Engaging students: right-triangle trigonometry

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Kelsie Teague. Her topic, from Precalculus: right-triangle trigonometry.

How has this topic appeared in popular culture?

In the famous T.V. show Numbers they do an episode using trig to find the angle of origin of the blood spatter. In forensic science they use trig every day to determine where the victim was originally injured. They can also use this to find the angle of impact, area/point of convergence, and area of origin. The following power point goes into more detail: http://cmb.physics.wisc.edu/people/gault/Blood%20Splatter%20Trig.pdf

If the blood was dropped by a 90-degree angle, the stain will appear to be an almost perfect circle.

We could get out some long paper and colored water and experiment with the idea of change of angle in the drop of blood and calculate the angles.

Angle of Impact =Sin (theta)= Width of drop/Length of drop.

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?

YouTube is a great website for engagement because you can find many videos to start the lesson off with some previous knowledge that they will be using for that days lesson. The following video would be a good way to engage the students when talking about right triangle trig.

It’s to a song that they probably have already heard and it’s teaching them something they already know. Since the students already have knowledge of this, the video isn’t teaching them the topic but refreshing their memory in a entertaining fashion.

When looking for a good video, I ran across many that would work for this lesson, but this one seemed like it would grab the student’s attention more and keep their attention.

The above video is also a good one, and it shows the lyrics in the description so you can make sure what they are saying is mathematically correct so it doesn’t give the students any misconceptions.

How could you as a teacher create an activity or project that involves your topic?

If I was teaching at a school that was close to a hill or a mountain outside I could take my students outside and have them figure out how far up the mountain they would have to walk to get to the top. We could use a tap measure to measure how high they had the protractor in the air and then we could look up the height and distance away of the mountain. They then could use the protractor to find the angle between themselves and the top of the mountain. We could then use this information inside the classroom to solve how far to travel up the mountain.

Similar to the above picture except they will know the height of the mountain. This would show the length of the hypotenuse of the right triangle. They will have to subtract the height they have the protractor at from the height of the mountain to be accurate since the height of the mountain is from the ground up.

Mathematics and College Football

For years, various algorithms (derisively called “the computers” by sports commentators) have been used to rank college football teams. The source of derision is usually quite simple to explain: most of these algorithms are too hard to explain in layman’s terms, and therefore they are mocked.

For both its simplicity and its ability to provide reasonable rankings, my favorite algorithm is “Random Walker Rankings,” published at http://rwrankings.blogspot.com. Here is a concise description of this ranking system (quoted from http://rwrankings.blogspot.com/2003_12_01_archive.html):

We’ve all experienced befuddlement upon perusing the NCAA Division I-A college football
Bowl Championship Series (BCS) standings, because of the seemingly divine inspiration that must have been incorporated into their determination. The relatively small numbers of games between a large number of teams makes any ranking immediately suspect because of the dearth of head-to-head information. Perhaps you’ve even wondered if a bunch of monkeys could have ranked the football teams as well as the expert coaches and sportswriters polls and the complicated statistical ranking algorithms.

We had these thoughts, so we set out to test this hypothesis, although with simulated monkeys (random walkers) rather than real ones.

Each of our simulated “monkeys” gets a single vote to cast for the “best” team in the nation, making their decisions based on only one simple guideline: They periodically look up the win-loss outcome of a single game played by their favorite team, and flip a weighted coin to determine whether to change their allegiance to the other team. In order to make this process even modestly reasonable, this random decision is made so that there is higher probability that the monkey’s allegiance and vote will go with the team that won the head-to-head contest. For instance, the weighting of the coin might be chosen so that 75% (say) of the time the monkey changes his vote to go with the winner of the game, meaning only a 25% chance of voting for the loser.

