))+((6+4)+(4+1)))

In other words, 70=1*1+4*4+6*6+4*4+1*1.

By the time we make our way down from the 1-4-6-4-1 row to 70, we see that:

The first one in the 1-4-6-4-1 row just is added once when making the 70.
The first four in the 1-4-6-4-1 row is added four times when making the 70.
The six in the 1-4-6-4-1 row is added six times when making the 70.
The second four in the 1-4-6-4-1 row is added four times when making the 70.
The second one in the 1-4-6-4-1 is added just once when making the 70.

In other words: 70=1^2+4^2+6^2+4^2+1^2.

The other teacher and I realized we could generalize this. But we were left unsatisfied. It was a Pascal’s Triangle argument, but I wanted to see the answer with an understanding of combinations. I wanted something even more conceptual. So my friend and I started thinking, and he had an awesome insight. And I want to record it here so I don’t lose it! It made me so happy — little mathematical endorphins exploding in my head!

Let’s assume we have a set of 2n letters, where n letters are A and n letters are B.

Blergity blerg, let’s just keep things concrete, and have 8 letters, where 4 are As and 4 are Bs. (We can generalize later, but I want to just see this happen!) Given 4As and 4Bs, there are \binom{8}{4} ways to arrange them to make different 8 letter words [1]. Great! That was the easy part.

Now we are going to construct a whole bunch of different sets of 8 letter words, in a particular way (using AAAABBBB), so that when we add up all those sets, we’re going to get all possible 8 letter arrangements of AAAABBBB.

How are we going to do this? We are going to create special of 4 letter words and concatenate them together to make 8 letter words. 

Set 1: We are going to create a 4 letter word with 0As (and thus by default, 4Bs) and a 4 letter word with 4As (and thus by default, 0Bs).

How many ways can we create 4 letter words with 0As? \binom{4}{0}. To be clear, this is just 1. The word is {BBBB}.

How many ways can we create 4 letter words with 4As? \binom{4}{4}. To be clear, this is just 1. The word is {AAAA}.

And when we concatenate them, we are going to have \binom{4}{0}\cdot\binom{4}{4} eight letter words. But we know \binom{4}{0}=\binom{4}{4}. So this is simply \binom{4}{0}\cdot\binom{4}{0}. And this is just 1, because the only eight letter word possible is {BBBBAAAA}.

This is a degenerate case, so it’s hard to really see what’s going on here. So let’s move on.

Set 2: We are going to create a 4 letter word with 1A (and thus by default, 3Bs) and a 4 letter word with 3As (and thus by default, 1Bs).

How many ways can we create 4 letter words with 1As? \binom{4}{1}. To be clear, this is just 4. The words are {ABBB, BABB, BBAB, BBBA}.

How many ways can we create 4 letter words with 3As? \binom{4}{3}. To be clear, this is just 4. The words are {AAAB, AABA, ABAA, BAAA}.

And when we concatenate them, we are going to have \binom{4}{1}\cdot\binom{4}{3} eight letter words. But we know \binom{4}{1}=\binom{4}{3}. So this is simply \binom{4}{1}\cdot\binom{4}{1}. And this is 16 eight letter words. (Each of the first four letter words can be paired with each of the second four letter words… so this is merely 4*4. Just to be clear, I’ll list the first few eight letter words out: ABBBAAAB, ABBBAABA, ABBBABAA, ABBBBAAA, BABBAAAB, BABBAABA, …

Set 3: We are going to create a 4 letter word with 2As (and thus by default, 2Bs) and a 4 letter word with 2As (and thus by default, 2Bs).

How many ways can we create 4 letter words with 2As? \binom{4}{2}. To be clear, this is just 6. The words are {AABB, ABAB, ABBA, BAAB, BABA, BBAA}.

How many ways can we create 4 letter words with 2As? \binom{4}{2}. To be clear, this is just 6. The words are {AABB, ABAB, ABBA, BAAB, BABA, BBAA}.

