And yes, today is Friday the 13th. But those suffering from triskaidekaphobia need not worry -- there's no reference to 13 in any of my math pages today. Lovers of the number 13, meanwhile, might have delayed last week's Chapter 13 Test to today, since there are many references to 13 in the thirteen questions on that test.

Meanwhile, I wish to continue my new "music" label. Since today is Friday the 13th, I should have a post on a New 13-Limit Scale -- the next logical step from the New 11-Limit Scale introduced in my Easter post.

In Kite's color notation, the otonal 13-limit is "emerald," while the utonal 13-limit is "ocher." We know that Mocha's SOUND command is based on utonal harmonics, so we use the color "ocher."

Here are all the ocher notes playable in Mocha:

Sound Degree Color Note Name

248 13 ocher G

235 26 ocher G

222 39 ocher C

209 52 ocher G

196 65 ocher-green Eb

183 78 ocher C

170 91 ocher-red A

157 104 ocher G

144 117 ocher F

131 130 ocher-green Eb

118 143 ocher-amber D

105 156 ocher C

92 169 deep ocher Bb

79 182 ocher-red A

66 195 ocher-green Ab

53 208 ocher G

27 234 ocher F

1 260 ocher-green Eb

Notice that we now have a name for the lowest playable note in Mocha -- Sound 1 (Degree 260) is now called "ocher-green Eb."

As we can see in the chart, the fundamental ocher note is ocher G, at Degree 13 (Sound 248). It's important to note the ocher G that is one octave lower, at Degree 26 (Sound 235). We compare this to the white G at Degree 27 (Sound 234). Thus ocher G lies above white G at an interval of 27/26 -- this is also known as the third-tone, since at 65 cents, it's about two-thirds of a semitone. We can call it the

*tridecimal*(that is, 13-limit) third-tone.

The note name G# implies that it's a semitone above G. Since our ocher G is a third-tone above G, we need to give it a name like "G 2/3-sharp." The musician Marc Sabat has already invented a symbol for "2/3-sharp" -- a capital "H," since this is sort of like 2/3 of a sharp # symbol. In other words, another way to say "ocher G" is "Gh." (I prefer using lowercase letters in ASCII. And of course, we should call it "third-sharp" rather than "2/3-sharp" if we want to match "third-tone.")

Notice that Degree 169 (Sound 92) is deep (that is, double) ocher Bb, which we write as Bbhh. As its name implies, Bb raised by two third-tones ought to sound more like B than Bb -- and in fact, red B appears one step above, at Degree 168 (Sound 93). Thus Bbhh and B7 differ by the comma 169/168 (about 10 cents).

The tridecimal neutral third, or ocher third, is 16/13. This means that the corresponding emerald neutral third must be 39/32, so that the ocher and emerald thirds combine to make a perfect fifth. As usual, there are more playable utonal (ocher) thirds than otonal (emerald) thirds. But notice that even though Degrees 16-13 form an ocher third, we can't build a perfect fifth (3/2) on Degree 16 because three doesn't divide 16 evenly. Thus the simplest ocher

*triad*is Degrees 48-39-32 -- and in fact, there are more playable emerald triads than ocher triads, despite the otonality of the emerald triad.

If we try to build a New 13-Limit Scale, we might start it on Degree 240, since this is the root of the lowest playable ocher triad (Degrees 240-195-160). This produces the following scale:

Key Sound Degree Note Ratio Color

1. 21 240 F+ 1/1 green F

2. 27 234 Fh 40/39 ocher F

3. 66 195 Ab+h 16/13 ocher-green Ab

4. 69 192 A 5/4 white A

5. 101 160 C+ 3/2 green C

6. 105 156 Ch 20/13 ocher C

7. 117 144 D 5/3 white D

8. 131 130 Eb+h 24/13 ocher-green Eb

9. 133 128 E 15/8 white E

0. 141 120 F 2/1 green F

This scale contains two ocher triads (rooted on green F and white A) and two emerald triads (rooted on ocher F and ocher-green Ab). There are other playable triads, including a major triad on green F.