The monkey starts by voting for a randomly chosen team. Each monkey then meanders around a network which describes the collection of teams, randomly changing allegiance from one team to another along connections representing games played between the two teams that year. This network is graphically depicted in the figure here, with the monkeys—okay, technically one is a gorilla—not so happily lent to us by Ben Mucha (inset). It’s a simple process: if the outcome of the weighted coin flip indicates that he should be casting his vote for the opposing team, the monkey stops cheerleading for the old team and moves to the site in the network representing his new favorite team. While we let the monkeys change their minds over and over again—indeed, a single monkey voter will forever be changing his vote in this scheme—the percentage of votes cast for each football team quickly stabilizes. We thereby obtain rankings each week of the season and at the end of the season, based on the games played to that point of the season, by looking at the fraction of monkeys that vote for each team…

The virtue of this ranking system lies in its relative ease of explanation. Its performance is arguably on par with the expert polls and (typically more complicated) computer algorithms employed by the BCS. Can a bunch of monkeys rank football teams as well as the systems in use now? Perhaps they can.

Using this algorithm, here’s the current ranking of college football teams as of today. (With great pride, I note that Stanford is ranked #4.) These rankings certainly don’t exactly match the latest AP poll or BCS rankings, but they’re also still reasonable and defensible.

Area of a circle (Part 4)

Math majors are completely comfortable with the formula $A = \pi r^2$ for the area of a circle. However, they often tell me that they don’t remember a proof or justification for why this formula is true. And they certainly don’t remember a justification that would be appropriate for showing geometry students.

In the previous three posts, I discussed various ways that calculus can be used to show that $A = \pi r^2$ . Still, most future high school teachers would like to know a justification for why $A = \pi r^2$ . After all, the definition of $\pi$ is

$\pi = \displaystyle \frac{ \hbox{Circumference} }{ \hbox{ Diameter}}$ , or $C = 2\pi r$

So there ought to be a reasonable explanation for why $\pi$ reappears in the formula for the area of a circle. Furthermore, this explanation should within the grasp of geometry students — so that the explanation should not explicitly use calculus. Even better, they’d prefer a hands-on classroom activity so that students could discover the formula for themselves.

The video below shows a completely geometric justification for why $A = \pi r^2$ that meets the above criteria. I have a couple of small quibbles with the narrated text — I’d prefer to say that the each rearrangement of pieces is approximately a parallelogram (as opposed to a rectangle), and that figures get closer and closer to a real parallelogram with area $A = \pi r^2$ .

In other words, I would avoid saying that we ultimately divide the circle into infinitely many wedges of infinitesimal width to get a perfect rectangle, as this promotes a misconception concerning the definition of a limit that they shouldn’t carry into a future calculus course.

However, the graphics are excellent in this video. In my mind, that more than counterbalances the preferred way that I would describe the process of taking a limit to students.

Pedagogically, I would recommend a hands-on activity along these lines. Let the students use a protractor to draw a 5- or 6-inch circle on a piece of paper. Then have them mark $18$ points on the circumference of the circle at every $20^o$ , and then draw the lines connecting these points and the center of the circle. Then have the students cut out these wedges and physically rearrange them as in the video. They should discover for themselves that the wedges approximately form a parallelogram, and they know how to find the area of a parallelogram.

After they do this activity, then I would show the above video to geometry students.

If anyone knows a video that (1) is as visually appealing as the one above and (2) correctly states the principle of limit for geometry students, please let me know.

Area of a circle (Part 3)

$A = \displaystyle \iint_R 1 \, dx \, dy$

This double integral may be computed by converting to polar coordinates. The distance from the origin varies from $r=0$ to $r=a$ , while the angle varies from $\theta = 0$ to $\theta = 2\pi$ . Using the conversion $dx \, dy = r \, dr \, d\theta$ , we see that

$A = \displaystyle \int_0^{2 \pi} \int_0^a r \, dr \, d \theta$

$A = \displaystyle \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_0^a \, d\theta$

$A = \displaystyle \int_0^{2\pi} \frac{a^2}{2} \, d\theta$

$A = \displaystyle 2 \pi \cdot \frac{a^2}{2}$

$A = \displaystyle \pi a^2$

We note that the above proof uses the fact that calculus with trigonometric functions must be done with radians and not degrees. In other words, we had to change the range of integration to $[0,2\pi]$ and not $[0^o, 360^o]$ .