And when we concatenate them, we are going to have \binom{4}{2}\cdot\binom{4}{2} eight letter words. And this is 36. (Each of the first four letter words can be paired with each of the second four letter words… so 6*6 eight letter words.)

Set 4: We are going to create a 4 letter word with 3As (and thus by default, 1B) and a 4 letter word with 1As (and thus by default, 3Bs). By the same logic as above, we are going to end up with \binom{4}{3}\cdot\binom{4}{3} eight letter words. This is just 4*4 eight letter words.

Set 5: We are going to create a 4 letter word with 4As (and thus by default, 0Bs) and a 4 letter word with 0As (and thus by default, 1B). By the same logic as above, we are going to end up with \binom{4}{4}\cdot\binom{4}{4} eight letter words. This is just 1*1 eight letter words.


Now look at all the different eight letter words created by this process, from Set 1, Set 2, Set 3, Set 4, and Set 5. We have captured every single possible eight letter word with four As and four Bs. Let’s check a few random words:

AABABBAB… okay this is in Set 4.
BAABAABB… okay this is in Set 3.
BBBABAAA… okay this is in Set 2.

Cool! I only have to look at the first four letters to decide which set it is going to be in!

But look at what we’ve done. We’ve shown that we can get all eight letter words in these five sets… so the number of eight letter words is:


If we simply write the squares out…


But we saw at the very start that the number of eight letter words is simply \binom{8}{4}

So the two are equal.

All the hard work is done, so I leave it as an exercise to the reader to generalize.

P.S. I take no credit for this amazingly wonderful letter rearrangement solution. I just bore witness as my friend figured it out, and I got giddier and giddier. I love it because it’s abstract, but still understandable to me. But it’s close to my threshhold of abstraction!


[1] If you don’t quite see this, imagine 8 blank slots.

___ ___ ___ ___ ___ ___ ___ ___

You choose four of them to put the As into. There are \binom{8}{4} ways to choose four of these slots. Put As into those four. By default the rest of the slots must be filled with Bs — they are forced! So there are \binom{8}{4} ways to create eight letter words with four As and four Bs.

Double Angle Formulae

I posted this on my Adv. Precalculus google classroom site. I don’t know if I’ll get any responses, but I loved the problem, so I thought I’d share it here.


I mentioned in class that I had stumbled across a beautiful different proof for the double angle formulae for sine and cosine, and I would post it to the classroom. But instead of *giving* you the proof, I thought I’d share it as an (optional) challenge. Can you use this diagram to derive the formulae? You are going to have to remember a tiiiiny bit of geometry! I already included one bit (the 2*theta) using the “inscribed angle theorem.”

If you do solve it, please share it with me! If you attempt it but get stuck, feel free to show me and I can nudge you along!



Below this fold, I’m posting an image of my solutions! But I say to get maximal enjoyment, you don’t look further, take out a piece of paper, and take a stab at this!


What felt like Forever… It was maybe 20 minutes?

On Friday in one of my Advanced Precalculus classes, kids were working on figuring out the double angle formulas for sine and cosine. They got \sin(2\theta)=2\sin(\theta)\cos(\theta) and \cos(2\theta)=\cos^2(\theta)-\sin^2(\theta).

And then… they got stuck.

You see, I showed them two alternative forms for the double angle formula for cosine (\cos(2\theta)=1-2\sin^2(\theta) and \cos(2\theta)=2\cos^2(\theta)-1). I showed them these forms. And I said: figure out where they came from.

All groups in a few minutes were on yellow cups (“our progress is slowing down, but we’re not totally stuck yet”). I didn’t want to give anything away, but I didn’t have any group have a solid insight that I could have them share with others. I let things remain a bit more, no luck, so then I said: “this looks related to something we’ve seen before… a trig identity… maybe that will be helpful. Bring in something you know to open up the problem for you.” Eventually kids realized they needed to bring in some outside information (namely: \sin^2(\theta)+\cos^2(\theta)=1).