I think it's a bit of a waste not to start the scale on Sound 1 (Degree 260), since it's the only possible scale that can start on the lowest note (as 260 has 13 as a factor). This produces the following scale:

Key Sound Degree Note Ratio Color

0. 1 260 Eb+h 1/1 ocher-green Eb

1. 21 240 F+ 13/12 green F

2. 27 234 Fh 10/9 ocher F

3. 66 195 Ab+h 4/3 ocher-green Ab

4. 69 192 A 65/48 white A

5. 101 160 C+ 13/8 green C

6. 105 156 Ch 5/3 ocher C

7. 117 144 D 65/36 white D

8. 131 130 Eb+h 2/1 ocher-green Eb

But this scale contains two ugly-looking (and ugly-sounding) intervals, 65/48 and 65/36. Also, the tonic of the scale isn't the root of any triad.

Here's one more attempt to create a New 13-Limit Scale. Kite, the inventor of this color notation, sometimes used the color "emerald" as a synonym of "greenish." Here's the reason for this -- the greenish fourth, 48/35, is very close to 11/8. This suggests that we could have a nearly symmetrical scale where the inversion of greenish is ocher:

Key Sound Degree Note Ratio Color

1. 5 256 E 1/1 white E

2. 27 234 Fh 128/117 ocher F

3. 51 210 G+7 128/105 greenish G

4. 69 192 A 4/3 white A

5. 93 168 B7 32/21 red B

6. 105 156 Ch 5/3 ocher C

7. 121 140 D+7 64/35 greenish D

8. 133 128 E 2/1 white E

The problem here is that the symmetry is broken by the lack of white B -- we're forced to use something like red B instead. And we can't lower this scale by a perfect fifth to white A, since this would be Degree 384, which is below Mocha's range.

For completion, let me post the New 7-Limit Scale, from which the greenish notes are derived. After all, I've posted 3-, 5-, 11-, and 13-limit under the new "music" label, but not the 7-limit:

The New 7-Limit Scale, starting on white G:

Key Sound Degree Note Ratio Color

0. 45 216 G 1/1 white G

1. 51 210 G+7 36/35 greenish G

2. 69 192 A 9/8 white A

3. 81 180 Bb+ 6/5 green Bb

4. 93 168 B7 9/7 red B

5. 99 162 C 4/3 white C

6. 117 144 D 3/2 white D

7. 121 140 D+7 54/35 greenish D

8. 135 126 E7 12/7 red E

9. 141 120 F+ 9/5 green F

0. 153 108 G 2/1 white G

Let's program these scales into Mocha:

http://www.haplessgenius.com/mocha/

10 N=1

20 FOR X=1 TO 8

30 READ A

40 SOUND 261-N*A,4

50 NEXT X

60 DATA 256,234,210,192,168,156,140,128

10 N=2

20 FOR X=0 TO 10

30 READ A

40 SOUND 261-N*A,4

50 NEXT X

60 DATA 108,105,96,90,84,81,72,70,63,60,54

This is what I wrote last year about today's lesson:

Lesson 14-3 of the U of Chicago text is on the tangent ratio, and Lesson 14-4 of the U of Chicago text is on the sine and cosine ratios. I have decided to combine all three trig ratios into one lesson.

David Joyce was not too thrilled to have trig in the geometry course. He wrote:

**Chapter 11**[of the Prentice-Hall text -- dw] covers right-triangle trigonometry. It's hard to see how there's any time left for trigonometry in a course on geometry, but at least it should be possible to prove the basic facts of trigonometry once the theory of similar triangles is done. The section of angles of elevation and depression need not appear, and the section of vectors omitted. (By the way, who ever calls the sum of two vectors the "resultant" of the two vectors?) The one theorem of the chapter (area of triangle = 1/2 bc sin A) is given for acute triangles.

As the trig functions for obtuse angles aren't covered, and applications of trig to non-right triangles aren't mentioned, it would probably be better to remove this chapter entirely.

