Engaging students: Deriving the double angle formulas for sine, cosine, and tangent

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 Daniel Adkins. His topic, from Precalculus: deriving the double angle formulas for sine, cosine, and tangent.

How does this topic extend what your students should have already learned?

A major factor that simplifies deriving the double angle formulas is recalling the trigonometric identities that help students “skip steps.” This is true especially for the Sum formulas, so a brief review of these formulas in any fashion would help students possibly derive the equations on their own in some cases. Listed below are the formulas that can lead directly to the double angle formulas.

A list of the formulas that students can benefit from recalling:

• Sum Formulas:
• sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
• cos(a+b) = cos(a)cos(b) – sin(a)sin(b)
• tan(a+b) = [tan(a) +tan(b)] / [1-tan(a)tan(b)]

• Pythagorean Identity:
• Sin2 (a) + Cos2(a) = 1

This leads to the next topic, an activity for students to attempt the equation on their own.

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

I’m a firm believer that the more often a student can learn something of their own accord, the better off they are. Providing the skeletal structure of the proofs for the double angle formulas of sine, cosine, and tangent might be enough to help students reach the formulas themselves. The major benefit of this is that, even though these are simple proofs, they have a lot of variance on how they may be presented to students and how “hands on” the activity can be.

I have an example worksheet demonstrating this with the first two double angle formulas attached below. This is in extremely hands on format that can be given to students with the formulas needed in the top right corner and the general position where these should be inserted. If needed the instructor could take this a step further and have the different Pythagorean Identities already listed out (I.e. Cos2(a) = 1 – Sin2(a), Sin2(a) = 1 – Cos2(a)) to emphasize that different formats could be needed. This is an extreme that wouldn’t take students any time to reach the conclusions desired. Of course a lot of this information could be dropped to increase the effort needed to reach the conclusion.

A major benefit with this also is that even though they’re simple, students will still feel extremely rewarded from succeeding on this paper on their own, and thus would be more intrinsically motivated towards learning trig identities.

How can Technology be used to effectively engage students with this topic?

When it comes to technology in the classroom, I tend to lean more on the careful side. I know me as a person/instructor, and I know I can get carried away and make a mess of things because there was so much excitement over a new toy to play with. I also know that the technology can often detract from the actual math itself, but when it comes to trigonometry, and basically any form of geometric mathematics, it’s absolutely necessary to have a visual aid, and this is where technology excels.

The Wolfram Company has provided hundreds of widgets for this exact purpose, and below, you’ll find one attached that demonstrates that sin(2a) appears to be equal to its identity 2cos(a)sin(a). This is clearly not a rigorous proof, but it will help students visualize how these formulas interact with each other and how they may be similar. The fact that it isn’t rigorous may even convince students to try to debunk it. If you can make a student just irritated enough that they spend a few minutes trying to find a way to show you that you’re wrong, then you’ve done your job in that you’ve convinced them to try mathematics for a purpose.

After all, at the end of the day, it doesn’t matter how you begin your classroom, or how you engage your students, what matters is that they are engaged, and are willing to learn.

Wolfram does have a free cdf reader for its demonstrations on this website: http://demonstrations.wolfram.com/AVisualProofOfTheDoubleAngleFormulaForSine/

References

My Favorite One-Liners: Part 6

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Sometimes, I’ll expect students to learn and master operations that cancel. For example, in Precalculus, I want my students to know the sum-to-product trigonometric identities

$\sin u + \sin v = \displaystyle 2 \sin\left( \frac{u+v}{2} \right) \cos \left( \frac{u-v}{2} \right)$,

$\sin u - \sin v = \displaystyle 2 \cos\left( \frac{u+v}{2} \right) \sin\left( \frac{u-v}{2} \right)$,

$\cos u + \cos v = \displaystyle 2 \cos\left( \frac{u+v}{2} \right) \cos \left( \frac{u-v}{2} \right)$,

$\cos u - \cos v = \displaystyle -2 \sin \left( \frac{u+v}{2} \right) \sin\left( \frac{u-v}{2} \right)$.

