1. Pappas Question of the Day
2. An Error on the Pappas Calendar?
3. Van Brummelen Chapter 5 The Modern Approach: Right-Angled Triangles
4. Exercise 5.1
5. Exercise 5.3
6. Exercise 5.4
7. Exercise 5.6
8. Exercise 5.7
9. Fourth of July Holiday Music
10. Conclusion
Pappas Question of the Day
Today on her Mathematics Calendar 2018, Theoni Pappas writes:
The radius of circle A = the diameter of circle B. How many times larger is circle A's area than circle B's?
Well, the area of circle A is simply (1/2)tau r^2. If circle B has a diameter of r (= the radius of A), then B's own radius must be (1/2)r. Thus its area must be (1/2)tau(1/2)^2 r^2, or (1/8)tau r^2. So the ratio of the areas must be:
(1/2)tau r^2 = 4
(1/8)tau r^2
Therefore the ratio of the areas is 4 -- and of course, today's date is the Fourth of July.
Notice that since Tau Day has passed, I will continue to use the constant tau in as many posts as possible until Pi Approximation Day, when I'll return to pi. But of course, opponents of tau can point out that today's problem is much easier with pi, since the area of a circle is just pi r^2. (And we'd be able to avoid fractional coefficients completely if we let r be the radius of circle B, not A.) But I pointed out in my Tau Day post why 1/2 in the area formula is desirable -- the area of a sector with central angle theta is (1/2)theta r^2.
It's also possible to avoid the pi vs. tau debate completely. All circles are similar (a Common Core Standard), and A's radius is exactly twice that of B's (since B's diameter is twice its radius). Then by the Fundamental Theorem of Similarity (Lesson 12-6 of the U of Chicago text), the ratio between their areas must be 2^2 = 4. There's no need for a circle constant in this problem at all.
Since this is the only day I'm posting this week, I wish to do a few more interesting Pappas problems that I see on her calendar this week. Friday's question is even more directly related to tau:
How many radians in a circle, rounded to the nearest radian?
There are precisely tau radians in a circle, where tau is approximately 6.28. Therefore rounded to the nearest whole radian, there are 6 radians in a circle -- and of course, Friday's date is the sixth.
An Error on the Pappas Calendar?
Meanwhile, tomorrow's question has nothing to do with circles or tau, but nonetheless it's a question that I want to discuss:
How many planes of symmetry does a cube have?
We know that reflections and symmetry are important concepts in Common Core Geometry. But the emphasis is on 2D reflections and symmetry, not 3D. Thus despite covering Lesson 9-5 (back in my January 24th post), I've never truly thought about this question at all.
We know that 3D problems are difficult to visualize. So let's try going down a dimension and solving the 2D analog of this question:
How many lines of symmetry does a square have?
This is easy -- a square has four lines of symmetry. Two of those symmetry lines bisect sides, while the other two bisect angles (that is, they contain the diagonals).
Sometimes it helps to think about on a coordinate plane. The vertices of the unit square could be given as (0, 0), (1, 0), (1, 1), and (0, 1).
However, there's another square that's easier to work with -- particularly in questions about symmetry, such as this one. We let the origin be the center of the square, not a vertex. If we let the side length be 2, then the vertices can be given as (1, 1), (-1, 1), (-1, -1), and (1, -1).
Now it's simple to name the four lines of symmetry -- the x-axis, y-axis, y = x, and y = -x. Let's locate the images of the vertex under each of these four lines of reflection. A reflection over the x-axis maps (1, 1) to (1, -1), the y-axis reflects (1, 1) to (-1, 1), and y = -x reflects (1, 1) to (-1, -1). The remaining mirror, y = x, reflects (1, 1) to itself.
Now we return to the original 3D problem. Recall that whereas a mirror in 2D is a line, a mirror in 3D is a plane. So we're counting the number of symmetry planes of a cube. Just as (+/-1, +/-1) (that is, plus-or-minus 1) labels four vertices of a square, the eight vertices of a cube are (+/-1, +/-1, +/-1).
And just as the x- and y-axis are symmetry lines of a square, we can readily find three symmetry planes of a cube -- the xy-plane, the xz-plane, and the yz-plane. And these three planes map the vertex (1, 1, 1) to (1, 1, -1), (1, -1, 1), and (-1, 1, 1) respectively. The name of the plane indicates which signs stay the same (that is, reflecting in the xy-plane keeps x and y, changes z).
But a square has two more symmetry lines -- the diagonals. So we must find the 3D analogs of these diagonals to find the remaining mirrors for the cube. Even though a cube has diagonals, we can't just declare the diagonals to be the mirrors we seek -- this is because diagonals are lines, but mirrors in 3D are planes.
The diagonals of a square are y = x and y = x. Now just as y = x is a line in 2D, we find that y = x is a plane in 3D. And just as in 2D, reflecting across y = x also switches x and y (and keeps z), so that the vertex (1, 1, 1) is mapped to itself, but (1, -1, 1) and (-1, 1, 1) are mapped to each other. Reflecting over the plane y = -x both switches and changes the sign of x and y, just as that line does in 2D.
This suggests that z = x, z = -x, z = y, and z = -y are also symmetry planes of the cube. Each reflection switches the two coordinates mentioned in the equation. If a negative sign appears in the equation, then those two coordinates switch sign as well as position.
This gives us nine planes so far. Three of the planes reflect (1, 1, 1) to itself (y = x, z = x, z = y), while there's one mirror reflecting (1, 1, 1) to each of the seven remaining vertices -- except (-1, -1, -1). No reflection changes all three signs -- which leads me to wonder whether there's a tenth possible mirror that indeed reflects (1, 1, 1) to (-1, -1, -1).
In Lesson 9-5 of the U of Chicago text, here is Question 11:
11. How many symmetry planes does a cube have?
which is identical to today's Pappas question. Fortunately, this is an odd-numbered question, so the answer is listed in the back of the text. The given answer is 9 -- which tells us that we're on the right track, and that we've determined all nine mirrors.