I was sure that was going to be enough. Totally certain. But after another 5 minutes of watching them struggle, I wasn’t so sure. I didn’t want to give anything more away, but I had to because we had to move forward. But what more could I give without giving the whole show away? Since many groups were trying some crazy stuff, I said: “this is a simple one or two step thing…” Why? I just wanted them to take fresh eyes and see what they could do thinking simply. They kept on saying I was trying to trick them, but I told them it wasn’t a trick!

And then, in the span of the next five minutes, all my groups got it.

But what was more interesting was that we had three different ways to do it. As kids moved on to the next set of questions (and I breathed a sigh of relief that they figured this out), I reflected on how awesome it was that they persevered and then came up with different approaches. So while they worked, I put up the three different approaches.


And with a few minutes to go at the end of class, I had everyone put everything away and I just pointed out the embarrassment of riches they came up with. And it was great to hear the audible reactions when kids who had one way saw the other ways and say things like “ooooh, I never would have thought of that!” or “that’s so clever!”

I had (have?) so many mixed feelings when I saw how difficult this question was for my kids. And I was hyperconscious about how much time we had to spend on this. But the ending made me feel like it was time well-spent.

Problem Solving with Trig

So I’m at #TMC17 and Rachel Kernodle nerdsniped me. Or rather, I asked to be nerdsniped. Her session is at a time when there were a lot of other amazing sessions I wanted to go to, so I wanted to know if hers was one where I could hear about it and get the gist of things instead of attending. After some internal debate, she said that since it involved working on a problem, and then using that problem solving to frame the session, the answer was maaaaybe not. But then she thought: maybe I can try the problem on you and see how it goes. As long as you’re willing to put in the time to problem solve. Of course I said yes.

First, you can see her session description, which then framed how I approached the problem:

triangle 2.png

And then this is what she gave me (but it was hand drawn):


From the session description, I knew I had to find the ratio of the side lengths, so I could find exact trig values for angles other than 30, 60, 90, 45.

Rachel also gave me a “hint page” which she told me to look at when I was stuck (and to time how long it took me before I opened it). Let’s just say I’m extremely stubborn, and so as long as I think I have the capability to solve something and I am not completely stuck, I knew I wasn’t going to open it. Turns out my stubbornness paid off, and I ended up solving it.

In this post, I wanted to write a little bit about my experience with the problem. Because now when I look at that triangle, I have an duh, there’s an obvious approach to use here and everything I know points at that obvious approach. And the answer feels really obvious too. It is funny that I’m almost embarrassed to post this because there are going to be people who see it right away, and I worry (irrationally) (math pun) that they are going to judge me for not seeing it as quickly as they did. Even though I know being good at math has nothing to do with speed. And that it was important to go through the steps I did!

It took me over an hour to solve this problem. I had to do a lot of play and make a lot of random leaps before I stumbled across the “obvious approach.”  And I needed to do that in order for me to mine it for lots of things. It was true problem solving. And I know I really deeply understand this because at first the problem looked flummoxing and interesting, and now it looks obvious and somewhat trite. That’s my metric of how I know I deeply understand something. There are still certain things that I teach that I don’t deeply understand: like how the cross product of two 3D vectors yields a third vector perpendicular to the original two. I have done the math, but it’s non-obvious to me why the crazy way we compute cross products give us something perpendicular.(When I only understand something by doing brute algebra, I rarely feel like I get it.)

I’m going to try to outline the messiness that was my thought process in this triangle problem, to show/archive the messiness that is problem solving.

  1. The first thing I noticed was 36 and 36 sum to 72. So I was like: obviously put two of those figures together, and just play around. Something nice will happen. I remember when seeing the problem that approach felt immediate, obvious, and would lead to the solution. I was like yes! I have an inroad! This is going to rock, and I’m going to solve it quickly! And I’ll even impress Rachel!


    That appraoch didn’t work. Nothing popped out. I saw 54s and 18s and 144s pop out. But those weren’t angles that helped me. But I did then realize something nice… 36 is a tenth of 360! So I was going to use a circle somehow in this solution. Obviously!