Yet most geometry books include trig because most state standards require it. And this most certainly includes the Common Core Standards:

#### Define trigonometric ratios and solve problems involving right triangles

CCSS.MATH.CONTENT.HSG.SRT.C.6

Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.

Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.

CCSS.MATH.CONTENT.HSG.SRT.C.7

Explain and use the relationship between the sine and cosine of complementary angles.

Explain and use the relationship between the sine and cosine of complementary angles.

CCSS.MATH.CONTENT.HSG.SRT.C.8

Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.*

Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.*

And all three of these standards appear in this lesson.

Some people may wonder, why do we use the same name "tangent" to refer to the "tangent" of a circle and the "tangent" in trigonometry? A Michigan math teacher, Mike Shelly, discusses the reasons at the following link:

On the other hand, the reason that "sine" and "cosine" have the same name is less of a mystery. In fact, the U of Chicago tells us that "cosine" actually means

*complement's sine*-- since the cosine of an angle is the sine of its complement. This is Common Core Standard C.7 above.
Last Friday's post was a whirlwind of ideas, and today's post continues these ideas. In the last two days, I linked to a variety of sources in search of answers to questions such as:

-- Should activities be taught during the trig unit?

-- Should a trig unit be taught during the Geometry class?

-- Should a Geometry class be taught during high school?

We searched high and low, from traditionalists to their opponents, seeking these answers. I fear that when I post links to all these competing sources, my own opinions are obscured. The blog readers know what David Joyce and the traditionalists believe, but not what I myself believe.

Well, here's my belief -- I answer all three of those questions in the affirmative. High school should have a Geometry class, Geometry class should have a trig unit, and a trig unit should have activities -- and I posted my activity for the trig unit of a high school Geometry course yesterday.

I also think back to the activity that sparked this debate -- proofs and the courtroom. We saw how the traditionalists objected to the courtroom activity on the grounds that it is too long.

I admit that I'm fascinated with the idea of using a courtroom to highlight Geometry proofs. I took Geometry back during the 1994-95 school year -- the year of the famous OJ Simpson trial. And so I often fantasized that my Geometry class was a courtroom -- the

*People's Court*. Actually, that TV show was off the air during that year. But it made a comeback in 1996, the first full year after the Simpson trial, as TV stations were trying to capitalize on the Simpson trial's popularity. (This was the same year that another famous courtroom show debuted --*Judge Judy*.)
So I might organize a

*People's Court*during my Geometry classes. When I would teach the lesson depends on what textbook I was using. If I had Michael Serra's text,*People's Court*would occur at the end of the year, around Chapter 13. With the U of Chicago text, court may occur in Chapter 3 (when the class first learns about proofs), and in many other texts, it may occur in Chapter 4 (where triangle congruence proofs appear).
One way to prevent the unit from taking too long is to assign each group a different medium-level proof -- then they present those proofs when the class actually reaches that unit! So one group may be assigned the Isosceles Triangle Theorem to put on trial a week later, while another is assigned some of the Parallelogram Theorems to put on trial a few months later. As long as all groups present before the end of the first semester, it works out in the end.

Here's one more connection between Ramanujan and trig. The U of Chicago text tells us how to find some trig values exactly, but not others. For example, cos(60) = 1/2, but cos(20), cos(40), and cos(80) aren't as easy to find. Well, the Indian genius found an interesting formula connecting the three cosines whose values we can't find. (All values are in degrees -- "cbrt" is cube root.)

cbrt(cos(40)) + cbrt(cos(80)) - cbrt(cos(20)) = cbrt(1.5(cbrt(9) - 2))

A 20-degree angle is not constructible and so its cosine can't be written exactly using square or cube roots -- of real numbers, that is. Complex numbers are a different issue:

cos(20) = (cbrt(

*a*) + cbrt(

*b*))/2

where

*a*and

*b*are the complex cube roots of 1 -- that is,

*a*= (1 +

*i*sqrt(3))/2,

*b*= (1 -

*i*sqrt(3))/2.

Now you can see why we have high school students memorize cos(60) and not cos(20).

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