Friday, June 16, 2017

Van Brummelen Chapter 1: Heavenly Mathematics

Table of Contents

1. Pappas Page of the Day
2. Introduction to Van Brummelen's Book
3. Chapter 1: Heavenly Mathematics
4. Exercise 1.1: sin 3
5. Exercise 1.2: Subtraction Formula for Sine
6. Exercise 1.3: Concavity of Sine
7. Exercise 1.11: Velocity Due to Rotation

Pappas Page of the Day

This is what Theoni Pappas writes on page 167 of her Magic of Mathematics:

"I have discovered such wonderful things that I was amazed -- out of nothing I have created a strange new universe." -- Janos Bolyai from a letter to his father, 1823.

This is the first page of a new section, "Geometries -- Old & New." By "old geometry," Pappas is referring to Euclidean geometry, and so the "new geometries" are non-Euclidean.

As the readers of this blog already know, non-Euclidean geometry is one of my favorite topics. I've devoted the past two summers to writing about non-Euclidean geometry. Notice that Pappas quotes Bolyai, one of the creators of one non-Euclidean geometry -- hyperbolic geometry. But the geometry I wrote about the past two years is the other type -- spherical geometry.

Usually, I write a few paragraphs summarizing the Pappas page of the day, as well as add my own knowledge of the topic. But I've already lined up another side-along reading book -- Glen Van Brummelen's Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry. This book extends spherical geometry by defining trigonometry on the sphere.

And so this will count as my explanation of the Pappas page -- a description and summary of Van Brummelen's book. Notice that last summer, I tried to alternate between my school posts and my spherical geometry posts. But this summer, I've changing it up so that my spherical geometry posts line up with Pappas pages. This is why my first two summer posts (June 1st and 9th) were both school posts, and now I'm following that up with multiple spherical trig posts.

Introduction to Van Brummelen's Book

Van Brummelen begins his book with a preface:

"Mathematical subjects come and go. If you glance at a textbook from a century ago you may recognize some of the contents, but some will be unfamiliar or even baffling. A high school text in analytic geometry, for instance, once contained topics like involutes of circles, hypocycloids, and auxiliary circles of ellipses: topics that most college students today will never see. But spherical trigonometry may be the most spectacular example of changing fashions in the 20th century."

These opening sentences sound a lot like the complaints of our traditionalists (most notably Katharine Beals) -- so many topics from 100 years ago are missing under the Common Core Standards. But according to Van Brummelen, the loss of spherical trig dates back to the 1950's -- well before the Common Core era.

According to the author, the loss of spherical trig is a shame. He writes:

"This paucity comes at a time when new applications of spherical trigonometry are being found. GPS devices have some of its formulas built in. It's amusing to see bibliographies of research papers in computer graphics and animation (for use in movies like those made by Pixar) referring to nothing older than last week, except for that stodgy old spherical trig text."

Of course, today Pixar releases its latest movie, Cars III. Who'd have thought that spherical trig could have been used to create the animation in that movie?

Van Brummelen writes about how to use his book:

"Mathematics teachers may wish to use some of the material in their classes. The core of this book is chapters 1, 2, 5, and 6, although chapter 1 can stand on its own. Chapters 3 and 4 provide an interesting historical contrast to the modern theory, but may be skipped if the instructor wishes a briefer journey; my own course covers chapters 1 through 6. The remaining chapters evolved from student projects. I can vouch personally that the first six chapters work well in a class setting with an enthusiastic group."

On the other hand, I mentioned in my May 15th post that Barron Trump, the president's son, attends a school that ends the Honors Geometry course with spherical geometry (not spherical trig). I've been thinking a lot about what such a course might look like. Indeed, I'll write more about some of my ideas in tomorrow's post.

Also, notice that the first two chapters of Van Brummelen aren't about spherical trig yet. Instead, they're more like review chapters -- Chapter 1 reviews the "trig" part, while Chapter 2 reviews the "spherical" part. Therefore in Chapter 2 tomorrow, we'll review much of the spherical geometry that we learned last year from Legendre's old book.

Chapter 1: Heavenly Mathematics

Chapter 1 of Glen Van Brummelen has the same title as the book, "Heavenly Mathematics." As I wrote above, this chapter actually reviews Euclidean trig, but the author does apply standard trig to answer a celestial question -- how far is it to the moon?

Van Brummelen begins:

"We're not ancient anymore. The birth and development of modern science have brought us to a point where we know much more about how the universe works. Not only do we know more; we also have reasons to believe what we know."

Here the author describes how much of science is based on evidence or proof -- and of course, Geometry class is all about proof. Then again, he laments that most trig students don't learn how to derive the sine addition formula.