But this is the Pappas question for tomorrow -- and tomorrow's date is the fifth, not the ninth. So according to Pappas, a cube has only five symmetry planes, even though we just found nine of them!
I decided to make a Google search to verify the answer. Interestingly enough, the first result isn't a Geometry site, but a Chemistry site (on crystalline substances):
https://learning.uonbi.ac.ke/courses/SGL201/scormPackages/path_2/156_symmetry_elements.html
Figure 1.9. The nine symmetry planes of the cube indicated by the dashed lines.
That settles it -- a cube has nine symmetry planes, not five. The U of Chicago text is correct, while Pappas is incorrect. We've just discovered an error on the Pappas calendar.
By the way, when I created the Lesson 9-5 worksheet (in my January 24th post), I included only Question 7 from that lesson, not Question 11:
7. How many symmetry planes does a box have?
Notice that a box (a rectangular prism) that isn't a cube doesn't have as many symmetry planes, just as in 2D, a rectangle that isn't a square doesn't have as many symmetry lines. In particular, the diagonals of a rectangle aren't symmetry lines unless the rectangle is a square.
Likewise a general box has only three symmetry planes. Notice that if two of the three dimensions of the box are equal -- for example, the box whose vertices are (+/-1, +/-1, +/-2), -- then two of the faces are squares, and planes perpendicular to the base and containing its diagonals (y = x and y = -x in the example given here) are indeed symmetry planes. In this case, the 2 * 2 * 4 box would have five symmetry planes -- maybe this is what Pappas has in mind when she gives the solution as 5?
There's one more thing I want to mention this week. Back on Sunday, July 1st, Google featured a mathematician -- the 17th century German mathematician Gottfried Leibniz.
Glen Van Brummelen mentions Leibniz in Chapter 6 of Heavenly Mathematics -- even though we're covering Chapter 5 today, let's move ahead to Chapter 6 to read more about Leibniz:
"It's surprisingly common, and rather eerie, when mathematical discoveries are made almost simultaneously by two or more people working on their own. The invention of calculus by Isaac Newton and Gottfried Wilhelm Leibniz is the most famous episode of this kind, followed by the birth of non-Euclidean geometry in the works of Janos Bolyai, Nicolai Lobachevsky, and Carl Friedrich Gauss in the early 19th century."
Of course, Van Brummelen proceeds to say more about non-Euclidean geometry -- which is logical considering that this is a spherical trig text. But the Newton vs. Leibniz Calculus Wars are worth mentioning here. It's important to point out that as far as notation is concerned, Leibniz won the Calculus Wars -- we still use his notation for the derivative today, dy/dx. The Doodle itself depicts Leibniz working with the binary system -- he worked on a forerunner of the modern computer (which he called a "step reckoner").
Once I saw the Google Doodle, I was considering posting on July 1st to celebrate Leibniz. But I'd already planned to make this a Fourth of July post instead. (And of course, by now Google is featuring a Fourth of July doodle instead. Apparently, here in my home state of California, we eat quinoa -- the mother of all grains.) I've never made a Fourth of July post before -- in past years, I've usually posted within a few days of the fourth (most notably last year, when I posted everyday during the first of week of July except the holiday).
Because today is the Fourth of July, there is no summer school in my new district. Students usually attend summer school from Monday to Thursday, but because of the holiday, summer school will be held on Friday this week. (This Friday is important -- this is the third week, so Friday will be the district-mandated final.)
Van Brummelen Chapter 5 The Modern Approach: Right-Angled Triangles
OK, let's get to the chapter that we're supposed to be covering today -- Chapter 5. Van Brummelen opens the chapter with the following:
"The word 'trigonometry' means 'triangle measurement,' which is how we've thought of the subject for the past several centuries. The term comes from Bartholemew Pitiscus's book 1600 book Trigonometria (figure 5.1), a variant of the phrase 'the science of triangles' that had been used for a number of decades previously."
Van Brummelen points out that he'll explain how to use spherical trig to solve spherical triangles -- which is what we think of when we hear the word "trig." He continues:
"A few of the formulas we'll see in this chapter go back to medieval or ancient astronomers, but much of what we're about to see was systematized in Scotland by a man who wasn't even a mathematician."
So no, we're not discussing Leibniz -- who lived a century later in another country. Instead, our focus for today is John Napier. The author describes him thusly:
"For Napier and most of his colleagues science was a hobby; it had not yet developed into a full-fledged profession. A landholder, Napier was widely known for his passionate commitment to Protestant causes against the Catholics."
In figure 5.2, Van Brummelen labels the parts of a right spherical triangle ABC. He adheres to the convention of letting C be the right angle and a, b, c be the sides opposite angles A, B, C. He converts the side lengths to angle measures in the usual way. (If the radius of the sphere is 1 and the angles are measured in radians, then the side length equals the angle measure.)
He continues:
"To generate new theorems we're going to have to convert Angles A and B from spherical to plane angles as well, and it's not quite as obvious how to do that. One way is to think of Angle A as the angle between the 'floor' plane OAC and the diagonal face OAB; similarly, Angle B is the angle between the vertical back wall OBC and the floor OAC."
Here O is the center of the sphere, by the way. Let's continue quoting Van Brummelen so we can visualize figure 5.2 fully:
"We'd like to piggyback on ordinary trigonometry to get some new results, so we need to express Angles A and B as angles between line segments rather than between planes. To make this transition, pick any point D on
Well, how do we ordinarily prove that a triangle is right triangle? That's right -- Pythagoras comes to the rescue. Here is Van Brummelen's proof:
OD^2 = OE^2 + ED^2 (since ODE is a right triangle)
= (OF^2 + EF^2) + (DF^2 - EF^2) (since OEF, DEF are right triangles)
= OF^2 + DF^2
So by the converse of the Pythagorean Theorem, Triangle OFD is a right-angled triangle as well as the other three faces of tetrahedron ODEF. Thus
By the way, here's one part that confused me a little. Notice that even though Angle A = DFE and Angle C = DEF (as both are right angles), it does not follow that Angle B = EDF. This is because in the angles of plane triangle DEF add up to 180, while the angles of spherical triangle ABC add up to more than 180. Thus Angle EDF must be less than Angle B.
