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# Mathematics

## Ptolemy's Table of Chords: Trigonometry in the Second Century

28 June 1994

### Introduction

Although certainly not the first trigonometric table1, Ptolemy's On the Size of Chords Inscribed in a Circle (2nd century CE) is by far the most famous. Based largely on an earlier work by Hipparchus (ca. 140 BC) it was included in Ptolemy's definitive Mathematike Syntaxis, better known by its Arabic name Almagest2. In this paper I will describe the geometric theorems used in the construction of this table and attempt to relate them to their contemporary trigonometric counterparts.

#### Equivalence of the Table of Chords and a table of sines

Given a circle whose diameter and circumference are divided into 120 and 360 parts respectively, Ptolemy was able to calculate the corresponding chord length for every central angle up to 180° in half-degree intervals. Given, in the diagram to the right that

 sin θ = AM = 2AM = AB = crd θ 2 OA 2OA diameter 120°

(where crd θ is the length of the chord described by the central angle subtending an arc of θ parts of the circumference), the Table of Chords as compiled by Ptolemy is equivalent to a table of sines for every angle up to 90° in quarter degree intervals.

### Special angles

Ptolemy began his discourse by calculating the chord lengths for the central angles corresponding to the sides of a regular inscribed decagon, hexagon, pentagon, square, and triangle. He determined the first three of these chords using the figure below with the following proof3.

Given
circle ABC with center D
DE = EC and EF = BE
Prove
CD is the side of a regular inscribed hexagon
DF is the side of a regular inscribed decagon
BF is the side of a regular inscribed pentagon
Statements   Reasons
(1) CF × DF + ED2 = EF2 (1)
 (2a + x)x + a2 = (a + x)2 2ax + x2 + a2 = a2 + 2ax + x2 a2 + 2ax + x2 = a2 + 2ax + x2
(2) CF × DF + ED2 = BE2 (2) BE = EF
(3) ED2 + DB2 = BE2 (3) Pythagoras' theorem
(4)
 CF × DF + ED2 = ED2 + DB2 CF × DF = DB2 CF × DF = DC2
(4) combine (3) with (2) and solve
(5) DF:DC is the golden ratio, a.k.a. the extreme and mean ratio
 √5 − 1 :1 = 0°37'5":1°3 2
(5) Ptolemy was rather unclear on how he arrived at this statement other than to cite Euclid (Euclid VI, 3). On inspection, however, if we let r = 2 then a = 1, x + a = √5, and x = √5 − 1. Thus DF:DC = x:r is indeed the golden ratio.
DC is the side of a regular inscribed hexagon DC is a radius
DF is the side of a regular inscribed decagon from Euclid, "the side of the hexagon and the side of a decagon which arre inscribed in the same circle… cut that line in the extreme and mean ratio (Euclid XII, 9)" (Ptolemy 20)
BF is the side of a regular inscribed pentagon also from Euclid, "the square on the side of a pentagon is equal to the square on the side of a hexagon together with the square on the side of a decagon, all inscribed in the same circle (Euclid XII, 10)" (Ptolemy 20)

Using these results, Ptolemy then calculated the chord lengths for the central angles.

DF is the side of a regular decagon
 DE2 + DB2 = BE2 30°2 + 60°2 = BE2 BE = 67°4'55" BE = EF DF = EF − DE DF = 67°4'55" − 30° DF = 37°4'55"

thus crd 36° = 37°4'55"

BF is the side of a regular pentagon
 DF2 + DB2 = BF2 (37°4'55")2 + 60°2 = BF2 BF = 70°32'3"

thus crd 72° = 70°32'3"
(This was reported as 70°32'4" in the Table of Chords.)

