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The Cord Table in the Ancient Greek TimesSeptember 28, 2000

Ryoji Tatsukawa1. Introduction

They say that Hipparcus systematically used trigonometry, calculated chords ofcircle and made a chord table, which corresponds to today's trigonometric ratiotable, for the first time in about 150 B.C. Unfortunately, it is not existing today,but they say, Ptolemy Ptolemaios , Claudios often cited it in his writings,( )

Almagest; however, Gerald Toomer restored it in 1973. Let us look at what kind oftable they used.

Note: The Almagest is a compilation of all the study about astronomy in the an-cient Greek times. It had been literally the greatest astronomical study, till helio-centric theory was advocated in the 17th century.

2. Ptolemy's cord table1 First of all, he divided a quadrant into 180 equal angles around its center( ) ﾟ

and its radius is divided into 120 equal parts at every 1/2 Secondly, he calculatedﾟ

the length of chord AA', crd. at every central angle . Being shown in the Table1,α α

he represented the fractions of chord using sexagesimal system.α

α

………………………

( ) ( )Fig.1 Tab.1

Ａ

Ｏ

A'

α

M 1° 1; 2,50 0;1,2,50

112°

1;34,15 0;1,2,50

2° 2; 5,40 0;1,2,50

212°

2;37, 4 0;1,2,48

3° 3; 8,28 0;1,2,48

312°

3;39,52 0;1,2,48

crd.ar=60

Averagefor every

increase1' increase

of choad

12°

0;31,25 0;1,2,50

The Cord Table in the Ancient Greek Times

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Note: The half length of the chord AA' corresponds to today's sine AM. This meansthat they used the length of the chord instead of a ratio with two sides of a righttriangle.

It was the 16th century when trigonometric ratio was dealt as a right triangle.Rheticus, Georg Joahim 1514 1576 represented it as length of side of a trian-( ― )

gle, not a ratio. A chord table had been made at every segment of radius, till thedecimal notation was established by Stevin, Simon 1548 1620 .( ― )

The trigonometric ratio was defined in the four quardrants, as the ratio with twosides of the right triangle, by Euler, Leonhard 1707 1783 in the 18th century,( ― )

and sine notation was also established by him.

2 Construction of regular inscribed polygons in a circle( )

The following lengths of chords are able to be calculated by using inscribedregular triangle, square and hexagon.

And we can also calculate the lengths of chords for 72 and 36 by usingﾟ ﾟ

well-known construction of regular hexagon.

Draw a circle centered at the・

Point O, of which diameter is AB, anda midperpendicular CO on AB.

With a radius CD, draw a circle・

centered at the point D which is amidpoint of OB. This circle meetsthe diameter AB at a point E.

The line segments EO, CE are the・

sides of a regular decagon and a reg-ular pentagon, respectively.

Proof:

( )Fig.2

crd.120°= 3 r = 60 3 ≒ 103;55'23"crd.60°= r = 60;

crd.90°= 2 r = 60 2 ≒ 1;24'51"

OE

A

C

BD

EB・EO + OD2

= (ED + OD)(ED－OD) + OD2

= ED2 = CD2= CO2 + OD2

The chord Table in Ancient Greek Times

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This means that the segment EB is divided in external and internal ratio at thepoint O. Because, the segment OB is a side of the inscribed regular pentagon, thenwe find that the segment EO is a side of inscribed hexagon in the circle O.

( )Proposition 9 in the books 13 of Euclid's Element

The segment CO is a side of regular hexagon and EO is a side of regular deca-gon, then we find that CE is a side of regular pentagon.

( )Proposition 10 in the Books 13 of Euclid's Element

Note: Setting CO = r in the Fig.2, we find

We can calculate crd.36° and crd.72° by using and .① ②

( 3 )

Proof: Setting AB =1, we getBC = crd. ,αAC = crd. 180 .( ﾟ－α)'

( )Fig.3

∴ EB・EO = CO2 = BO2

CE2 = CO2 + EO2

OD =2r

.