The author now produces new results by applying trig functions to the four right triangles that are the faces of tetrahedron ODEF. We pick any vertex, such as D, and identifies a ratio consisting of two of the three line segments that may be interpreted as a trig function:
sin a = DE/OD.
Now we insert the third line segment into the ratio, as follows, and interpret the two new ratios as trigonometric expressions:
sin a = DE/DF * DF/OD = sin A sin c.
We find three more identities by choosing the three other vertices of ODEF:
sin b = tan a cot A
cos A = tan b cot c
cos c = cos a cos b.
We may now switch A and B, as well as a and b, to generate three new theorems:
sin b = sin B sin c
sin a = tan b cot B
cos B = tan a cot c.
There are three theorems left to find -- those that refer to both A and B. Again, we can't just look at Angle EDF because this isn't equal to Angle B. Here is Van Brummelen's construction:
We need to add Angle B to our diagram, so we adapt the process that we used to construct Angle A (figure 5.3). Choose a point G on
Each of the three planes containing O now contains several similar triangles, drawn separately in figure 5.4, These triangles will unlock the new identities.
(Let me at least describe figure 5.4 to you, the reader. One plane contains right triangle OEG with the altitude to the hypotenuse
The idea is to start with some trig ratio, say cos c = OF/OD, and interpose line segments as we did before. Next, use the similar triangles of figure 5.4 to convert the new ratios into trig functions of other known angles. For instance:
cos c = OF/OD = OF/OE * OE/OD = EF/EG * EH/DE = EF/DE * EH/EG
= cot A cot B.
Van Brummelen tells us that two more identities can be found the same way:
cos A = cos a sin B
cos B = cos b sin A
We have finally arrived at the ten fundamental identities of a right-angles spherical triangle:
Column I:
sin b = tan a cot A
cos c = cot A cot B
sin a = cot B tan b
cos A = tan b cot c
cos B = cot c tan a
Column II:
sin a = sin A sin c
cos A = sin B cos a
cos B = cos b sin A
sin b = sin c sin B
cos c = cos a cos b.
At this point, Van Brummelen applies the locality principle. He reminds us that as a spherical triangle gets smaller, it approaches a planar triangle. Also, as a gets small, sin a approaches a -- and in fact, tan a also approaches a as a gets small. This provides us with the following analogs:
Spherical: sin A = sin a/sin c -- Planar: sin A = a/c
Spherical: cos A = tan b/tan c -- Planar: cos A = b/c
Spherical: tan A = tan a/sin b -- Planar: tan A = a/b.
Of course, we can't do the reverse and generate spherical trig ratios from plane trig ratios, since it's not obvious whether to use sine or tangent. (We can't know, for example, that tan A = a/b must become tan A = tan a/sin b, with the tangent of a and the sine of b.)
The author also finds a planar equivalent for cos c = cos a cos b. It turns out that we can use the approximation cos a = 1 - a^2/2 to find something interesting:
cos c = cos a cos b
1 - c^2 = (1 - a^2/2)(1 - b^2/2)
c^2 = a^2 + b^2 + a^2 b^2 /2
As a and b approach 0, a^2 b^2 is small compared to a^2 or b^2, so we can drop this term:
c^2 = a^2 + b^2
And voila-- we obtain the Pythagorean Theorem. In other words, cos c = cos a cos b is actually the new spherical Pythagorean Theorem.
At this point, Van Brummelen proceeds to apply our knowledge to the sky and sea. He writes:
"In figure 5.5 it is a day in late May, and the Sun has traveled lambda = 60 degrees along the ecliptic since it passed Aries roughly two months ago, at the spring equinox. Our goal is to find the right ascension alpha and the declination delta."
The figure is set up so that lambda = 60 is the hypotenuse of a spherical right triangle, with alpha (along the equator) and delta (opposite Aries) at its legs. The angle at Aries is the angle of the ecliptic, epsilon = 23.44 degrees.
First, the author applies the first identity of column II, which we've seen in a previous chapter:
sin delta = sin lambda sin epsilon
delta = sin^o-1(sin 60 * sin 23.44) = +20.15.
(By the way, I write inverse sine as sin^o-1 to emphasize that this is the inverse of the composite, not the inverse of multiplication -- the reciprocal. Too many students believe that sin^-1(x) = 1/sin x.)
Next he applies the first identity in column I -- we arrive at:
sin alpha = tan delta cot epsilon
alpha = sin^o-1(tan 20.15 * cot 23.44) = 57.81 degrees = 3 hours 51 minutes.
Van Brummelen warns us that the sine function is not 1-1, so we must be careful with inverse sine. So he tries plugging in lambda = 130 (early August) into the formula and obtains sin alpha = 0.8668, when the correct value of alpha is 119.91, not 60.09 degrees. He suggests that if there's a choice between inverse sine and inverse cosine, prefer the latter, as cosine is 1-1 on [0, 180].
Here is Van Brummelen's next application:
"A ship leaves Halifax (position, 44.67N, 63.58W), starting due east and continuing on the great circle. Find its position and its direction after it has sailed 1000 nautical miles."
As the author reminds us, the ship would begin heading due east, but its direction would alter gradually southwards. After all, "due east" means along a parallel of latitude -- but a parallel of latitude is not a great circle. So we already know that the correct answer will not be "due east" for the direction or "44.67N" for the latitude.
The author draws a triangle with A at the North Pole, B at the ship's final location, and C at the ship's initial location (Halifax). As usual a, b, c are the sides opposite these angles. We already know what two of the sides are -- b is 90 - 44.67 = 45.33 (latitude difference between the pole and Halifax) and a is 1000/60 = 16.67 (as one nautical mile = one minute of arc = 1/60 degree).