These can be helpful for solving trigonometric equations. For example, to solve $\cos 3x + \cos 7x = 0$, we have

$\cos 3x + \cos 7x = 0$

$\displaystyle 2 \sin\left( \frac{3x+7x}{2} \right) \cos \left( \frac{3x-7x}{2} \right) = 0$

$2 \sin 5x \cos (-2x) = 0$

$2 \sin 5x \cos 2x = 0$

$\displaystyle x = \frac{n\pi}{5} \qquad \hbox{or} \qquad \displaystyle x = \left( \frac{n}{2} + \frac{1}{4} \right)\pi$ for integers $n$.

However, I also want my students to know the product-to-sum trigonometric identities

$\cos x \cos y = \displaystyle \frac{1}{2} [\cos(x+y) + \cos(x-y) ]$,

$\sin x \sin y = \displaystyle \frac{1}{2} [\cos(x-y) - \cos(x+y) ]$,

$\sin x \cos y = \displaystyle \frac{1}{2} [\sin(x+y) + \sin(x-y) ]$.

These are useful when computing certain definite integrals (especially related to Fourier series). For example, if $m \ne n$ are both integers, then

$\displaystyle \int_0^{2\pi} \cos mx \cos nx \, dx = \displaystyle \int_0^{2\pi} \frac{1}{2} \left[\cos([m+n]x) + \cos([m-n])x) \right] \, dx$

$= \left[ \displaystyle \frac{\sin([m+n]x)}{2(m+n)} + \frac{\sin([m-n]x)}{2(m-n)} \right]^{2\pi}_0$

$= \displaystyle \frac{\sin(2[m+n]\pi) - \sin 0}{2(m+n)} + \frac{\sin(2[m-n]\pi) - \sin 0}{2(m-n)}$.

$=0$

This integral and other similar integrals are necessary to find the formula for the coefficients in a Fourier series.

In other words, sometimes I’ll want my students to convert a product into a sum. Other times, I’ll want my students to convert a sum into a product.

To help this sink in, I’ll tell my students, “To quote the great philosopher: Sometimes you gotta know when to hold ’em, know when to fold them.”

However, when I made this joke recently, a student innocently asked, “What great philosopher said that?” I turned the question back to my class, but not one of my class of millennials knew the answer. One person came close with his answer of “Willie” — wrong answer but correct genre and time frame. (Somebody else answered Socrates.)

So that my students actually learn something important in my class, here’s the cultural reference:

The antiderivative of 1/(x^4+1): Part 7

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

As we’ve seen in this series, the answer is

$\displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} ( x\sqrt{2} - 1 ) + \frac{\sqrt{2}}{4} \tan^{-1}( x \sqrt{2} + 1) + C$

It turns out that this can be simplified somewhat as long as $x \ne 1$ and $x \ne -1$. I’ll use the trig identity

$\tan(\alpha + \beta) = \displaystyle \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

When I apply this trig identity for $\alpha = \tan^{-1} ( x\sqrt{2} - 1 )$ and $\beta = \tan^{-1} ( x\sqrt{2} + 1 )$, I obtain

$\tan \left[ \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) \right] = \displaystyle \frac{x \sqrt{2} - 1 + x \sqrt{2} + 1}{1 - (x\sqrt{2} - 1)(x\sqrt{2} + 1)}$

$= \displaystyle \frac{2x \sqrt{2}}{1 - (2x^2 - 1)}$

$= \displaystyle \frac{2x \sqrt{2}}{2 - 2x^2}$

$= \displaystyle \frac{x \sqrt{2}}{1 - x^2}$.

So we can conclude that

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + n\pi$

for some integer $n$ that depends on $x$. The $+n\pi$ is important, as a cursory look reveals that $y = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ and $y = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ have different graphs. (The vertical lines in the orange graph indicate where the right-hand side is undefined when $x = 1$ or $x = -1$.

The two graphs coincide when $-1 < x < 1$ but differ otherwise. However, it appears that the two graphs differ by a constant. Indeed, if I subtract $\pi$ from the orange graph if $x < -1$ and add $\pi$ to the orange graph if $x > 1$, then they match:

So, evidently

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < -1$,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $-1 < x < 1$,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> 1$.