"The goal of this chapter is twofold. Firstly, we will revisit topics in plane trigonometry in order to prepare for our passage to the sphere. But our second purpose takes precedence: to explore and learn without taking anything on faith that we cannot ascertain with our own eyes and minds. This is how mathematics works, and by necessity it was how ancient scientists worked. They had no one to build on. Our mission is as follows: Accepting nothing but the evidence of our senses and simple measurements we can take ourselves, determine the distance to the Moon."

I admit that sometimes it's difficult to appreciate how difficult it is to find this distance. We can type in "distance to the moon" into Google and obtain a value within seconds, but that doesn't show us where that number came from. It's not as if anyone took a yardstick or tape and actually measured this distance!

Van Brummelen begins with an easier task -- to find the size of the earth. He suggests a method discovered by the medieval scientist al-Biruni, who was actually trying to determine how to find the direction to Mecca, to aid Muslim prayer. (Yes, we are currently in the holy month of Ramadan, and so the qibla is definitely relevant right as we speak.)

The method involves climbing to the top of a mountain and constructing similar triangles. The author provides us with an equation:

cos theta = OT/OE = r / (r + 305.1 m)

where r = OT is the radius of the earth, OE is the radius plus the height of the mountain, and theta is the angle of depression to the horizon as viewed from the mountaintop, which al-Biruni found to be about 34 minutes of arc (as in 34/60 degrees). The author solves the equation to obtain r = 6238, but van Brummelen writes that there's a problem:

"Our goal was to work without relying on anyone or anything, and at the end we likely relied on Texas Instruments to tell us the value of cos 0.56667 degrees."

And so Van Brummelen proceeds to calculate exact values to create a trig table, just as the ancient Greek mathematician Claudius Ptolemy would have done. Some are easy to find -- Lesson 14-4 of the U of Chicago text gives us the sines of 30, 45, and 60 degrees. The author tells us that the next simplest value to find is sin 36, since this is related to the regular pentagon (though we don't show this in high school Geometry). Then, the sine addition formula is used to find more values -- he states and proves this formula:

Theorem: If alpha, beta < 90, then sin (alpha + beta) = sin alpha cos beta  + cos alpha sin beta.

Given: Van Brummelen's proof refers to a "Figure 1.9," so let me describe it here. There is a right triangle OCE, and its hypotenuse OC is the leg of another right triangle OCD, whose hypotenuse OD is equal to 1. Angle COE has measure alpha, and angle DOC has measure beta. We drop a perpendicular from D to OE and label the point of intersection G. Last, we drop a perpendicular from C to GD and label the point of intersection F.

Proof: In figure 1.9, since OD = 1, the quantity we're after is GD = sin (alpha + beta). It is conveniently broken into two parts, GF and FD. Now from Triangle OCD we know that OC = cos beta and CD = sin beta. So in Triangle OCE we know now the hypotenuse. Thus sin alpha = EC/cos beta, so EC = sin alpha cos beta. Since EC = FG, we're halfway there: we've found one of the two line segments comprising GD.

We can find FD by noticing first that Triangle OCE is similar to Triangle DCF. This statement is true because Angle FCO = alpha, so Angle FCD = 90 - alpha, and the two triangles share two angles, so they must share the third. So Angle FDC = alpha...and we already know the hypotenuse CD of Triangle DCF. So cos alpha = FD/sin beta, which gives FD = cos alpha sin beta, and finally we have sin (alpha + beta) = sin alpha cos beta + cos alpha sin beta. QED

Van Brummelen tells us that perhaps for the first time in our mathematics education, we have a reason to believe the sine addition law. He tells us that using this law, we can use the sines of 30 and 45 degrees to find the sin 75, and sin 36 twice to find sin 72. A similar-looking formula for subtraction then gives us sin 3 from the sines of 75 and 72 degrees -- and from there, we can find the sine of any angle that's a multiple of three degrees.

But unfortunately, much to Ptolemy's dismay, we can't find the sine of 1 degree this way. The best he could do is use the following theorem, which Van Brummelen states but doesn't prove:

Theorem: If beta < alpha < 90, then alpha/beta > sin alpha/sin beta.

According to the author, Ptolemy would use a half-angle formula with sin 3 to find the sines of 3/2 and 3/4 of a degree, and then interpolate these using the inequality above to find sin 1. The upper and lower bounds produced by this formula give sin 1 to within five decimal digits.