We know two sides of this right triangle, so to find the third, let's use the Pythagorean Theorem -- the spherical one, that is:
c = AB = cos^o-1(cos 45.33 * cos 16.67) = 47.66.
This is the latitude difference between the pole and the ship, so latitude is 90 - 47.66 = 42.34N. As for longitude, this is the angle A at the North Pole:
cos A = tan b cot c
A = cos^o-1(tan 45.33 * cot 47.66) = 22.81
So the longitude is 63.58 - 22.81= 40.77W, which is well on its way from Halifax to the Azores. And finally, the heading is the angle at B:
cos B = cos b sin A
B = cos^o-1(cos 45.33 * sin 22.81) = 74.18
The heading is 74.18 degrees east of south.
At this point Van Brummelen suddenly changes course (in the chapter, not on that ship):
"If you tried to solve either of the two problems above on your own, you likely made two discoveries: firstly, with ten identities at our disposal there are often many paths to the solution, and part of the challenge lies simply in recalling all of the identities. More on this later. Secondly, the arithmetic frequently requires that we multiply and divide messy trigonometric quantities."
And this was a huge problem in the pre-computer age for mathematicians like John Napier. (Even Leibniz and his step reckoner were a century away, much less modern calculators.) And so to make multiplication and division easier, Napier invented that for which he's most famous -- logarithms -- and his invention that uses logarithms, Napier's bones.
sin delta = sin lambda sin epsilon
log(sin delta) = log(sin lambda) + log(sin epsilon).
As the author explains:
"The form of this equation explains why many logarithm tables (including Napier's) did not display pure logarithms, but rather logarithms of sines. Solving the problem then reduces to looking up and adding log(sin lambda) and log(sin epsilon), finding their sum within the table's entries, and reading backward to get delta."
This sounds amazing from a modern perspective. Consider an Algebra II student -- to that student, logs are something to learn about in one chapter and trig is something to learn in another chapter. Yet back in Napier's day, logs and spherical trig were taught side-by-side:
"Their effect was so powerful that the famed mathematical astronomer Pierre Simeon de Laplace praised them two centuries after Napier's death by saying: 'by shortening the labours, [logarithms] doubled the life of the astronomer.'"
Now we arrive at the final issue -- remembering the formulas. (Compare this to Lesson 10-6 of the U of Chicago text, "Remembering Formulas.") Van Brummelen repeats the chart:
Column I:
sin b = tan a cot A
cos c = cot A cot B
sin a = cot B tan b
cos A = tan b cot c
cos B = cot c tan a
Column II:
sin a = sin A sin c
cos A = sin B cos a
cos B = cos b sin A
sin b = sin c sin B
cos c = cos a cos b.
And now the author points out the patterns in these lists. In column I, the identities are all of the form "co/sine equals co/tangent times co/tangent" while those in column II consist entirely of co/sines. He also tells us that within each sub-column, the variables follow the sequence a, A, B, b, c.
In fact, Napier would draw a circle divided into five equal sectors. Starting in the upper-left corner and moving clockwise, the sectors read a, b, co-A, co-c, co-B.
Napier's Rule I: The sine of any circular part is equal to the product of the tangents of the two parts adjacent to it.
Napier's Rule II: The sine of any circular part is equal to the product of the cosines of the two parts opposite to it.
As usual, co-A means 90 - A, so that the "sine of co-A" means "the cosine of A."
But in the 19th century, some texts criticized Napier's Rules as an artificial mnemonic device. By the way, this same debate shows up 200 years later in Nix the Tricks. This is a book written by Tina Cardone -- yes, the same Cardone who came up with the "Day in the Life" MTBoS challenge. Now Cardone dislikes "tricks" such as "cross-multiply" and FOIL -- she argues that these tricks hinder the learning of mathematics. Well, to some authors, Napier's Rules were the first tricks that needed to be nixed, centuries before Cardone's book.
Yet Van Brummelen argues that Napier's Rules aren't merely a mnemonic device -- instead, they can be derived from a special figure called the "pentagramma mirificum" (the miracle pentagon). This requires a lot of explanation to describe the diagram in the book, so let's begin.
We start with any right triangle ABC labeled in the usual manner (right angle at C, legs a, b, and the hypotenuse is c). Then we extend side AB to form arc RABS, side CA to form arc CAXU, and side CB to form arc CBVT. Draw two new arcs SVWU and RXWT with poles A and B respectively. The resulting pentagon ABVWX is self-polar, and the five triangles that make up the "points" of the pentagram are all right angles (the intersection between a line drawn from a pole and its equator).
The author continues:
"But there is much more to discover. For instance, we know that AS = 90 since it connects pole A with equator SVWU, so SB =
[Here Van Brummelen actually writes c with a bar (or vinculum) to denote 90 - c, but here on Blogger I'll just use the Strikethrough command.]
"Now consider what would have happened if we had begun the construction of the pentagram with Triangle BVS rather than ABC. The two arcs departing from S would be the same as they are now, as would the hypotenuse CBVT. RXWT, the equator of pole B, would also be identical. The last arc of the pentagram, CAXU, requires a short argument: CV and UV rise perpendicularly from it and meet at V, which implies that V is a pole of CAXU. We have arrived at a powerful conclusion: if we had started with Triangle BVS (or any of the other five corner triangles) rather than Triangle ABC, we would have ended up with exactly the same diagram.
"This symmetry gives us much information. For starters, two adjacent segments of any of the arcs in the figure sum to 90, just as the segments in RABS do. This allows is to fill in quickly the lengths of all the segments in the figure, except those on RXWT and SVWU -- and they're not far behind. Consider AW drawn through the pentagon. Since (by symmetry) W is a pole of RABS, Angle RAW = 90, so Angle UAW =
"Now all that is left is to identify are the angles of the four new triangles. W is a pole of RABS, so Angle RWS = RABS = c + 2
Finally, the author tells us that applying a formula in columns I or II to one of the triangles and then moving to another triangle produces another formula in the column. So in column II, we start with the identity sin a = sin A sin c in the original triangle and move to the next triangle clockwise. This leads to the identity sin
"Thus our diagram explains all of the astonishing symmetries we've seen. It is surely entitled to bear the name 'pentagramma mirificum.'"