DC is the side of a regular hexagon
 DC = AC = 120° = 60° 2 2

thus crd 60° = 60°

Likewise since the square of the side of an inscribed square is twice the square of the radius and the square of the side of an inscribed equilateral triangle is three times the square of the radius, we get

crd 090° = √(2 × 60°2) = 084°51'10"
crd 120° = √(3 × 60°2) = 103°55'23"

Given these angles, Ptolemy then showed how it was possible to derive other chord lengths using the fact that the inscribed angle that subtends the diameter of a circle is 90°. Therefore, by application of Pythagoras theorem,

crd 108° = √(120°2 − crd2 72°) = 097°4'56"0
crd 144° = √(120°2 − crd2 36°) = 114°7'37"4

The chords of the special angles are summarized in Table 1 below. For the remaining chords we need to create new mathematical tools.

Table 1: Chords of the special angles
angle crd
36° 37°4'55"
60° 60°
72° 70°32'3"
90° 84°51'10"
108° 97°4'56"
120° 103°55'23"
144° 114°7'37"
180° 120°

### Ptolemy's theorem

In a cyclic quadrilateral the product of the diagonals is equal to the sum of the products of the pairs of opposite sides.

Given
pick a point E such that ∠ABE = ∠DBC
Prove
AC × BD = AB × CD + AD × BC
Statements   Reasons
(1) ∠ABD = ∠EBC (1) add ∠EBD to ∠ABE and ∠DBC
(2) ∠BDA = ∠BCE (2) since they subtend the same arc
(3)
 BC = BD CE DA
(3) △ABD and △BCE are similar
(4) ∠ABE = ∠DBC (4) given
(5) ∠BAE = ∠BDC (5) since they subtend the same arc
(6)
 BA = BD AE DC
(6) △ABE and △BCD are similar
(7)
 (AB × CD = BD × AE) +(BC × AD = BD × CE) AB × CD + AD × BC = BD × AE + BD × CE = BD × (AE + CE)
(7) cross-multiply (3) and (6) and add them
AC × BD = AB × CD + AD × BC AE + CE = AC

With this theorem, Ptolemy produced three corollaries from which more chord lengths could be calculated: the chord of the difference of two arcs, the chord of half of an arc, and the chord of the sum of two arcs. I will now present these corollaries and the subsequent proofs given by Ptolemy. I will also derive a formula from each corollary that can be used to calculate the additional chords. (Ptolemy did not supply any formulae.) Furthermore, I will show that the three corollaries are equivalent to the trigonometric identities for the sine of the difference of two angles, the sine of half an angle, and the sine of the sum of two angles respectively.

#### Corollary 1: Chord of the difference of two arcs

Given
AC and AB chords of known length
[α and β as shown but also
let AC = crd θ and AB = crd φ]
Prove
BC can be found
[find a formula for the chord of the difference of two arcs and show its equivalence to the identity for the sine of the difference of two angles]
Statements   Reasons
(1) AD is a diameter, AC and AB are known (1) given
(2) CD and BD can be found (2) Pythagoras' theorem
BC can be found substitution into Ptolemy's theorem AB × CD + AD × BC = AC × BD
(4)
(4) solve Ptolemy's theorem for BC and apply Pythagoras' theorem to BD and CD
 crd(θ − ϕ) = ⎡⎢⎣ crd θ√(120°2 − crd2ϕ) − ⎤⎥⎦ ÷ 120° crd ϕ√(120°2 − crd2θ)
in chord notation (note θ = 2α and ϕ = 2β)
(6)
(6) divide Ptolemy's theorem by AD2
(7) sin β cos α + sin(α − β) = sin α cos β (7) definition of sine and cosine
sin(α − β) = sin α cos β − sin β cos α in modern notation

By successive application of this theorem to the chords summarized in Table 1, it is possible to calculate all the chord lengths for the angles between 6° and 180° in 6° intervals. Thus

crd 12° = crd(72° − 60°) = 12°32'36"
crd 6° = crd(18° − 12°) = 6°16'50"
and so on…

These values are within 1" of those found in the Table of Chords. When there is a discrepancy, it is usually due to rounding errors. It appears that either Ptolemy's computers (persons hired to do the menial calculations) did not carry their work out beyond the seconds place or they did not believe in rounding up ever. This was true for many of the values I calculated.