By the way, once a table of trig values has been produced, we can use it not only to find the size of the earth, but also the distance to the moon. Claudius Ptolemy may have used parallax -- the idea that two people in different locations would see the moon in two different spots in the sky. The value he came up with was 395,167 km or 245,545 miles. This is not that far off from the distance as given by Google -- 238,900 miles. (Note: Some of my posts dated April 2016 discuss these calculations in more detail, except here we focus more on calculating the actual trig values.)

Exercise 1.1: sin 3

Believe it or not, Van Brummelen's book contains exercises. Well, in the preface, he does admit that he uses this as a textbook in his college classes! Let's try some of these exercises here on the blog.

1. Using only a basic pocket calculator (no scientific calculators, although you may take square roots), determine the value of sin 3 in the most efficient way that you can. Include in your work the computation of any sine values you need along the way.

Let's use the method Van Brummelen suggests in the text above -- find the sines of 75 and 72 degrees first, then use the subtraction formula. And to find the sine of 75 degrees, we use the addition formula with the sines of 30 and 45 degrees:

sin 75 = sin 45 cos 30 + cos 45 sin 30
           = (sqrt(2)/2) (sqrt(3)/2) + (sqrt(2)/2) (1/2)
           = (sqrt(6) + sqrt(2))/4
           = 0.96593

I've seen some calculators display exact values not just for fractions, but square roots as well. (I've mentioned such calculators in my November 29th post.) With these calculators, if we enter sin 75, the calculator actually displays (sqrt(6) + sqrt(2))/4. Of course, I didn't ask my middle school students to find sin 75, but it did come up back when I was tutoring high school students a few years ago -- and of course they were confused by the appearance of that value.

Before we leave 75 degrees, notice that we must find cos 75 as well, since both sin 75 and cos 75 appear in the subtraction formula for 3 degrees. Since cos 75 = sin 15, it's easiest just to use the subtraction formula with 45 and 30 degrees:

sin 15 = sin 45 cos 30 - cos 45 sin 30
           = (sqrt(2)/2) (sqrt(3)/2) - (sqrt(2)/2) (1/2)
           = (sqrt(6) - sqrt(2))/4
           = 0.25882

Now let's try 72 degrees. Van Brummelen already shows us how to find sin 36 degrees -- he uses a regular pentagon with each side equal to 1. Here are the labels -- A, B, and C are consecutive vertices of the pentagon, D and E are where AC intersects the diagonals drawn from B (in the order A-D-E-C), and F is the midpoint of DE (and of AC).

The author finds BD by noting that triangles ABC and ADB are similar. He obtains 0.61803, which he notes is the golden ratio. This is often denoted by the symbol "phi" -- and it's exact value is easily found to be (sqrt(5) - 1)/2. (This is not to be confused with (sqrt(5) + 1)/2 = 1/phi = 1 + phi. This ratio is often written as a capital "Phi.")

He then finds the other two values DF and BF using decimals. But let's try to keep the exact values as long as possible. So we'll perform his calculations in terms of phi:

DF = AC/2 - AD
       = (1 + phi)/2 - phi
       = (1 - phi)/2

BF^2 + DF^2 = BD^2
BF^2 + (1 - phi)^2/4 = phi^2
BF^2 = phi^2 - (1 - phi)^2/4
BF^2 = (-1 + 2phi + 3phi^2)/4
BF^2 = (2 - phi)/4                     (using the known fact that phi^2 = 1 - phi)
BF = sqrt(2 - phi)/2

If we write this in terms of sqrt(5) instead of phi, we obtain BF = sqrt((5 - sqrt(5))/2)/2. At this point, we want to avoid the nested radicals and so we write it as a decimal, 0.58779.

At this point, Van Brummelen suggests that we use addition formulas to find sin 72. But notice that this isn't necessarily -- earlier, he points out that Angle BDF = 72. So Triangle BDF is a right triangle with a 72 degree angle, and so we can just find sin 72 directly using the values we already found!

sin 72 = BF/BD
           = (sqrt(2 - phi)/2)/phi
           = 0.58779/0.61803
           = 0.95107

cos 72 = DF/BD
           = ((1 - phi)/2)/phi
           = 0.19098/0.61803
           = 0.30901

We see that sin 72 requires nested parentheses, but cos 72 does not. Let's try to write cos 72 exactly in terms of sqrt(5):

cos 72 = (1 - phi)/(2phi)
           = (phi^2)/(2phi)    (taking full advantage of the known fact that phi^2 = 1 - phi)
           = phi/2
           = (sqrt(5) - 1)/4

This value is simple enough to appear on symbolic calculators along with sin 75 and cos 75. But unfortunately, sin 72 requires nested radicals and so it doesn't appear on our calculator.