Exercise 5.1
1. Using figure 5.2, derive the identities sin b = tan a cot A, cos A = tan b cot c, and cos c = cos a cos b, in the same manner as we derived sin a = sin A sin c.
Let's describe figure 5.2 again so we can do this correctly:
In figure 5.2, Van Brummelen labels the parts of a right spherical triangle ABC. He adheres to the convention of letting C be the right angle and a, b, c be the sides opposite angles A, B, C. He converts the side lengths to angle measures in the usual way. (If the radius of the sphere is 1 and the angles are measured in radians, then the side length equals the angle measure.)
"To generate new theorems we're going to have to convert Angles A and B from spherical to plane angles as well, and it's not quite as obvious how to do that. One way is to think of Angle A as the angle between the 'floor' plane OAC and the diagonal face OAB; similarly, Angle B is the angle between the vertical back wall OBC and the floor OAC."
"We'd like to piggyback on ordinary trigonometry to get some new results, so we need to express Angles A and B as angles between line segments rather than between planes. To make this transition, pick any point D on
OK, so now let's prove the identities:
sin b = EF/OE
= EF/DE * DE/OE
= cot A tan a
cos A = EF/DF
= EF/OF * OF/DF
= tan b cot c
cos c = OF/OD
= OF/OE * OE/OD
= cos b cos a. QED
Exercise 5.3
3. Demonstrate Geber's Theorem (cos A = cos a sin B) geometrically using figures 5.3 and 5.4.
This is easier for me than it is for you, since unlike you, I can see figures 5.3 and 5.4. Just scroll above to earlier in this post when I described those two figures.
Proof:
From figure 5.3, we see that Angle A = DFE, where DEF is a right triangle (right angle at E). Thus:
cos A = EF/DF
= EF/OF * OF/DF (algebra)
= EG/OE * OH/GH (using first SL/LL and third LL/SL diagrams in figure 5.4)
= OH/OE * EG/GH (algebra)
= OE/OD * EG/GH (using second LL/SL in figure 5.4)
= cos a sin B (using figure 5.3). QED
Here I give hints as to whether I used the ratio Short Leg:Long Leg (SL/LL) or vice versa (LL/SL) in the diagrams (figure 5.4) that you can't see.
Exercise 5.4
4. From the given data, solve each of the following right spherical triangles. [Brink 1942, 15]
Notice the reference to an old 1942 text. Traditionalists should be happy to see that we're solving problems from the Golden Age of Textbooks.
(a) A = 72.72, c = 109.8
Recall that to solve a triangle means to find all of the missing sides and angles (except Angle C, of course, since this will always be 90). The method to do it is to find the correct formula to use, which must be one of the ten in the list.
Let's check the list and isolate the formulas involving both A and c:
cos c = cot A cot B
cos A = tan b cot c
sin a = sin A sin c
We could, for example, use the first formula to find B and then use a different formula involving, say, B and c to find b. But in order to minimize rounding error, I prefer to use only the A and c formulas:
cos c = cot A cot B
B = cot^o-1(cos c/cot A)
B = tan^o-1(1/(tan A cos c)) (since there's no cot button, we use 1/tan theta)
B = tan^o-1(1/(tan 72.72 cos 109.8))
B = tan^o-1(-0.918357)
Now finding the arctan of a negative angle returns a negative value on the calculator -- which makes no sense as the angle of a triangle can't be negative. Since the period of tan is 180, we add 180 to the negative value to find the measure of (obtuse) angle B. (That's right -- a right triangle can have an obtuse angle in spherical, unlike Euclidean, geometry.)
B = tan^o-1(-0.918357) + 180
B = 137.44
cos A = tan b cot c
b = tan^o-1(cos A/cot c)
b = tan^o-1(cos A tan c)
b = tan^o-1(cos 72.72 tan 109.8) + 180
b = 140.48
sin a = sin A sin c
a = sin^o-1(sin 72.72 sin 109.8)
a = sin^o-1(0.898414)
a = 63.95
But hold on a minute here -- recall what Van Brummelen wrote about the sine function. The answer could be 63.95 or it could be 180 - 63.95. To double-check, we must follow Van Brummelen's advice and use one more formula involving cosine, such as:
cos c = cos a cos b
We only need to check the signs here -- b, c are both obtuse, so cos b, cos c are both negative. Thus cos a must be positive, and so a is acute. Therefore a = 63.95.
The solution to the triangle is a = 63.95, b = 140.48, B = 137.44.
(b) a = 51.45, b = 78.73
Let's check the list and isolate the formulas involving both a and b:
sin b = tan a cot A
sin a = cot B tan b
cos c = cos a cos b
sin b = tan a cot A
A = cot^o-1(sin b/tan a)
A = tan^o-1(tan a/sin b)
A = 51.99
sin a = cot B tan b
B = cot^o-1(sin a/tan b)
B = tan^o-1(tan b/sin a)
B = 81.14
cos c = cos a cos b
c = cos^o-1(cos a cos b)
c = 83
The solution to the triangle is A = 51.99, B = 81.14, c = 83.
(c) a = 63.48, B = 80.57
Let's check the list and isolate the formulas involving both a and B:
sin a = cot B tan b
cos B = cot c tan a
cos A = sin B cos a
sin a = cot B tan b
b = tan^o-1(sin a/cot B)
b = tan^o-1(sin a tan B)
b = 79.48
cos B = cot c tan a
c = cot^o-1(cos B/tan a)
c = tan^o-1(tan a/cos B)
c = 85.33
cos A = sin B cos a
A = cos^o-1(sin B cos a)
A = 63.87
The solution to the triangle is A = 63.87, b = 79.48, c = 85.33.