  2. So I drew this:
    and I was like, I have something here! But after looking around, I was getting less. You can see I was trying to draw in some other lines lightly and play around — I thought maybe creating other triangles within these triangles would work. But nothing seemed to pop out. At one point, I thought I had possibly created an equilateral triangle in this (even though I saw one of the angles was 72! I was clearly desperate!). I started to get dejected at this point. I knew the circle had something to do with it…
  3. But seeing that 54s and 18s and 36s and 72s kept appearing, I thought maybe algebraically I should play around with the numbers (adding in 180 also, since I can draw a straight line wherever) to see if algebraically I could get a 30, 60, or 45. I tried adding and subtracting numbers from the set {18, 36, 54, 72, 180} looking for 30, 60, or 45. I figured if I could somehow do that, then I could find a diagram that would have angles I could get side relationships from. And then like a domino effect, I could get others. I don’t know. But after like 2 seconds, I got bored with this and didn’t see it as very efficient. My intuition was strongly saying I was going in the wrong direction. So I stopped:


  4. At this point, I was pretty dejected. I was slightly losing interest in the problem, thinking it was too hard for me. I tried to “force” a 60 degree angle in a diagram of that original blasted triangle. Hope! And then hope dashed!

  5. Damnit! I know the circle had something to do with it. It is just too nice to abandon the circle! Maybe…

    At first I drew all ten vertices for a 10-gon. I started connecting them in different ways. I thought I could exploit the chord-chord theorem in geometry, but that wasn’t good. I tried in that second diagram to extract part of the circle diagram and investigate it more. And the third was just more of the same. At one point, I was like e^{i\theta}=\cos\theta+i\sin\theta and was thinking I could somehow think of this as a problem on the complex plane, where each vertex was e^{ni\pi/5} and then look at the real parts for the x-coordinate and the imaginary parts for the y-coordinate. Clearly my mind was whirring, and I was going anywhere and everywhere. I actually thought maybe this complex plane thing seems ugly but it will be so elegant. But then I realized I didn’t know where to go if I labeled each of the points on the complex plane. Done and done and doneAt this point I put the problem away. Nothing was working.

  6. But after a minute, I couldn’t let it go! I wanted to solve it!!! So I went back. I thought I was getting too complicated, so I went simple.


    Nope. Didn’t help. But for some reason, this diagram and looking at the 72 reminded me of something I hadn’t thought of before. This is the leap that helped me get to the answer. And I can’t quite explain why this diagram sparked this leap. Which sucks because this is that moment that led to the rest of the problem for me! But I immediately remembered something about 72s and pentagons. And it hit me.

  7. So I drew what this connection was. My brain was whirring, and I was somewhere good…  

    I remembered the 72 degree angle appeared in a star. And this star was related to a pentagon. And that the pentagon had something about the golden ratio tied up in it. So I knew that maybe the golden ratio was involved in the answer. And where does the golden ratio appear? When there are similar triangles and proportions. I had my new approach and my inroad that I thought would work. Two triangles next to each other failed. Circles failed. But star/pentagon might work!

  8. So I looked at the original triangle and tried to figure out where I could find a similar triangle. And so I drew one line and created a similar triangle. I labeled the two legs as having length “1.”

    Initially, I was thinking I could do something with the law of sines. Because if you think about it, this is the ASS case — where you have that 36 degrees (circled), the side I labeled 1 (circled), and the other side I labeled y (circled). But you note that last side could be in two different places, which is why there are two ys circled. I still think there is something fun that I could do with this. But as I was doing this, I realized I was making things more complicated.

    I knew that the golden ratio came out of a proportion. So I abandoned the law of sines for the proportion. I simply set up a proportion with the two similar triangles. I first found “?” by doing 1/y=y/?. So ? was y^2This was exciting. I knew the golden ratio came out of solving a quadratic. Yeeeeee! At this point, my excitement was growing because I was fairly confident I was almost at the solution.