So as long as $x \ne 1$ and $x \ne -1$, this constant $-\pi$, $0$, or $\pi$ can be absorbed into the constant $C$:

$\displaystyle \int \frac{1}{x^4 + 1} dx = \displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + C$.

However, a picture may be persuasive but is not a proof, and there are some subtle issues with this simplification. I’ll discuss these further details in tomorrow’s post.

How I Impressed My Wife: Part 2a

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

In this series, I’ll explore different ways of evaluating this integral.I begin by adjusting the range of integration:

$Q = Q_1 + Q_2 + Q_3$,

where

$Q_1 = \displaystyle \int_0^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$,

$Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$,

$Q_3 = \displaystyle \int_{3\pi/2}^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$.

I’ll begin with $Q_3$ and apply the substitution $u = x - 2\pi$, or $x = u + 2\pi$. Then $du = dx$, and the endpoints change from $3\pi/2 \le x 2\pi$ to $-\pi/2 \le u \le 0$. Therefore,

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 (u+2\pi) + 2 a \sin (u+2\pi) \cos (u+2\pi) + (a^2 + b^2) \sin^2 (u+2\pi)}$.

Next, we use the periodic property for both sine and cosine — $\sin(x + 2\pi) = \sin x$ and $\cos(x + 2\pi) = \cos x$ — to rewrite $Q_3$ as

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{du}{\cos^2 u + 2 a \sin u \cos u + (a^2 + b^2) \sin^2 u}$.

Changing the dummy variable from $u$ back to $x$, we have

$Q_3 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$.

Therefore, we can combined $Q_3 + Q_1$ into a single integral:

$Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{0} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$+ \displaystyle \int_0^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$Q_3 + Q_1 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Next, we work on the middle integral $Q_2$. We use the substitution $u = x - \pi$, or $x = u + \pi$, so that $du = dx$. Then the interval of integration changes from $\pi/2 \le x \le 3\pi/2$ to $-\pi/2 \le u \le \pi/2$, so that

$Q_2 = \displaystyle \int_{\pi/2}^{3\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{du}{\cos^2 (u+\pi) + 2 a \sin (u+\pi) \cos (u+\pi) + (a^2 + b^2) \sin^2 (u+\pi)}$.

Next, we use the trigonometric identities

$\sin(u + \pi) = \sin u \cos \pi + \cos u \sin \pi = \sin u \cdot (-1) + \cos u \cdot 0 = - \sin u$,

$\cos(u + \pi) = \cos u \cos \pi - \sin u \sin \pi = \cos u \cdot (-1) - \sin u \cdot 0 = - \cos u$,

so that the last integral becomes

$Q_2 = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{(-\cos u)^2 + 2 a (-\sin u)(- \cos u) + (a^2 + b^2) (-\sin u)^2}$

$= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{du}{\cos^2 u + 2 a \sin u \cos u + (a^2 + b^2) \sin^2 u}$

$= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

On the line above, I again replaced the dummy variable of integration from $u$ to $x$. We see that $Q_2 = Q_1 + Q_3$, and so

$Q = Q_1 + Q_2 + Q_3$

$Q = 2 Q_2$

$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

I’ll continue with the evaluation of this integral in tomorrow’s post.

Proving theorems and special cases (Part 12): The sum and difference formulas for sine

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

3. Theorem 1. $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha + \sin \beta$

Theorem 2. $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

For angles that are not acute, these theorems can be proven using a unit circle and the following four lemmas:

Lemma 1. $\cos(x - y) = \cos x \cos y + \sin x \sin y$

Lemma 2. $\cos(x + y) = \cos x \cos y - \sin x \sin y$

Lemma 3. $\sin(\pi/2 - x) = \cos x$

Lemma 4. $\cos(\pi/2 - x) = \sin x$

Specifically, assuming Lemmas 1-4, then:

$\sin(\alpha + \beta) = \cos(\pi/2 - [\alpha + \beta])$ by Lemma 4

$= \cos([\pi/2 - \alpha] - \beta)$

$= \cos(\pi/2 - \alpha) \cos \beta + \sin(\pi/2 - \alpha) \sin \beta$ by Lemma 1

$= \sin \alpha \cos \beta + \cos \alpha \sin \beta$ by Lemmas 3 and 4.