(d) a = 69.72, c = 78.42
Let's check the list and isolate the formulas involving both a and c:
cos B = cot c tan a
sin a = sin A sin c
cos c = cos a cos b
cos B = cot c tan a
B = cos^o-1(cot c tan a)
B = cos^o-1(tan a/tan c)
B = 56.32
sin a = sin A sin c
A = sin^o-1(sin a/sin c)
A = 73.24 (Here I assume that A is acute, since no negative sign appears in any of the equations.)
cos c = cos a cos b
b = cos^o-1(cos c/cos a)
b = 54.61
The solution to the triangle is A = 73.24, B = 56.32, b = 54.61.
(e) A = 52.4, B = 122.27
Let's check the list and isolate the formulas involving both A and B:
cos c = cot A cot B
cos A = sin B cos a
cos B = cos b sin A
cos c = cot A cot B
c = cos^o-1(cot A cot B)
c = 119.1
cos A = sin B cos a
a = cos^o-1(cos A/sin B)
a = 43.81
cos B = cos b sin A
b = cos^o-1(cos B/sin A)
b = 132.37
The solution to the triangle is a = 43.81, b = 132.37, c = 119.1. (I know -- it looks weird for the hypotenuse c not to be the longest side, but that's what happens in spherical geometry!)
Exercise 5.6
(a) Is there a right spherical triangle in which b = 30 and B = 100? Explain. [Seymour/Smith 1948, 175]
Well, we can try solving the triangle and find out. Again, let's check the list and isolate the formulas involving both b and B:
sin a = cot B tan b
cos B = cos b sin A
sin b = sin c sin B
sin a = cot B tan b
a = sin^o-1(cot B tan b)
a = sin^o-1(tan b/tan B)
a = sin^o-1(-0.1018)
a = -5.84
And there we have it -- a has a negative length, which is a contradiction. Therefore there is no right spherical triangle in which b = 30 and B = 100.
In fact, we observe that in the equation sin a = cot B tan b, tangent and cotangent are positive for acute angles and negative for obtuse angles. Thus in order for a to be positive, B and b must be either both acute or both obtuse (or both right). Legs shorter than a quadrant must be opposite acute angles and legs longer than a quadrant must be opposite obtuse angles.
(b) Show that no isosceles right triangle can have its hypotenuse greater than 90 nor its acute angle less than 45. [Moritz 1913, 63]
Proof:
By the definition of isosceles triangle, a = b. And the Isosceles Triangle Theorem, which holds in both spherical and Euclidean geometry, tells us that A = B. Let's check out the formulas now. The first part of the proof asks about the length of the hypotenuse, which looks like a job for the so-called Spherical Pythagorean Theorem:
cos c = cos a cos b
cos c = cos a cos a
cos c = cos^2 a
Assume that c > 90. Then cos c would be negative, as the cosine of an obtuse angle is negative. But then cos^2 a would be negative, and hence cos a would be imaginary, a contradiction. Thus c < 90.
sin b = tan a cot A
sin a = tan a cot A
cot A = sin a/tan a
cot A = cos a (since sin a/cos a = tan a)
Now assume that A is less than 45. Then cot A would be greater than 1 (since cot A = 1/tan A, and when A < 45, tan A < 1). But this would make cos a > 1, a contradiction. Therefore A > 45. QED
Exercise 5.7
7. (a) From the relation cos c = cos a cos b show that if a right spherical triangle has only one right angle, the three sides are either all acute, or one is acute and two obtuse. [Moritz 1913, 20]
Proof:
We only need to look at the signs here. We know that the cosine of an acute angle is positive and the cosine of an obtuse angle is negative. Thus "the three sides are all acute" corresponds to "positive times positive is positive," and "one side is acute and two obtuse" corresponds to the rules to multiply positive and negative numbers. ("One positive and two negative" -- think about those mnemonic smiley faces with one plus sign and two minus signs.) QED
(b) Prove that a side and the hypotenuse of a right spherical triangle are of the same or opposite quadrants accordingly as the angle included between them is less than or greater than 90. [Muhly/Saslaw 1950, 150]
Proof:
We can let a be the leg, and c is always the hypotenuse, so the angle between them is B. So let's look at an identity linking a, c, and B:
cos B = cot c tan a
Tangent and cotangent, like cosine, are positive for acute angles and negative for obtuse angles. So if B is less than 90, then cos B is positive, so cot c and tan a must have the same sign (quadrant). And if B is greater than 90, then cos B is negative, so cot c and tan a must be in the opposite quadrant. (In multiplication, same signs are positive, opposite signs are negative.) QED
Fourth of July Holiday Music
The most musical time of the year, of course, is
I often wondered whether there is a second musical season of the year. Even though the two major Christian holidays are Christmas and Easter, we don't associate Easter with many songs at all -- except possibly hymns sung at church services on the holiday.
In fact, if there's a second musical day at all, it's Independence Day. My local library has a book of American patriotic songs, and I check it out at this time each year. Some of the songs in this book are well-known, such as "My Country 'Tis of Thee" and "America the Beautiful."
And so just as I converted holiday music to Mocha on Christmas Eve, let's convert patriotic music into the EDL scales. I know what you're saying -- we're supposed to create music in a new scale, not convert music to the new scale. But once again, I'd be composing songs if I had an audience for them, namely a class to teach. Besides, we learn more about the power of the new scales and can imagine what's possible in them by converting existing songs to these scales.
The two songs I'll convert in this post are not as commonly known, but I believe that these songs are representative of what's possible in the scales.
Both of the chosen songs are written in major keys. We know that 18EDL and 20EDL are the closest to the major scale, so we'll look at these scales more closely. If I'd chosen a song in a minor key, we might choose 12EDL instead -- but most patriotic music is major. The span of each song is a ninth -- that is, one note beyond an octave. We saw back in my Christmas Eve post that ninths and tenths are common spans, since larger spans are difficult to sing. (In other words, I didn't choose "The Star Spangled Banner" with its wide twelfth/tritave span.)