Area of a circle (Part 2)

A circle centered at the origin with radius $r$ may be viewed as the region between $f(x) = -\sqrt{r^2 - x^2}$ and $g(x) = \sqrt{r^2 - x^2}$ . These two functions intersect at $x = r$ and $x = -r$ . Therefore, the area of the circle is the integral of the difference of the two functions:

$A = \displaystyle \int_{-r}^r \left[g(x) - f(x) \right] \, dx= \displaystyle \int_{-r}^r 2 \sqrt{r^2 - x^2} \, dx$

This may be evaluated by using the trigonometric substitution $x = r \sin \theta$ and changing the range of integration to $\theta = -\pi/2$ to $\theta = \pi/2$ . Since $dx = r \cos \theta \, d\theta$ , we find

$A = \displaystyle \int_{-\pi/2}^{\pi/2} 2 \sqrt{r^2 - r^2 \sin^2 \theta} \, r \cos \theta d\theta$

$A = \displaystyle \int_{-\pi/2}^{\pi/2} 2 r^2 \cos^2 \theta d\theta$

$A = \displaystyle r^2 \int_{-\pi/2}^{\pi/2} (1 + \cos 2\theta) d\theta$

$A = \displaystyle r^2 \left[ \theta + \frac{1}{2} \sin 2\theta \right]_{-\pi/2}^{\pi/2}$

$A = \displaystyle r^2 \left[ \left( \displaystyle \frac{\pi}{2} + \frac{1}{2} \sin \pi \right) - \left( - \frac{\pi}{2} + \frac{1}{2} \sin (-\pi) \right) \right]$

$A = \pi r^2$

Area of a circle (Part 1)

In this series of posts, I’ll discuss several ways that the area of a circle can be found using calculus. I’ll also discuss a straightforward classroom activity by which students can discover for themselves why $A = \pi r^2$ .In the first few weeks after a calculus class, after students are introduced to the concept of limits, the derivative is introduced for the first time… often as the slope of a tangent line to the curve. Here it is: if $y = f(x)$, then

$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$

From this definition, the first few rules of differentiation are derived in approximately the following order:

1. If $f(x) = c$ , a constant, then $\displaystyle \frac{d}{dx} (c) = 0$ .

2. If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$ .

3. If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$ .

4. If $f(x) = x^n$ , where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$ . This may be proved by at least two different techniques:

The binomial expansion $(x+h)^n = x^n + n x^{n-1} h + \displaystyle {n \choose 2} x^{n-2} h^2 + \dots + h^n$
The Product Rule (derived later) and mathematical induction

5. If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$ . In other words, taking the derivative of a polynomial is easy.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let $A(r) = \pi r^2$ . Notice I’ve changed the variable from $x$ to $r$ , but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.) What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$ . Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$ . Does this remind you of anything? (Students answer: the volume of a sphere.) What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$ . Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known on television as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.)

In the spirit of a cliff-hanger, I offer the following thought bubble before presenting the answer.

By definition, if $A(r) = \pi r^2$ , then

$A'(r) = \displaystyle \lim_{h \to 0} \frac{ A(r+h) - A(r) }{h} = 2\pi r$

The numerator may be viewed as the area of the ring between concentric circles with radii $r$ and $r+h$ . In other words, imagine starting with a solid red disk of radius $r +h$ and then removing a solid white disk of radius $r$ . The picture would look something like this:

Notice that the ring has a thickness of $r+h -r = h$ . If this ring were to be “unpeeled” and flattened, it would approximately resemble a rectangle. The height of the rectangle would be $h$ , while the length of the rectangle would be the circumference of the circle. So

$A(r + h) - A(r) \approx 2 \pi r h$

and we can conclude that

$A'(r) = \displaystyle \lim_{h \to 0} \frac{ 2 \pi r h}{h} = 2\pi r$

By the same reasoning, the derivative of the volume of a sphere ought to be the surface area of the sphere.