#### Corollary 2: Chord of half an arc

Given
semicircle ABCD with diameter AC
BC chord of known length
arc BC bisected at D
DF ⟂ AC
let AE = AB
[½α as shown, but also let BC = crd θ so that BD = DC = crd ½θ]
Prove
CF = ½(AC − AB)
[find a formula for the chord of half an arc and show its equivalence to the identity for the sine of half an angle]
Statements   Reasons
(2) DC = DE (2) BD = DC since ∠BAD = ∠DAC
(3) △DEC is isosceles (3) definition of isosceles and (2)
(4) EF = CF (4) altitude to the base of an isosceles triangle bisects the base
(5) EC = AC − AB (5) AE = AB
CF = ½(AC − AB) (4) and (5)
(7) △ACD and △DCF are similar (7) ∠ADC = ∠DFC = 90°
∠ACD = ∠FCD
(8)
 AC = CD CD CF

CD2 = AC × CF
(8) similar triangles

cross-multiplication
(9) CD2 = ½AC(AC − AB) (9) substitute (6) into (8)
(10) CD2 = ½AC[AC − √(AC2 − BC2)] (10) Pythagoras' theorem on chord AB
crd ½θ = √[7200° − 60°√(120°2 − crd2θ)] in chord notation (note θ = 2α)
(12)
 ⎡⎢⎣ CD ⎤2⎥⎦ = ½ ⎡⎢⎣ 1 − AB ⎤⎥⎦ AC AC
(12) divide (9) by AC2
 sin ½α = ⎡⎢⎣ 1 − cos α ⎤½⎥⎦ 2
in modern notation

This theorem makes it possible to calculate chords in ever smaller increments. Thus…

crd 3° = crd(½ × 6°) = 3°8'28"
crd 1½° = crd(½ × 3°) = 1°34'15"
crd ¾° = crd(½ × 1½°) = 0°47'8"
and so on…

#### Corollary 3: Chord of the sum of two arcs

Given
circle ABCDE with center F
AFD and BFE diameters
AB and BC chords of known length
[α and β as shown, but also let AB = crd θ and BC = crd φ]
Prove
AC can be found
[give a formula for the chord of the sum of two arcs and show its equivalence to the identity for the sine of the sum of two angles]
Statements   Reasons
AB is known
BD can be found from which
DE can be found
(1) given
given
Pythagoras' theorem
Pythagoras' theorem
(2) BE is a diameter
BC is known
CE can be found
(2) given
given
Pythagoras' theorem
(3) CD can be found (3) substitution into Ptolemy's theorem BC × DE + CD × BE = BD × CE
AC can be found substitution into Ptolemy's theorem AB × CD + AD × BC = AC × BD
(5) BE = AD (5) both are diameters
(6) DE = AB (6) application of Pythagoras' theorem to △BDE and △ABD and (5)
(7)
 AD(AB × CD + AD × BC = AC × BD) −AB(AD × CD + AB × BC = BD × AC × AD − BD × AB × CE) AD2 × BC − AB2 × BC = BD × AC × AD BC(AD2 − AB2) = BD(AC × AD − AB × CE)
(7) combine Ptolemy's theorem in (3) and (4) with (5) and (6) then eliminate CD
(8) BD = √(AD2 − AB2)
(8) Pythagoras' theorem and (5)
(9)
(9) substitute (8) into (7) and solve
 crd(θ + ϕ) = ⎡⎢⎣ crd θ√(120°2 − crd2ϕ) + ⎤⎥⎦ ÷ 120° crd ϕ√(120°2 − crd2θ)
in chord notation (note θ = 2α and ϕ = 2β)
(11)
(12) ∠BEC = ∠BDC = β (12) since they subtend the same arc
(13)
 sin β[1 − sin 2α] = cos α[sin(α + β) − sin α cos β] sin β cos 2α = cos α[sin(α + β) − sin α cos β] sin β cos α = sin(α + β) − sin α cos β sin(α + β) = sin α cos β + sin β cos α
(13) definition of sine and cosine
Pythagoras' theorem (identity)
divide both sides by cos α
in modern notation