The first song from the book that I'll convert is "Boys, Keep Your Powder Dry." This song is written in the key of G major, in 2/4 time. The span of this song is familiar, because we observed that many Christmas songs have the same span -- the lowest note in the melody is D (the fifth note of the scale) and the highest note is E (the sixth note of the scale). By the way, other simple songs such as "The Farmer in the Dell" and "Frere Jacques" also has this same span.
Fortunately, I find that we can take advantage of this span in Mocha. First, recall that the simplest playable major triad is on Degrees 15:12:10. So if we place the tonic on Degree 15, then we can have the lowest note of the melody be Degree 20 (the fifth note of the scale below the tonic) and the highest note be Degree 9 (a major sixth above the tonic). Degrees 18 and 16 also fall on notes of the major scale (the sixth and seventh).
The only deviations from the major scale, then, are the second and fourth. The closest available second is Degree 13, which is 15/13 (about 248 cents) above the tonic, while the best possible fourth is Degree 11, which is 15/11 (about 537 cents) above the tonic.
As for the rhythm, the song is in 2/4 time. Most of the notes are eighth notes, and there are also dotted eighth notes (followed by sixteenth notes of course), quarter notes, and dotted quarter notes (followed by eighth notes of course).
The song has two verses, so we'll have the song repeat with a FOR V=1 TO 2 loop and RESTORE the DATA lines, which must contain both the Degrees and the note lengths.
http://www.haplessgenius.com/mocha/
10 N=7
20 FOR V=1 TO 2
30 FOR X=1 TO 78
40 READ D,T
50 SOUND 261-N*D,T
60 NEXT X
70 RESTORE
80 NEXT V
90 END
100 DATA 20,2,15,3,15,1,15,2,12,2,15,4,15,2
110 DATA 20,2,15,3,20,1,15,2,12,2,15,6
120 DATA 12,2,13,3,13,1,12,2,11,2,10,4,13,2
130 DATA 12,2,13,3,15,1,18,3,16,1,20,6
140 DATA 20,2,13,3,13,1,12,2,12,2,11,4,13,2
150 DATA 13,2,12,3,13,1,15,3,12,1,13,6
160 DATA 20,2,15,3,13,1,12,3,11,1,10,2,9,4
170 DATA 11,2,12,3,12,1,13,3,13,1,15,6
180 DATA 20,2,13,3,13,1,12,2,12,2,11,4,13,2
190 DATA 13,2,12,3,13,1,15,3,12,1,13,6
200 DATA 20,2,15,3,13,1,12,3,11,1,10,2,9,4
210 DATA 11,2,12,3,12,1,13,3,13,1,15,6
Notice that lines 140-170 are identical to lines 180-210.
By the way, the original score contains a raised fourth in the third line (Line 120), intended to be a leading tone to the ensuing perfect fifth. The fourth in this song (Degree 11) is already partly raised, and so we don't change it at all. In the first line, N can be any value from 1 to 13 -- here I choose N=7, which corresponds to the key of greenish G (as the score is written in G major).
Here are the lyrics to the song in case you want to sing along with Mocha:
BOYS, KEEP YOUR POWDER DRY
1st Verse:
Can'st tell who lost the battle,
Oft in the conncils field?
Not they who struggle bravely,
Not they who never yield,
Not they who are determined
To conquer or to die
And harken to this caution:
"Boys, keep your powder dry."
Refrain:
Not they who are determined
To conquer or to die
And harken to this caution:
"Boys, keep your powder dry."
2nd Verse:
The foe awaits you yonder,
He may await you here;
Have brave hearts, stand with courage,
Be strangers all to fear;
And when the charge is given,
Be ready at the cry:
Look well each to his priming
"Boys, keep your powder dry."
(Repeat Refrain)
Our other song to convert is "God of Our Fathers." This is a religious song, but on the day we read Chapter 4 of Van Brummelen, I wrote about the Islamic religion of the medieval mathematicians. So let's celebrate the Protestant mathematicians Napier and Leibniz with a Christian song here.
This song is written in the key of Eb major. Its span sticks to the octave from tonic to tonic, except that one note below the tonic (the seventh note of the scale) appears. The key signature is 4/4, and so we're basically doubling the note lengths from the previous song (instead of 1, 2, 3, 4, and 6, our note lengths will be 2, 4, 6, 8, and 12, from eighth note to dotted half note).
Since the span has changed, using Degrees 20 to 9 doesn't work. Instead, we must decide between 18EDL and 20EDL for this song. Neither has a just major triad on the tonic -- instead, we find that 18EDL has a perfect fifth but a supermajor third. Meanwhile 20EDL has a just major third but a dissonant 20/13 raised fifth. In this particular song, I notice that fifths are more important than thirds, and so we'll select 18EDL.
We can use 19 for the note below the tonic (the seventh note of the scale), since 19/18 is essentially a type of semitone (the fawn diatonic semitone). But there is no major seventh below the highest note of the song -- all we have is 9/5, a type of minor seventh. The only fourth is also very wide (18/13, about 563 cents), and the only sixth is neutral (18/11, about 853 cents). Just as in the previous song, though, we take advantage of this as the original score contains raised fourths and lowered sevenths.
According to the score, the song has two verses. However, in each verse the melody repeats twice, and so we can code it as if it's really four verses.
NEW
10 N=7
20 FOR V=1 TO 4
30 FOR X=1 TO 40
40 READ D,T
50 SOUND 261-N*D,T
60 NEXT X
70 RESTORE
80 NEXT V
90 SOUND 261-N*18,32
100 DATA 18,8,18,6,16,2,14,12,13,4
110 DATA 16,4,18,4,18,4,19,4,18,16
120 DATA 14,8,14,6,14,2,12,12,11,4
130 DATA 10,4,12,4,12,4,13,4,12,16
140 DATA 12,8,12,6,12,2,12,12,12,4
150 DATA 10,4,12,4,12,4,13,4,12,16
160 DATA 9,8,10,6,11,2,12,12,12,4
170 DATA 13,4,14,4,16,4,16,4,18,16
For example, in the fourth line (Line 130) the score gives a major seventh, while in the sixth line (Line 150), the score gives a minor seventh. But the only available seventh in Mocha is the minor seventh, and so Lines 130 and 150 are written identically.