Also,

$\sin(\alpha - \beta) = \cos(\pi/2 - [\alpha - \beta])$ by Lemma 4

$= \cos([\pi/2 - \alpha] + \beta)$

$= \cos(\pi/2 - \alpha) \cos \beta - \sin(\pi/2 - \alpha) \sin \beta$ by Lemma 2

$= \sin \alpha \cos \beta - \cos \alpha \sin \beta$ by Lemmas 3 and 4.

However, we see that what I’ve called Lemma 3, often called a cofunction identity, can be considered a special case of Theorem 2. However, this is not circular logic since the cofunction identities can be proven without appealing to Theorems 1 and 2.

Inverse functions: Arcsine and SSA (Part 14)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like $\sin \theta = 0.8$. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has exactly one solution:

Solve $\triangle ABC$ if $a = 15$, $c = 10$, and $\alpha = 30^\circ$.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case. We also note that the circle would have intersected the black dashed line had the dashed line been extended to the left. This will become algebraically clear in the solution below.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{15} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1}{3} = \sin \gamma$

At this point, the natural inclination of a student is to pop out the calculator and find $\sin^{-1} \frac{1}{3}$.

This is incorrect logic that, as we’ll see tomorrow, nevertheless leads to the correct conclusion. This is incorrect logic because there are two angles between $0^\circ$ and $180^\circ$ with a sine of $1/3$. There is one solution in the first quadrant (the unique answer specified by arcsine), and there is another answer in the second quadrant — which is between $90^\circ$ and $180^\circ$ and hence not a permissible value of arcsine. Let me demonstrate this in three different ways.

First, let’s look at the graph of $y = \sin x$ (where, for convenience, the units of the $x-$axis are in degrees). This graph intersects the line $y = \frac{1}{3}$ in two different places between $0^\circ$ and $180^\circ$. This does not violate the way that arcsine was defined — arcsine was defined using the restricted domain $[-\pi/2,\pi/2]$, or $[-90^\circ, 90^\circ]$ in degrees.

Second, let’s look at drawing angles in standard position. The angle in the second quadrant is clearly the reflection of the angle in the first quadrant through the $y-$axis.

Third, let’s use a trigonometric identity to calculate $\sin \left( \pi - \sin^{-1} \displaystyle \frac{1}{3} \right)$:

$\sin \left( \pi - \sin^{-1} \displaystyle \frac{1}{3} \right) = \sin \pi \cos \left( \sin^{-1} \displaystyle \frac{1}{3} \right) - \cos \pi \sin \left( \sin^{-1} \displaystyle \frac{1}{3} \right)$

$=0 \cdot \cos \left( \sin^{-1} \displaystyle \frac{1}{3} \right) + 1 \cdot \sin \left( \sin^{-1} \displaystyle \frac{1}{3} \right)$

$= \displaystyle \frac{1}{3}$

Fourth, and perhaps most convincingly for modern students (to my great frustration), let’s use a calculator:

All this to say, blinding computing $\sin^{-1} \frac{1}{3}$ uses incorrect logic when solving this problem.

Tomorrow, we’ll examine what happens when we try to solve the triangle using these two different solutions for $\gamma$.

Calculators and complex numbers (Part 3)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta)$

where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series.

Why is this important? When students first learn to multiply complex numbers like $1+i$ and $2+i$, they are taught to just distribute (or, using the nomenclature that I don’t like, FOIL it out):

$(1+i)(1+2i) = 1 + 2i + i + 2i^2 = 1 + 3i - 2 = -1 + 3i$.

The trigonometric form of a complex number permits a geometric interpretation of multiplication, given in the following theorem.

Theorem. $\left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right] = r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2])$.

Proof. As above, we distribute (except for the $r_1$ and $r_2$ terms):

$\left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right]$

$= r_1 r_2 (\cos \theta_1 \cos \theta_2 + i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + i^2 \sin \theta_1 \sin \theta_2$

$= r_1 r_2 (\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i[ \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2])$

$= r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2])$.

When actually doing this in class, the big conceptual jump for students is the last step. So I make a big song-and-dance routine out of this:

Cosine of the first times cosine of the second minus sine of the first times sine of the second… where have I seen this before?