By successive application of this theorem to the chords found with the first two corollaries it is possible to calculate all the chord lengths for the angles between 0° and 180° in 1½° increments. Thus…

crd 19½° = crd(18° + 1½) = 20°19'20"
crd 21° = crd(18° + 3°) = 21°52'6"
crd 22½° = crd(21 + 1½°) = 23°24'40"
and so on…

Again these values are within 1" of those calculated by Ptolemy.

With things as they stand now, we still cannot calculate the chords for two-thirds of the values in our intended table. However, if we knew the values of crd ½° and crd 1° we could then apply corollary 3 repeatedly to the chords already known and finish the table. If the trisection of an angle were geometrically possible, we could use crd 1½° to find crd ½° algebraically and then apply corollary 2 to find crd 1°. Given the well-known impossibility of this trisection, Ptolemy decided instead to approximate the value of crd 1° by means of "a little lemma which, even if it may not suffice for determining chords in general, can yet in the case of very small ones, keep them indistinguishable from chords rigorously determined" (Ptolemy 28). This lemma, attributed to Aristarchus, appears with its proof below.

### Aristarchus' inequality

Given
circle ABCD
BA and BC chords of unequal length (BA < BC)
∠ABC bisected by BD
DFH ⟂ AC at F
DFH = DE = DG
[let α and β be angles on separate inscribed right triangles such that chord BC and AB are opposite angle α and β respectively, thus α > β]
Prove
 BC < arc BC BA arc BA
 ⎡⎢⎣ and that sin α < α ⎤⎥⎦ sin β β
Statements   Reasons
(1) CD = AD (1) ∠ABD = ∠DBC
(2) CE > EA (2) (Euclid VI, 3)
(3) DE > DF
DH > DF
(3) (Euclid III, 3 and 26)

DH = DE
(4)
 area △DEF < area sectorDEH area △DEA area sectorDEG
(4) area △DEF < area sector DE
area △DEA > area sector DEG
(5)
 area △DEF = EF area △DEA EA
(5) similar triangles
(6)
 ∠FDE = area sectorDEH ∠EDA area sectorDEG
(6) area of a sector ∝ angle describing it
both sectors are on the same circle
(7)
 EF < ∠FDE EA ∠EDA FA < ∠FDA EA ∠EDA CA < ∠CDA EA ∠EDA CE < ∠CDE = ∠CDB EA ∠EDA ∠BDA
(7) (4), (5), and (6)

"componendo…

doubling the antecedents…

separando" (Ptolemy 29–30)
(8)
 CE = BC and ∠CDB = arcBC EA BA ∠BDA arcBA
(8) (Euclid VI, 3 and 33)
 BC < arcBC BA arcBA
(7) and (8)
 sin α < α sin β β
 arcBC < α and see section 1 arcBA β

#### Approximation of small chords

Given
circle ABC
two chords AB and AC such that AC > AB
Find
crd 1°
crd ½°
Statements   Reasons
(1) let AB = crd 3/4° and AC = crd 1°
then arcAC = 4/3arcAB
AC < 4/3AB = 4/3(0°47'8")
thus crd 1° < 1°2'50"
(1) substitution into and solution of Aristarchus' inequality
(2) let AB = crd 1° and AC = crd 11/2°
then arcAC = 3/2arcAB
AB > 2/3AC = 2/3(1°34' 15")
thus crd 1° > 1°2'50"
(2) repeat (1) with different values
crd 1° ≈ 1°2'50" (1) and (2)
crd ½° ≈ 0°31'25" corollary 2

#### Sixtieths

Ptolemy carried his work out further by dividing the interval between successive chords into thirtieths. This effectively allows for the calculation of any chord between 0° and 180° in one second intervals. While not rigorously produced, the values of the sixtieths are, in Ptolemy's words, "accurate as far as the sense are concerned" (Ptolemy 32).