The last verse ends with the word "Amen," played as two whole notes. Here, we use Line 90 for the final "Amen," written as a single note of length 32 (a double whole note or breve). After all, the way Mocha beeps, we can't always distinguish between one long note and two not-as-long notes anyway.
Here are the lyrics (which I'll also write as four verses):
GOD OF OUR FATHERS
1st Verse:
God of our fathers,
Whose almighty hand
Leads forth in beauty
All the starry band
Of shining worlds
In splendor through the skies,
Our grateful songs
Before Thy throne arise.
2nd Verse:
Thy love divine hath
Led us in the past.
In this free land
By Thee our lot is cast.
Be Thou our ruler,
Guardian, guide, and stay,
Thy word our law,
Thy paths our chosen way.
3rd Verse:
From war's alarms,
From deadly pestilence,
Be Thy strong arm
Our ever sure defense;
Thy true religion
In our hearts increase,
Thy bounteous goodness
Nourish us in peace.
4th Verse:
Refresh Thy people
On their toilsome way.
Lead us from night
To never ending day;
Fill all our lives
With love and grace divine,
And glory laud,
And praise forever Thine!
Amen.
Using N=7, the song is in the key of red E, while N=8 would put it in the key of white D. This is the closest we can get to the key in which the score is written, Eb.
By the way, I keep using absolute Kite's color notation when I write "red E," "white D," or "blue A" (sorry, but I just had to make that red, white, and blue today). But you might wonder how those notes compare to the usual standard, A440. That is, if I play A at 440 Hz, is that a "white A"?
So far, "white A" means Degrees 12, 24, 48, 96, and 192 as played on Mocha. I have no idea how those A's compare to A440. (Last week, we saw how on the Atari, A440 lies between 16-bit white A and 8-bit white A -- or "blue A" and "white A.")
Well, frequency as measured in hertz is otonal -- raising the frequency raises the pitch. And so it's logical to reserve the color "white" for frequencies that are 3-smooth (that is, containing only 2 and 3 as prime factors). For example, 256 Hz (2^8, which sounds as a C) would be white C. The resulting white A would be A432, since 432 = 2^4 * 3^3.
One advocate of using C256 and A432 is the controversial politician Lyndon Larouche. He's also a crank when it comes to music, since he believes that C256 and A432 are in accord with "the natural frequency of the earth," whatever that means.
But regardless of how crankish Larouche is both politically and musically, his musical notation is compatible with Kite's color notation. So if we're given frequencies in hertz, then we should call C256 "white C" and A432 "white A." Concert A -- that is, A440 -- works out to be jade-yellow A (assuming that j4 = P4, as we usually do).
The A440 standard (which is based on pure 12EDO) is incompatible with Kite's colors (which is obviously just intonation). Indeed, in 12EDO with A440, G works out to be almost exactly 392 Hz, but 392 Hz in Kite's colors is "deep blue Ab" -- that is, it's not even spelled as G.
In Larouche notation, the white keys all differ by Pythagorean (white) intervals. The black keys are tuned so that tritones between a white and black key are exactly 600 cents (that is, sqrt(2)). Thus white-black intervals are not just intervals, but black-black intervals are also Pythagorean.
Before we leave Larouche, let me point out that according to his writings, one of his favorite philosophers is Leibniz -- taking us back full circle to last weekend's Google Doodle.
Conclusion
Let's end today with a link to Fawn Nguyen's latest post:
http://fawnnguyen.com/st-cloud-minnesota/
Yesterday, Nguyen writes about her upcoming visit to her childhood home in Minnesota (to eat Sloppy Joes, according to the Google Doodle). In doing so, she writes about growing up as a young Vietnamese immigrant in a new country. Her story definitely belongs here in my Fourth of July post, since it's all about how a poor girl from a large family can move thousands of miles and live the American dream. (The move must have been tough -- she leaves Vietnam, where the climate is tropical, and goes to Minnesota, where the climate is, um, not tropical.)
Nguyen writes:
I will sit and watch the news with the uncle. I have no idea what they are saying, but I just like seeing white people’s faces and listening to how fast they talk. The best part is there’s always something on TV, there’s no curfew. I have two favorite shows, The Price is Right and Happy Days. You don’t have to understand very much English to watch The Price because prices are numerical, and English numbers look the same as Vietnamese numbers, except Americans are weird to write $50 instead of 50$. They claim to read from left to right too.
Hey, The Price Is Right is one of my favorite game shows, too! But Nguyen's story takes place during the show's earliest seasons in the 1970's, when Bob Barker (with dark hair) was still the host. Also, traditionalists might appreciate this part of the story. Nguyen writes that she doesn't need to know much English to understand numerical prices or math, while Common Core often requires students to explain their answers, which can be difficult for English learners like the young Nguyen.
She continues:
I like Happy Days because it’s a show with cute boys, Chachi and Fonzie. (My family calls me Fawnzie. My name morphed from Phương to Fawn to Fawnzie. More recently, my son Gabriel probably sensed that I was stressed in our conversation and said, “Mom, I need you to be Fawnzie right now.” And I knew what he meant.)
That's interesting -- so her name "Fawn" isn't really a color (of a 19-limit musical interval), but is actually the Anglicized version of her birth name Phuong.
Fawn Nguyen -- um, Phuong Nguyen -- concludes:
I get to visit St. Cloud this August; it’ll be my first time back since I left in 1979. I’ll be facilitating a full-day workshop, and St. Cloud will just be 70 miles away. I’m flushed with nostalgia and gratitude — going back to my first home in America.
And with that, I wish everyone a happy Fourth of July!
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