Pedagogically, I find that the above discussion helps reinforce the definition of a derivative at a time when students are most willing to forget about the formal definition in favor of the various rules of differentiation.

In the above work, we started with the formula for the area of the circle and then confirmed that its derivative matched the expected result. However, the above logic can be used to derive the formula for the area of a circle from the formula $C(r) = 2\pi r$ for the circumference. We begin with the observation that $A'(r) = C(r)$ , as above. Therefore, by the Fundamental Theorem of Calculus,

$A(r) - A(0) = \displaystyle \int_0^r C(t) \, dt$

$A(r) - A(0) = \displaystyle \int_0^r 2\pi t \, dt$

$A(r) - A(0) = \displaystyle \left[ \pi t^2 \right]_0^r$

$A(r) - A(0) = \pi r^2$

Since the area of a circle with radius $0$ is $0$ , we conclude that $A(r) = \pi r^2$ .

Pedagogically, I don’t particularly recommend this approach, as I think students would find this explanation more confusing than the first approach. However, I can see that this could be useful for reinforcing the statement of the Fundamental Theorem of Calculus.

By the way, the above reasoning works for a square or cube also, but with a little twist. For a square of side length $s$ , the area is $A(s) = s^2$ and the perimeter is $P(s) = 4s$ , which isn’t the derivative of $A(s)$ . The reason this didn’t work is because the side length $s$ of a square corresponds to the diameter of a circle, not the radius of a circle.

But, if we let $x$ denote half the side length of a square, then the above logic works out since

$A(x) = s^2 = (2x)^2 = 4x^2$

and

$P(x) = 4s = 4(2x) = 8x$

Written in terms of the half-sidelength $x$ , we see that $A'(x) = P(x)$ .

Graphical statistics

Source: http://www.xkcd.com/539/

Importance of the base case in a proof by induction

In Precalculus, Discrete Mathematics or Real Analysis, an arithmetic series is often used as a student’s first example of a proof by mathematical induction. Recall, from Wikipedia:

Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step, known as the base case, is to prove the given statement for the first natural number. The second step, known as the inductive step, is to prove that the given statement for any one natural number implies the given statement for the next natural number. From these two steps, mathematical induction is the rule from which we infer that the given statement is established for all natural numbers.

The simplest and most common form of mathematical induction infers that a statement involving a natural number n holds for all values of n. The proof consists of two steps:

The basis (base case): prove that the statement holds for the first natural number $n$ . Usually, $n=0$ or $n=1$ .

The inductive step: prove that, if the statement holds for some natural number $n$ , then the statement holds for $n+1$ .

The hypothesis in the inductive step that the statement holds for some $n$ is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for $n+1$ .

As an inference rule, mathematical induction can be justified as follows. Having proven the base case and the inductive step, then any value can be obtained by performing the inductive step repeatedly. It may be helpful to think of the domino effect. Consider a half line of dominoes each standing on end, and extending infinitely to the right. Suppose that:

The first domino falls right.

If a (fixed but arbitrary) domino falls right, then its next neighbor also falls right.

With these assumptions one can conclude (using mathematical induction) that all of the dominoes will fall right.

Mathematical induction… works because $n$ is used to represent an arbitrary natural number. Then, using the inductive hypothesis, i.e. that $P(n)$ is true, show $P(k+1)$ is also true. This allows us to “carry” the fact that $P(0)$ is true to the fact that $P(1)$ is also true, and carry $P(1)$ to $P(2)$ , etc., thus proving $P(n)$ holds for every natural number $n$ .