    Then I labeled the part of the leg that wasn’t ? as 1-y^2 (since the whole leg length was 1). Finally I looked at the third triangle in the diagram that wasn’t similar to the original triangle. It was isosceles and has legs of y and 1-y^2 so I set them equal and solved and not-quite-the-golden-ratio came out! (There was a mistake I made where I set y^2=1-y^2 and got y=\sqrt{2}/2. But I then found it and rewrote the equation y=1-y^2. This was the most depressing part of it. Because I couldn’t find my error because I was so tired. I went through my work multiple times and nothing. But taking some time away and then looking with fresh eyes, it was like: doh!)

    And so that was the end. I found if the original triangle had leg lengths of 1, the base was going to have a length of \sqrt{5}/2-1/2.

    I was so proud. I was on cloud nine. I was telling everyone! SO COOL!!! 

It probably took me in total 90 minutes or so from start to finish. So many false starts at the beginning, and one depressing transcription error that I couldn’t find.

The point of this post isn’t to teach someone the solution to the problem. I could have written something much easier. (See we can draw this auxiliary line to create similar triangles. We use proportions since we have similar triangles. Then exploit the new isosceles triangle by setting the leg lengths equal to each other.) But that’s whitewashing all that went into the problem. It’s like a math paper or a science paper. It is a distillation of so freaking much. It was to capture what it’s like to not know something, and how my brain worked in trying to get to figure something out. To show what’s behind a solution.



Multiple Representations for Trigonometric Equations

I have to say that we’re doing some pretty neat stuff for trig this year in precalculus. I’m working with two other teachers and totally writing everything we’re doing from scratch. I had about 3 days to teach solve some basic trigonometric equations. They are basic. Like 2\sin(x)+5=4.7. But we’ve put a lot of thought into what we’re teaching, how we’re teaching it, and why we’re teaching it — and more complicated trig equations just didn’t make the cut. [1]

Besides not-a-lot-of-time, the other bugaboo I was contending was how to deal with inverse trig. Long story short, I’ve found a way to teach inverse trig which makes me fairly happy in my advanced precalculus class. But because of our time constraints, I decided that we could get my standard precalculus kids solving trig equations without understanding the theory behind the restricted domain of inverse trig functions. :) Why? They learned years ago in geometry that if they have a triangle like the one below


they could get an angle, like angle A, by writing: \sin(A)=\frac{3}{5}. And then using the inverse of sine, they could get A=\sin^{-1}(\frac{3}{5})\approx 36.87^oThey know about the inverse trig functions already. So I wanted to exploit that fact.  And if organically a question about what the calculator was doing when spitting out an answer, and why it only gave one answer, I promised myself I would address it. (This year, no question like that arose.)


A quick last note, before I shared how I approached these few days in class, I decided to totally eliminate the use of the term “reference angle.” Kids would discover the relationships among the solutions of trig equations on their own. No need for new terminology here. Just logic.

Day 1: Three important “do nows”


This led to a great discussion. Every group decided the “top left equation” was going to be the easiest. And every group decided that the log and tangent equations were going to be the hardest. When I pressed them on why, they said it’s because they forgot logarithms from last year, and that tangent was just kinda tricky. They could “undo” a square root or a square, but they didn’t really know how to “undo” a logarithm or tangent function.

Next I threw up this slide. I just wanted to remind kids that sometimes there are more than one solution to equations — even simple equations they know. I also wanted them to see that they knew something about the tangent equation. They knew it had infinitely many solutions — even though they might (right now) know what those solutions are!


Finally, I wanted to do a serious review of special angles and their relationship with the unit circle. So I had kids spend 5 minutes solving these basic trig equations.


Obviously I put the unit circle on there as a prompt to get them thinking. And YES, that last trig equation, with the 3/7ths, was done on purpose. I asked kids after they got stuck on it if there were some of these they would not want to appear on a pop quiz. They all recognized that the 3/7th one was bad because it wasn’t one of the coordinates associated with the special angles.

This laid the groundwork for the packet.

[docx editable version: 2017-04-24 Basic Trigonometric Equations]

Kids had good conversations and were able to solve equations like \sin(x)=0.3 and \cos(x)=-0.8 using the unit circle/protractor, a detailed graph of the sine and cosine waves, and using their calculators to get fairly precise answers.