The idea is for my students to search deep into their mathematical memories until they recall the appropriate trig identity.

For the original multiplication problem, we see that

$1+i = \sqrt{2} \left( \cos 45^\circ + i \sin 45^\circ \right)$

$1 + 2i = \sqrt{5} \left( \cos[\tan^{-1} 2] + i \sin[\tan^{-1} 2] \right) \approx \sqrt{5} \left( \cos 63.435^\circ + i \sin 63.435^\circ \right)$

Therefore, the product of $1+i$ and $1+2i$ will be a distance of $\sqrt{2} \cdot \sqrt{5} = \sqrt{10}$ from the origin, and the angle from the positive real axis will be $45^\circ + \tan^{-1} 2 \approx 45^\circ + 63.435^\circ = 108.435^\circ$. Indeed,

$-1 + 3i \approx \sqrt{10} \left( \cos 108.435^\circ + i \sin 108.435^\circ \right)$.

For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

Two ways of doing an integral (Part 2)

A colleague placed the following problem on an exam, expecting the following solution:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = \sin^{-1} \left( \displaystyle \frac{x-2}{2} \right) + C$

However, one student produced the following solution (see yesterday’s post for details):

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = 2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right) + C$

As he couldn’t find a mistake in the student’s work, he assumed that the two expressions were equivalent. Indeed, he differentiated the student’s work to make sure it was right. But he couldn’t immediately see, using elementary reasoning, why they were equivalent. So he walked across the hall to my office to ask me if I could help.

Here’s how I showed they are equivalent.

Let $\alpha = \displaystyle \sin^{-1} \left( \frac{x-2}{2} \right)$ and $\beta = \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right)$. Then

$\displaystyle \sin(\alpha - 2\beta) = \sin \alpha \cos 2\beta - \cos \alpha \sin 2\beta$.

Let’s evaluate the four expressions on the right-hand side.

First, $\sin \alpha$ is clearly equal to $\displaystyle \frac{x-2}{2}$.

Second, $\cos 2\beta = 1 - 2 \sin^2 \beta$, so that

$\cos 2\beta = \displaystyle 1 - 2\left( \frac{\sqrt{x}}{2} \right)^2 = \displaystyle 1 - \frac{x}{2} = \displaystyle -\frac{x-2}{2}$.

Third, to evaluate $\cos \alpha$, I’ll use the identity $\displaystyle \cos \left( \sin^{-1} x \right) = \sqrt{1 - x^2}$:

$\cos \alpha = \displaystyle \sqrt{1 - \left( \frac{x-2}{2} \right)^2 } = \displaystyle \frac{\sqrt{4x-x^2}}{2}$

Fourth, $\sin 2\beta = 2 \sin \beta \cos \beta$. Using the above identity again, we find

$\sin 2\beta = \displaystyle 2 \left( \frac{ \sqrt{x} }{2} \right) \sqrt{ 1 - \left( \frac{ \sqrt{x} }{2} \right)^2 }$

$= \sqrt{x} \sqrt{1 - \displaystyle \frac{x}{4}}$

$= \displaystyle \frac{\sqrt{4x-x^2}}{2}$

Combining the above, we find

$\sin(\alpha - 2 \beta) = \displaystyle -\left( \frac{x-2}{2} \right)^2 - \left( \frac{\sqrt{4x-x^2}}{2} \right)^2$

$\sin(\alpha - 2 \beta) = \displaystyle \frac{-(x^2 - 4x + 4) - (4x - x^2)}{4}$

$\sin(\alpha - 2 \beta) = -1$

$\alpha - 2 \beta = \displaystyle -\frac{\pi}{2} + 2\pi n$ for some integer $n$

Also, since $-\pi/2 \le \alpha \le \pi/2$ and $0 \le -2\beta \le \pi$, we see that $-\pi/2 \le \alpha - 2 \beta \le 3\pi/2$. (From its definition, $\beta$ is the arcsine of a positive number and therefore must be nonnegative.) Therefore, $\alpha - 2\beta = -\pi/2$.

In other words,

$\sin^{-1} \left( \displaystyle \frac{x-2}{2} \right)$ and $2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right)$

differ by a constant, thus showing that the two antiderivatives are equivalent.

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.