Now we can finally use the subtraction formula with the trig values for 75 and 72 to find sin 3:

sin 3 = sin 75 cos 72 - cos 75 sin 72
         = (0.96593)(0.30901) - (0.25882)(0.95107)
         = 0.05233

There is a little rounding error in our calculation, of course -- the true value is closer to 0.05234. We can find an exact value avoiding nested radicals by using phi, sqrt(2), and sqrt(6):

sin 3 = sin 75 cos 72 - cos 75 sin 72
         = ((sqrt(6) + sqrt(2))/4)(phi/2) - ((sqrt(6) - sqrt(2))/4)((sqrt(2 - phi)/2)

There's no way to make this simpler, except maybe to write it all with a denominator of 8.

Exercise 1.2: Subtraction Formula for Sine

2. The sine subtraction law is sin(alpha - beta) = sin alpha cos beta - cos alpha sin beta
(a) Derive this result by replacing beta with -beta in the addition law.

Okay, this is easy, as sin (-beta) = -sin beta and cos (-beta) = cos beta. So we have:

sin(alpha - beta) = sin alpha cos (-beta ) + cos alpha sin (-beta)
                           = sin alpha cos beta - cos alpha sin beta

(b) Now attempt the more interesting task: prove it geometrically using figure E-1.2.

Let me describe the figure. We begin with right triangle OCA, with hypotenuse OC and Angle COA, with measure alpha. We choose point B on side CA and drop a perpendicular to side OC, and we label the intersection as D. We label as OBD is a right triangle with hypotenuse OB = 1 and Angle BOD, with measure beta. Then OBA is a right triangle with hypotenuse OB = 1 and Angle BOA, with measure alpha - beta (so AB is the side we want to find). Finally, we drop a perpendicular from D to side CA, and we label the intersection as E.

At first, it appeared that to find AB, we must determine AC and BC and then subtract them. But both of these are tricky -- and it turns out that this is wrong. Let's determine some easier values first:

BD = sin beta
OD = cos beta

Now we notice that Triangles OCA and BCD are similar. This is because both are right triangles with Angle C in common, hence they are similar by AA~. It also follows that the measure of the third angle, CBD, must be alpha. Now this leads to another triangle similar to OCA, namely BDE. This is because both are right triangles with an angle alpha, hence they are similar by AA~. Since Angle CBD has measure alpha, we can find its cosine:

cos CBD = BE/BD
cos alpha = BE/sin beta
BE = cos alpha sin beta

Notice that cos alpha sin beta looks like part of the formula -- the part that we subtract. This suggests that our goal is not to subtract BC from AC, but rather BE from AE!

But how do we find AE? This is the trickiest part of all. We know that sin alpha = AC/OC, but we can use the Side-Splitting Theorem to conclude that sin alpha = AE/OD, since OA and DE are parallel (as both are perpendicular to AC). We know that the Side-Splitting Theorem involves ratios and trig involves ratios, yet we rarely see them use together this way!

So we write:

sin alpha = AE/OD
                = AE/cos beta
AE = sin alpha cos beta

AB = AE - BE
      = sin alpha cos beta - cos alpha sin beta

which is exactly what we wanted to prove. QED

Exercise 1.3: Concavity of Sine

3. (a) Show by construction that 2 sin A > sin 2A.
[Wentworth 1894, p,8]

Hmmm, Wentworth -- where have we heard that name before? That's right -- traditionalist Katharine Beals likes to mention Wentworth as an example of how rigorous math texts were in the late 19th century as opposed to today:

In this link above, Beals compares Wentworth to the Singapore math texts. Most traditionalists are fond of the Singapore curriculum, and Beals is -- up to a point, at least. She likes Singapore math only for sixth grade and below. At that point, she wants to switch to full-blown Algebra I, and the only text she finds rigorous enough for her daughter is Wentworth.

Beals likes to point out how quickly Wentworth gets into difficult problems. In Chapter 5, Wentworth is already on polynomial long division -- and some of these questions are very abstract, with multiple variables (including in the exponents).

The Wentworth text that Van Brummelen mentions here probably isn't the same text -- according to Beals, her text is called New School Algebra and dated 1898, four years after Van Brummelen's. Most likely the 1894 text is trig, not algebra. But notice that this difficult-looking trig question appears on page eight of the old text!

I know, this isn't supposed to be yet another traditionalist post. But I can't help it -- Van Brummelen wrote about how old texts used to teach spherical trig and then I saw the name Wentworth, and I just had to bring up Beals. (By the way, her daughter is about to start her junior year. I wonder what level of math Beals is teaching her daughter now -- spherical trig, perhaps?)