CD2 = CO2 + OD2 = r2 +2r 2

=54

r2∴ CD =

52

r

EO = ED－OD = CD－OD =5－1

2r ……………①

CE2 = EO2 + CO2 =5－12

r2

+ r2

∴ CE =10－2 5

2r ……………②

crd.36°=5－12

×60 ≒ 37;4'55", crd.72°=10－2 5

2×60 ≒ 70;32'3"

C

α

A BO

CB2 + CA2 = AB2

∴ (crd.α)2 +

=

crd.(180°－α

(crd.180°)2

2

(crd.a )2 + crd.(180°－a) 2 = (crd.180°)2

The Cord Table in the Ancient Greek Times

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Note: Using this formula, we can calculate crd.108 and crd.44 .ﾟ ﾟ

4 Lemma: For the inscribing quadrangle ABCD in a circle, we have( )

・ ・ ・ ( )AB CD + BC AD = AC BD. Ptolemy's theoremProof:Take a point E on the segment BDso that DAC = EAB.∠ ∠

ACD = ABE∠ ∠

ADC ABE∴ △ ∽△

AB : BE = AC : CD∴ ・ ・ ①AB CD = AC BE ………

BAC = EDA,∠ ∠

ACB = ADB∠ ∠

ABC AED∴ △ ∽△

BC : AC = ED : AD∴ ・ ・ ②AD BC = AC ED………

+ : AB CD + AD BC① ③ ・ ・

= AC (BE + ED)・

( )= AC BD Fig.4・

(crd.108°)2 + crd.(180°－108°) 2 = (crd.180°)2

(crd.108°)2 = (crd.180°)2－(crd.72°)2

= (2r)2 －(10－2 5

2r)2 =

6 + 2 54

r2

crd.108°=5 +12

r =5 +12 ×60

= 30 ( 5 + 1) ≒ 97;4'55"

(crd.144°)2 + crd.(180°－144°) 2 = (crd.180°)2

(crd.144°)2 = (crd.180°)2－(crd.36°)2

= (2r)2 －(5－12

r)2 =10 + 2 5

4r2

crd.144°=10 + 2 5

2r =

10 + 2 52

×60

≒ 114;7'36"

A

B C

D

E

・

・

The chord Table in Ancient Greek Times

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5 Theorem( )

Theorem 1: When chord AB and BD meet at an end point B of diameter BC, ifBC = 1, AOC = and DOC = , the following addition theorem is established:∠ α ∠ β

α β α β α βcrd.180° crd.( ) = crd. crd.(180° ) crd.(180° ) crd.・ － ・ － － － ・

Proof: By using Ptolemy's theorem,we have

AD BC = AC BD AB DC.・ ・ － ・

AD = crd.( ), BC = crd.180°,α β－

AC = crd. , BD = crd.(180° ),α β－

βDC = crd.Then, we get

crd.180° crd.( )・ －α β

= crd. crd.(180° )α β・ －

crd.(180° ) crd. .－ － ・α β

( )Note: This chord addition theorem corresponds Fig.5to the following sine addition theorem:

Proof:

By using Ptolemy's theorem, wehave

In AED,△

α β∠ ∠ －AED = ABD =

By using and , we get① ②

( )Fig.6

A

B

D

CO

α β

A

B

D

C

E

O

α

β

sin(a－b) = sina cosb －cosa sinb

AB = cosa, AC = sinaBD = cosb, DC = sinb

AD・BC = AC・BD－AB・DC.∴ AD= sina cosb－cosa sinb …①

∠ADE =12∠

R

∴ AD = sin (a－b) ……………②

sin(a－b) = sina cosb －cosa sinb.

The Cord Table in the Ancient Greek Times

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Note: If we take the point D on arc BEC, we get the following formula:

Note: We can calculate crd.12 .ﾟ

Theorem 2:Setting AOB= and DOA= in the Fig.7, we have the following formula:∠ ∠α β

crd.180° crd.(180° BD) = crd.(180° AB) crd.(180° AD) crd.AB crd.AD.・ － － ・ － － ・

Proof:

By using Ptolemy's theorem, we getAE CD = AC DE + AD CE.・ ・ ・

Note:

( )Fig.7

sin(a + b) = sina cosb + cosa sinb.

crd.180°crd.(72°－60°) = crd.72°crd.(180°－60°)－crd.(180°－72°)crd.60°2r・crd.12°= crd.72°crd.120°－crd.108°crd.60°

=10－2 5

2r× 3 r －

5 + 12

r×r

crd.12°=30－6 5 － 5－1

4r

=30－6 5 － 5－1

4×60

= ( 30－6 5 － 5－1)×15≒ 12;32'36"

A

B

D

E

O

Cα

β

AC = crd. (180°－α),DE = crd.(180°－β),AD = crd.β, CE = crd.α

AE = 2r, CD = crd. 180°－(α+β) ,

crd.180°crd. 180°－(a + b)= crd.a・ crd.(180°－b)－crd.b・ crd.a

cos(a + b）= cosa cosb－sina sinb

The chord Table in Ancient Greek Times

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Proof: Setting ACB = and DCA = in the Fig.8, we get this formula.∠ ∠α β

By using Ptolemy's theorem, we get

Theorem 3:

Proof: Setting COB = in the Fig.9, we get this formula. The bisector of∠ α

COB meets the circle O at a point D and a point E satisfies AC = AE.∠

So we find that CD = DE = DB.Let F be a foot of the perpendicularthat we drop to AB from the pointD.