When students first encounter mathematical induction (in either Precalculus, Discrete Mathematics, or Real Analysis), the theorems that students are asked to prove usually fall into four categories:

Calculating a series (examples below).
Statements concerning divisibility (for example, proving that $4$ is always a factor of $5^n-1$ ).
Finding a closed-form expression for a recursively defined sequence (for example, if $a_1 = 4$ and $a_n = 3a_{n-1}$ if $n \ge 2$ , proving that $a_n = 4 \times 3^{n-1}$ )
Statements concerning inequality (for example, proving that $n! > 4^n$ if $n \ge 9$ )

Here’s a common first (or maybe second) example of mathematical induction applied to an arithmetic series.

Theorem. $1^2 + 2^2 + \dots + (n-1)^2 + n^2 = \displaystyle \frac{n(n+1)(2n+1)}{6}$

Proof. Induction on $n$ .

$n = 1$ : The left-hand is simply $1$ , while the right-hand side is $\displaystyle \frac{(1)(2)(3)}{6}$ , which is also equal to $1$ . So the base case works.

$n$ : Assume that the statement holds true for the integer $n$ .

$n+1$ . If I replace $n$ by $n+1$ in the statement of the theorem, then the right-hand side becomes

$\displaystyle \frac{(n+1)[(n+1)+1][2(n+1)+1]}{6} = \displaystyle \frac{(n+1)(n+2)(2n+3)}{6}$

I find it helpful to describe this to students as my target. In other words, as I manipulate the left-hand side, my ultimate goal is to end up with this target. Once I have done that, then I have completed the proof.

If I replace $n$ by $n+1$ in the statement of the theorem, then the left-hand side will now end on $n+1$ instead of $n$ :

$1^2+ 2^2 + \dots + (n-1)^2 + n^2 + (n+1)^2$

Notice that we’ve seen almost all of this before, except for the extra term $(n+1)^2$ . So we will substitute using the induction hypothesis, carrying the extra $(n+1)^2$ along for the ride.

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 + (n+1)^2 = \displaystyle \frac{n(n+1)(2n+1)}{6} + (n+1)^2$

Now our task is, by hook or by crook, using whatever algebraic tricks we can think of to convert this last expression into the target. Most students are completely comfortable doing this, although they typically multiply out the term $n(n+1)(2n+1)$ unnecessarily. Indeed, many early proofs by induction are simplified by factoring out terms whenever possible — in the example below, $(n+1)$ is factored on the third step — as opposed to multiplying them out. In my experience, proofs by induction often serve as a stringent test of students’ algebra skills as opposed to their skills in abstract reasoning.

In any event, here’s the end of the proof:

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 + (n+1)^2 = \displaystyle \frac{n(n+1)(2n+1)}{6} + (n+1)^2$

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 + (n+1)^2 = \displaystyle \frac{n(n+1)(2n+1) + 6(n+1)^2}{6}$

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 + (n+1)^2 = \displaystyle \frac{(n+1)[n(2n+1) + 6(n + 1)]}{6}$

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 + (n+1)^2 = \displaystyle \frac{(n+1)(2n^2 + n + 6n + 6)}{6}$

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 + (n+1)^2 = \displaystyle \frac{(n+1)(2n^2 + 7n + 6)}{6}$

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 + (n+1)^2 = \displaystyle \frac{(n+1)(n+2)(2n+3)}{6}$

A fair amount of algebra was needed to prove the $n+1$ case. However, the first step — the base case — was especially easy. Indeed, in most proofs by induction seen by students, the base case is often quite trivial… to the point that students often wonder why the base case is needed in the first place.

I first saw this next example in Calculus, by Tom M. Apostol. This next fallacious example illustrates what can happen if the base case is ignored. The statement of this “theorem” doesn’t match the formula for an arithmetic series, and so clearly something is wrong with the following “proof.”