OK, that's enough about Beals -- let's just try to solve the Wentworth problem. As it turns out, this is related to the sine addition formula, so we can use the same diagram that we used for that problem (the proof I gave before all of the exercises). Instead of calling COE "alpha" and DOC "beta," we're going to call both of those angles alpha (or just A, since Wentworth doesn't use Greek letters).

Now sin A appears twice in this diagram -- it is both CE/OC and CD/OD (= CD, since OD = 1). We use both of these in our problem:

2 sin A = sin A + sin A
            = CE/OC + CD

Now CD is clearly greater than FD, since CD is the hypotenuse of right Triangle CDF and FD is merely a leg. So we write:

            = CE/OC + CD
            > CE/OC + FD

So we see the trick here -- we want to make the RHS smaller so that we can use the > sign. We can make the term CE/OC smaller either by decreasing the numerator or increasing the denominator. We will do the latter, by replacing OC with OD, which is greater since OD is the hypotenuse of right Triangle OCD and OC is merely a leg:

           > CE/OC + FD
           > CE/OD + FD
           = CE/1 + FD
           = CE + FD

Notice that CE and GF are equal as these are opposite sides of a rectangle, so we obtain:

          = CE + FD
          = GF + FD
          = GD
          = sin 2A

So we have 2 sin A > sin 2A, which is what we wanted to prove. QED

(b) Given two angles A and B (A + B being less than 90), show that sin (A + B) < sin A + sin B.

This is basically the same problem as part (a), except that we generalize to beta (or B) as the measure of Angle DOC. Oh, and this time we start with sin (A + B) and reverse the steps:

sin (A + B) = GD
                   = GF + FD
                   = CE + FD
                   = CE/OD + FD
                   < CE/OC + FD
                   < CE/OC + CD
                   = sin A + sin B

Take that, Beals -- I just solved a Wentworth problem! Of course, I assume Beals won't be impressed unless I tried teaching this to high school trig students -- and close to the first day of school, since this is supposed to be a Page 8 problem. Never mind!

Exercise 1.11: Velocity Due to Rotation

Let's try one more of Van Brummelen's exercises. I don't want simply to do the problems in order, so we'll skip to Exercise 11:

11. The latitude of New York City is 40.75 degrees. Find the velocity of New York in space due to the rotation of the Earth on its axis.
[Welchons/Krickenberger 1954, p. 43]

I assume that we're allowed to use either the value of the Earth's radius that we already measured, 6238 km, or a more accurate value, (as opposed to having to use trig to find it again).

Notice that the radius of the circle of latitude 40.75 is in fact cos 40.75 times as long as the radius of the sphere (or a great circle such as the Equator). To see why it's cos 40.75, we consider a circle as a cross section of the sphere, and we see that the radius is just the x-coordinate of a point on the circle, and we know that x is cosine (while y is sine).

Also, I assume that we can use a calculator to find cos 40.75. Actually, 40.75 isn't one of the values for which a cosine can be determined exactly. (It is possible to find cos 40.5, however -- we can use addition with 45 and 36 to find cos 81 and then a half-angle formula.)

Notice that we actually need the circumference, not the radius, of the small circle. And so we'll multiply cos 40.75 by the approximate circumference given in the text, 40,000 km (which is closely related to how the meter was originally defined):

40,000(cos 40.75) = 30,303 km

The length of time it takes for the earth to rotate on its axis is about one day (which is closely related to how the day was originally defined, of course). So the velocity is 30,303 km/day. We can convert this easily to km/h to obtain about 1262 km/h, or about 785 mph.

I don't live in New York, of course. Since I live closer to the equator, I'm actually moving faster. In previous posts, I've given my latitude as 34 degrees N. My actual latitude is a bit closer to 33.75 degrees -- which incidentally is a value for which a cosine can determined.

The text gives a half-angle formula for sine -- sin(alpha/2) = sqrt((1 - cos alpha)/2). As it turns out, the formula for cosine replaces the - sign with +, cos(alpha/2) = sqrt((1 + cos alpha)/2). We will apply this formula twice, beginning with 135 degrees:

cos 135 = -sqrt(2)/2
cos 67.5 = sqrt((1 - sqrt(2)/2)/2)
              = 0.38268
cos 33.75 = sqrt((1 + sqrt((1 - sqrt(2)/2)/2))/2)
                = 0.83147

40,000(cos 33.75) = 40,000(0.83147)
                              = 33259

And 33,259 km/day is about 1386 km/h, or about 861 mph. I'm moving mighty fast now!

No comments:

Post a Comment