EF = FB

DOB DEF△ ∽△

②……………By using and we have① ②,

( )Fig.9

A

B

D

E

O

α

β

β

C

( Fig.8 )

AC = cosa, AB = sinaDE = cosb, AD = sinbAE = 1, CD = cos(a + b)

AB = CE

AE・CD = AC・DE－AD・CE.∴ cos(a + b）= cosa cosb－sina sinb

A B

C

D

EO Ｆ

α

(crd.a2

)2 =12

(crd.180°)・ crd.180°－crd.(180°－a)

EB = AB－AE = AB－AC ……①

OB・EB = DB2

DB2 =12

AB(AB－AC)

DB = crd.a2

, AB = crd.180°,

AC = crd.(180°－a)

The Cord Table in the Ancient Greek Times

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Note: When CAB = , theorem 3 corresponds to the following formula for half∠ α

angle.

We can calculate the following chords by using this formula.

Similarly, we can calculate the following chords.

6 Interpolation formula( )

Proof: Let E be the point at which the bisector of the angle ABC meets the side AC,

∴ (crd.a2

)2 =12

(crd.180°)・ crd.180°－crd.(180°－a)

sin2a2

=12

(1－cosa )

=64－ 30－6 5 － 5－1 2

16r2

crd.168°=4r

64－( 30－6 5 － 5－1)2

crd.6°=2r

8－ 64－( 30－6 5 － 5－1)2

≒ 6;16'49"

crd.3°=2

2r 4－ 8 + 64－ 30－6 5 － 5－1

2

≒3;8'28"

crd.112°=

22

r4－ 2 4 + 8 + 64－( 30－6 5 － 5－1)2

≒1;34'14"

crd.34 °

= r 2 － 2 +2

24 + 8 + 64 － ( 30 － 6 5 － 5－1)2

≒0;47'7"

crd.bcrd.a

<ba

for 0 < a < b <p2

(crd.168°)2 = (crd.180°)2－(crd.12°)2

(crd.6°)2 = (crd.180°)(crd.180°－crd.12°) = r(2r－crd.168°)

The chord Table in Ancient Greek Times

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we find AD = CD.AB : BC = AE : EC

AE < EC∴

Let DF be a perpendicular to AC,then we find,

AD > ED > FD.The circle which has a point D as a

center and a radius DE, meets theline AD at the point G and the ex-tension of DF at the point H.

EF : AE = EDF : ADE△ △

（ ）(sectorEDH) : sectorGDE= EDH : GDE∠ ∠

EF : AE < EDF : ADE∴ ∠ ∠

AF : AE < ADF : ADE∠ ∠

( )( )componendo Fig.10AC : AE < ADC : ADE (multiplying both sides by 2)∠ ∠

EC : AE < EDC : ADE (dividendo)∠ ∠

BC : AB < EDC : ADE∠ ∠ ∠

Note: We can calculate crd.1 as follows:ﾟ

A

B

C

D

H

EG

αβ

Ｆ

= BC: AB ∴ crd.b : crd.a < b : a

34°< 1°<

32° ∴

crd.34°

crd.1°<

34

1,

crd.32°

crd.1°<

321

∴23

crd.32°

< crd.1°<43

crd.34°

1;2'50"< crd.1°<1;2'5023

"

crd.1°≒1;2'50"

The Cord Table in the Ancient Greek Times

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Reference Books

ギリシャ数学史 ヒース著1. T. L.平田 寛・菊池 俊彦・大沼 正則訳 １９５９ 共立出版

A Manual of Greek Mathematics Thomas L. Heath Oxford 1931ギリシャの数学 彌永 昌吉・伊藤 俊太郎・佐藤 徹著 １９７３ 共立出版2.中世の数学 伊藤 俊太郎編 １９８７ 共立出版3.１８世紀の数学 小堀 憲著 １９７９ 共立出版4.数学史２ コールマン・シュケービッチ著 山内 一次訳 １９７１ 東京図書5.数学と数学の記号の歴史 大矢 真一著 昭和５３（１９７８） 裳華房6.

ユークリッド原論 中村幸四郎・寺坂英孝・伊藤俊太郎・池田惠美訳 共立出版7. 1971I.L.Heiberg et H.Menge, Euclidis Elementa opera omnia, Lipsiae 1883-1916

The chord Table in Ancient Greek Times