“Theorem.” $1 + 2 + \dots + (n-1) + n = \displaystyle \frac{(2n+1)^2}{8}$

“Proof.” Induction on $n$ .

$n = 1$ : Let’s just ignore the base case, it’s unimportant.

$n$ : Assume that the statement holds true for the integer $n$ .

$n+1$ . If I replace $n$ by $n+1$ in the statement of the theorem, then the right-hand side — my target — becomes

$\displaystyle \frac{[2(n+1)+1]^2}{8} = \displaystyle \frac{(2n+3)^2}{8}$

On the left-hand side, we use the induction hypothesis:

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{(2n+1)^2}{8} + (n+1)$

Now our task is, by hook or by crook, using whatever algebraic tricks we can think of to convert this last expression into the target.

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{(2n+1)^2}{8} + (n+1)$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{(2n+1)^2 + 8(n+1)}{8}$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{4n^2+4n+1+8n+8}{8}$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{4n^2+12n+9}{8}$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{(2n+3)^2}{8}$

So that’s the end of the “proof.”

Clearly, something went wrong with the above proof. What went wrong, obviously, is that we didn’t check the base case. If $n=1$ , then the left-hand side is $1$ . However, the right-hand side is $\displaystyle \frac{[(2)(1) + 1]^2}{8} = \displaystyle \frac{9}{8}$ . So the base case is false.

So what happened?

We correctly showed that, if the case $n$ is true, then the case $n+1$ is also true. The catch, of course, is that the case $n$ is never true. Using the domino analogy, we showed that if a domino falls, then the next domino will fall. But the first domino never falls.

All this to say… yes, it’s important to check that the base case actually works.

Engaging students: Computing trigonometric functions using a unit circle

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Angel Pacheco. His topic, from Precalculus: computing trigonometric functions using a unit circle.

How can this topic be used in your students’ future courses in mathematics or science?

The first course to bring in the unit circle is Pre-Calculus. It is used in a lot in calculus when it comes to finding certain values of trigonometric functions. Knowing how the unit circle works, it allows the students to solve a lot of trigonometric functions on their own. Once students reach college level mathematics, they will learn that the unit circle is a key element to trigonometry. Trigonometry is a huge part of all the calculus courses.

Science contains a lot of trigonometry, mainly physics. The law of sine and cosine allows the students to determine the angle an object is or even how far it is. Being able to use the unit circle to solve for functions, it allows them to use it any subject whether it be a science or a math class. Students or scientists that know how to solve trigonometric functions using the unit circle allows them to compute certain things on paper as opposed to relying on a calculator to do all the work.

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?

Technology can be used to introduce and also evaluate their content. There are different ways to use technology. One example is using Khan Academy videos to show students how it works or how Khan explains. Students having to look at a video can have them engage on the topic. My personal favorite is to create an exciting video and put it on YouTube. I have noticed that parodies are a popular trend so creating a parody with the unit circle with a popular song will be effective to engaging the students to this topic. The next thing I would use for technology is graphing calculators. I think if the students see that the calculator gives them the same answer as the values they learned from the unit circle, they would be amazed on how the concept of the unit circle is. My classmates and I were in complete shock when we realized how the unit circle worked. My former teacher also had a clock based off of the unit circle so we had to learn it in order to read the time.

How could you as a teacher create an activity or project that involves your topic?

The link below shows a sample lesson that allows the students work in groups to solve trigonometric functions and create a table that shows the solution to certain problems. The students will have a calculator with them that can be used for checking their answers to see if they are on the right track with the assignment. Also, having access to the computers to research particular things that they need for explaining will be acceptable. In my opinion, I feel that there is some tweaking that I recommend making it more effective. I would like to have a website that visually shows the unit circle. If possible, I would like for the students to have a worksheet that allows them to know which steps to follow to ensure that they are on the right track. A great form of assessment will be a quiz following this activity. I feel asking them to draw the unit circle and also solve certain trigonometric functions to see if they understand it. I would also like to like to bring in all six of the functions and show the relation with the unit circle.