Their nightly work was simply to finish the sine and cosine questions in Part 1 (questions #1-4).

Day 2: Expanding Understanding

I started with an awesome “do now.”


I thought this was going to be a quick 4-5 minute discussion. But kids took 3-4 minutes just to really talk in their groups. And I had them share their thinking. It led to kids talking about “efficiency” and “conceptual understanding” themselves! They all pretty much though the unit circle was the best way to solve it — even with the annoyance of the protractor — because they liked the conceptual understanding it provided. They thought the calculator did the work quickly, and was more accurate, but it annoyingly only gave one of the solutions (so you had to use logic and the unit circle to figure out the second solution), and you could easily forget the meaning of what you were doing. I was so proud of what they were saying. Super awesome metacognition! All in all, this was probably 7-8 minutes.

Then I let them loose on the tangent questions in the packet (Part I #5 and 6). They initially had to solve \tan(x)=1.1 using a protractor. Every single group remembered tangent represented slope. Most groups reasoned that if \tan(x)=1, they would get 45^o and 225^o as their solutions. And since this slope was slightly greater than 1, the angles would be slightly different, just a few degrees higher. It was lovely. (And exactly what I hoped would happen, which is why I chose to use 1.1 in the equation.) But one group literally drew a line with a slope of 1.1 and measured the angles associated with that. I wasn’t surprised that a group did that, but I expected a few more to do so. (I had this group share their thinking with the rest of the class, at the end of the period.)

Then kids spent the rest of the class working on select questions in Part II (8, 9, 11) and Part III (13, 14, 15).

For nightly work, kids finished any of those problems (#8, 9, 11, 13, 14, 15) that they didn’t finish up in class.

Day 3: Polishing Things Off

I started with a question that I wanted to reinforce after the previous class:




We did a bit of review of some unrelated Algebra II ideas to help set them up for our next unit on polynomials. And then…

… to work! I had kids discuss problem 13 in their groups first (since I could see that being a place where a kid, at home, might get trapped… and I wanted them to use each other to get unstuck). And then they compared their answers to the other nightly work questions — and used a solution sheet I gave them to see if they were correct. Then I set them loose on using Desmos to do Part IV. The rest of the period was spent working on finishing up the problems that weren’t assigned in the packet (the ones they skipped).

Pretty much all groups were working together amazingly, and when I went around to check in on different groups, everyone was getting all the questions correct. The biggest problem was actually finding a good window in Desmos! If that’s the biggest problem, I’m golden.


What I loved:

Okay, so I’m going to toot my own horn here. Although the packet may “look” simple, I have to say the only way to see why it’s so awesome is to actually do it. The choice of having kids solve \sin(x)=0.3 and then immediately solve \sin(x)=-0.3 was on purpose, to generate good conversations with kids about reference angles without using that term. The choice of \tan(x)=1.1 was done specifically to exploit their understanding of \tan(x)=1. And the fact that they’re constantly looking at the same question through three different lenses (unit circle, wave, calculator) is deliciously sweet. And then — at the very end — they get to see the solution a fourth way, by using Desmos to graph these equations to find a solution? SO COOL. Because the very last thing we had done in this class was learning transformations of sine and cosine graphs! [2]

This packet, and associated “do nows” and conversations, did what I was hoping for. It highlighted multiple representations. It had kids thinking deeply about the meaning of sine, cosine, and tangent. It had kids develop a way to understand multiple solutions to trig equations by simply using logic and what they know. It had kids recognize that the more they understand trigonometry, the more ways they have to solve a trig problem. And no kid got derailed because they didn’t understand inverses deeply.


[1] I could argue a case for these type of equations, as well as a case against them. But considering our goals and what we’ve already done with trig, I think we’re making the right decision. Why? Because our goal isn’t solving algebraic equations writ large, and I could see solving something like 2\sin^2(2x-180)=5 being useful for that. But for getting a deeper understanding of the trigonometric functions? I see less value. (Not no value, mind you, but less…)

[2] We did this in a deliciously marvelous way. I hope to blog about it!