### The Table of Chords

A section of the Table of Chords is shown in Table 2 below.

Table 2: Page 1 from the Table of Chords5
Arcs Chords Sixtieths   Arcs Chords Sixtieths
00½ 00 31 25 0 1 2 50   12½ 12 32 36 0 1 2 28
01½ 01 02 50 0 1 2 50   12½ 13 03 50 0 1 2 27
0 01 34 15 0 1 2 50   13½ 13 35 04 0 1 2 25
02½ 02 05 40 0 1 2 50   13½ 14 06 16 0 1 2 23
0 02 37 04 0 1 2 48   14½ 14 37 27 0 1 2 21
03½ 03 08 28 0 1 2 48   14½ 15 08 38 0 1 2 19
0 03 38 52 0 1 2 48   15½ 15 39 47 0 1 2 17
04½ 04 11 16 0 1 2 48   15½ 16 10 56 0 1 2 15
0 04 42 40 0 1 2 47   16½ 16 42 03 0 1 2 13
05½ 05 14 04 0 1 2 47   16½ 17 13 09 0 1 2 10
0 05 45 27 0 1 2 46   17½ 17 44 14 0 1 2 07
06½ 06 16 49 0 1 2 45   17½ 18 15 17 0 1 2 05
0 06 48 11 0 1 2 43   18½ 18 46 19 0 1 2 02
07½ 07 19 33 0 1 2 42   18½ 19 17 21 0 1 2 00
0 07 50 54 0 1 2 41   19½ 19 48 21 0 1 1 57
08½ 08 22 15 0 1 2 40   19½ 20 19 19 0 1 1 54
0 08 53 35 0 1 2 39   20½ 20 50 16 0 1 1 51
09½ 09 24 51 0 1 2 38   20½ 21 21 11 0 1 1 48
0 09 56 13 0 1 2 37   21½ 21 52 06 0 1 1 45
10½ 10 27 32 0 1 2 35   21½ 22 22 58 0 1 1 42
10½ 10 58 49 0 1 2 33   22½ 22 53 49 0 1 1 39
11½ 11 30 05 0 1 2 32   22½ 23 24 39 0 1 1 36
11½ 12 01 21 0 1 2 30

A random sample of sines produced from the Table of Chords were compared with those generated by a pocket calculator accurate to ten places. The results are summarized in Table 3 below.

Table 3: A comparison of the Table of Chords with a ten-place calculator
θ crd θ (crd θ)/120° sin (θ/2)
16½° 17°13'9" 0.1434930556 0.1434926220 0.0000004336
49° 49°45'48" 0.4146944444 0.4146932427 0.0000012017
64° 63°35'25" 0.5299189815 0.5299192642 0.0000002827
83½° 79°54'21" 0.6658819444 0.6658816660 0.0000002784
110½° 98°35'32" 0.8216018519 0.8216469379 0.0000450860
126° 106°55'15" 0.8910069444 0.8910065242 0.0000004202
155° 117°9'20" 0.9762962963 0.9762960071 0.0000002892
176½° 119°56'39" 0.9995347222 0.9995335908 0.0000011314

As the table shows, Ptolemy's results agree with modern calculator values to five or six decimal places. (See the Postscript for more on the accuracy of the Table of Chords.)

The remainder of the Almagest consists of astronomical calculations: the position of the sun, moon, and planets at various times relative to the fixed stars. The Table of Chords played an important role in their compilation.