Source(s): http://alex.state.al.us/lesson_view.php?id=27478

Why do we teach students about radians?

Throughout grades K-10, students are slowly introduced to the concept of angles. They are told that there are $90$ degrees in a right angle, $180$ degrees in a straight angle, and a circle has $60$ degrees. They are introduced to $30-60-90$ and $45-45-90$ right triangles. Fans of snowboarding even know the multiples of $180$ degrees up to $1440$ or even $1620$ degrees.

Then, in Precalculus, we make students get comfortable with $\pi$ , $\displaystyle \frac{\pi}{2}$ , $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{\pi}{4}$ , $\displaystyle \frac{\pi}{6}$ , and multiples thereof.

We tell students that radians and degrees are just two ways of measuring angles, just like inches and centimeters are two ways of measuring the length of a line segment.

Still, students are extremely comfortable with measuring angles in degrees. They can easily visualize an angle of $75^o$ , but to visualize an angle of $2$ radians, they inevitably need to convert to degrees first. In his book Surely You’re Joking, Mr. Feynman!, Nobel-Prize laureate Richard P. Feynman described himself as a boy:

I was never any good in sports. I was always terrified if a tennis ball would come over the fence and land near me, because I never could get it over the fence – it usually went about a radian off of where it was supposed to go.

Naturally, students wonder why we make them get comfortable with measuring angles with radians.

The short answer, appropriate for Precalculus students: Certain formulas are a little easier to write with radians as opposed to degrees, which in turn make certain formulas in calculus a lot easier.

The longer answer, which Precalculus students would not appreciate, is that radian measure is needed to make the derivatives of $\sin x$ and $\cos x$ look palatable.

Source: http://mathworld.wolfram.com/CircularSector.html

1. In Precalculus, the length of a circle arc with central angle $\theta$ in a circle with radius $r$ is

$s = r\theta$

Also, the area of a circular sector with central angle $\theta$ in a circle with radius $r$ is

$A = \displaystyle \frac{1}{2} r^2 \theta$

In both of these formulas, the angle $\theta$ must be measured in radians.

Students may complain that it’d be easy to make a formula of $\theta$ is measured in degrees, and they’d be right:

$s = \displaystyle \frac{180 r \theta}{\pi}$ and $A = \displaystyle \frac{180}{\pi} r^2 \theta$

However, getting rid of the $180/\pi$ makes the following computations from calculus a lot easier.

2a. Early in calculus, the limit

$\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

is derived using the Sandwich Theorem (or Pinching Theorem or Squeeze Theorem). I won’t reinvent the wheel by writing out the proof, but it can be found here. The first step of the proof uses the formula for the above formula for the area of a circular sector.

2b. Using the trigonometric identity $\cos 2x = 1 - 2 \sin^2 x$ , we replace $x$ by $\theta/2$ to find

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \frac{2\sin^2 \displaystyle \left( \frac{\theta}{2} \right)}{ \theta}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \sin \left( \frac{\theta}{2} \right) \cdot \frac{\sin \displaystyle \left( \frac{\theta}{2} \right)}{ \displaystyle \frac{\theta}{2}}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0 \cdot 1$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0$

3. Both of the above limits — as well as the formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ — are needed to prove that $\displaystyle \frac{d}{dx} \sin x = \cos x$ and $\displaystyle \frac{d}{dx} \cos x = -\sin x$ . Again, I won’t reinvent the wheel, but the proofs can be found here.

So, to make a long story short, radians are used to make the derivatives $y = \sin x$ and $y = \cos x$ easier to remember. It is logically possible to differentiate these functions using degrees instead of radians — see http://www.math.ubc.ca/~feldman/m100/sinUnits.pdf. However, possible is not the same thing as preferable, as calculus is a whole lot easier without these extra factors of $\pi/180$ floating around.