### Conclusion

Hipparchus' earlier 12-book treatise on the construction of a Table of Chords disappeared sometime after the fourth-century because it was superseded by the far more comprehensive Almagest. The Almagest reigned supreme as the treatise in practical trigonometry for approximately one-thousand years. During the tenth-century, the Islamic mathematician Abû'l-Wefâ computed the values for the sines and tangents of an angle in quarter-degree intervals and essentially reproduced the Table of Chords in contemporary form. In the sixteenth-century, the Teutonic mathematician George Joachim Rhaeticus had, over the course of twelve years and with the help of hired computers, calculated the values of all six trigonometric functions to ten places and the sine function to fifteen places in ten second intervals. With the ubiquity of programmable calculators and personal computers, computational ability has advanced to the point where it is within the economic means of large segments of the earth's population to reproduce the life work of the ancients on demand. Technology has rendered the work of such mathematicians superfluous in much the same way the Almagest obliterated all twelve volumes of Hipparchus.

### Postscripts

#### How accurate is the Table of Chords?

All 360 values from Ptolemy's Table of Chords were compared to their "actual" values calculated in Google Sheets.

 absolute error = Ptolemy's value − spreadsheet value

The graph below shows that the values in the Table are generally a tiny bit larger than they should be. The root-mean-square of the error calculated this way is 0.000136°, implying that the Table is accurate to three decimal places — not the five or six I stated in the main body of the paper. Since we don't compute chords anymore and we don't use the sexagesimal convention to divide the diameter into 120° parts, this number may not be that helpful. The graph below also shows an increasing trend in the absolute error caused, no doubt, by the increase in the value of the entries as one reads down the table. As the angle gets bigger, the chords get bigger. As the chords get bigger, the error gets bigger.

Relative error may be a more useful way to test Ptolemy's work.

The graph below shows that the relative error is greatest for small angles and seems to stay nearly constant after 60°. This makes sense as the precision of Ptolemy's degree-minute-second notation is always the same — to the nearest second. A one second error is bigger relative to a small angle than a large one. The root-mean-square of the error calculated this way is 0.00000737 or 7.37 parts per million.

### Footnotes

1. The earliest reputed trigonometric table, fifteen secants from 30° to 45°, can be found in the famous Babylonian tablet, Plimpton 322 (ca. 1900–1600 BC).
2. Commentators identified Mathematike Syntaxis (mathematical composition or compilation) by the superlative "ta megiste" (the greatest). This was subsequently transliterated by Arab scholars as al-magiste, then Almagestum by Latin speaking European scholars, then eventually Almagest by English speaking scholars.
3. In the sexagesimal notation used by Ptolemy, the degrees symbol (°) refers to a unit of measure, the minutes symbol (') to 1/60 of the unit, and the seconds symbol (") to 1/3,600 of the unit. Thus 0°37'5" represents 0 + 37/60 + 5/3,600 = 0.618055556. The notation applies equally to the lengths of arcs (angular measure) and line segments (linear measure). Only the angular usage of this notation has survived to the present.
4. For some reason, crd 144° was reported as 114°7'47" in the Table.
5. The sexagesimal system is carried out one place further in the sixtieths column. Thus 0°1'2"50''' represents 0 + 1/60 + 2/3,600 + 50/216,000 = 0.0174537037.

### Sources

• Bunt, Lucas N.H., Jones, Philip S. and Bedient, Jack D. The Historical Roots of Elementary Mathematics. New York, NY: Dover, 1988.
• Eves, Howard. An Introduction to the History of Mathematics. sixth edition. Fort Worth, TX: Harcourt Brace Jovanovich, 1990.
• Eves, Howard. Great Moments in Mathematics (Before 1650): The Dolciani Mathematical Expositions, Number Five. Washington, DC: The Mathematical Association of America, 1980.
• Kline, Morris. Mathematical Thought from Ancient to Modern Times. Oxford, UK: Oxford University Press, 1972.
• Ptolemy, Claudius. Mathematical Composition (Almagest). Translated from the Greek text of Heiberg by R. Catesby Taliaferro. Annapolis, MD: St. John's University, 1938.
• Ptolemy, Claudius. Ptolemy's Almagest. Translated and annotated by G.J. Toomer. London: Gerald Duckworth and Co., 1984.