Development of trigonometry Hipparchus & Ptolemy Serge G. Kruk/Laszl´ o´ Liptak´

Oakland University

Development of trigonometryHipparchus & Ptolemy – p.1/31 Overview

• Aristotle: the movement of the heavens are prefect and can therefore only be described by spheres (circles)

• Three names: Hipparchus, Menelaus, and Ptolemy

• Trigonometry is the applied mathematics of astronomy

• Take-away: trig functions as examples of “good” definitions

Development of trigonometryHipparchus & Ptolemy – p.2/31 Hipparchus 190–120 BCE

• Few writings left, only comments of later writers

• Tables of chords

• Distance of Earth to Sun (490R) 1 • Distance of Moon to Earth (67 3R)

• Duration of lunar month

• Duration of solar year

• Suggested position on Earth via latitude and longitude

Development of trigonometryHipparchus & Ptolemy – p.3/31 Hipparchus 190–120 BCE

• Chord of angle α will be denoted by crd(α)

• Depends on the radius!

crd( 180− α )

R crd( α )

α

Development of trigonometryHipparchus & Ptolemy – p.4/31 In modern terms

crd(α) = 2R sin(α/2)

crd(180◦ α) = 2R cos(α/2) − Common procedure

• Given a table of crd(α) based on radius R1

• To find crd(α) on circle of radius R2 R 2 crd(α) R1

Development of trigonometryHipparchus & Ptolemy – p.5/31 Choose a circumference of 60 360 • · 60 360 6,0,0 · 3438 2π ≈ 6;17 ≈

We are back to Babylonian base 60

Chords depend on radii

• Hipparchus picked a radius of 3438 units, why?

Development of trigonometryHipparchus & Ptolemy – p.6/31 We are back to Babylonian base 60

Chords depend on radii

• Hipparchus picked a radius of 3438 units, why? Choose a circumference of 60 360 • · 60 360 6,0,0 · 3438 2π ≈ 6;17 ≈

Development of trigonometryHipparchus & Ptolemy – p.6/31 Chords depend on radii

• Hipparchus picked a radius of 3438 units, why? Choose a circumference of 60 360 • · 60 360 6,0,0 · 3438 2π ≈ 6;17 ≈ We are back to Babylonian base 60

Development of trigonometryHipparchus & Ptolemy – p.6/31 from right-angle triangle 2 2 crd(180◦ α) = 4R crd (α) • − − and (proof in book) p 2 crd (α/2) = R 2R crd(180◦ α) • − −

How did he compute chords? Note

• crd(60◦) = R = 34380 (equilateral triangle)

Development of trigonometryHipparchus & Ptolemy – p.7/31 and (proof in book) 2 crd (α/2) = R 2R crd(180◦ α) • − −

How did he compute chords? Note

• crd(60◦) = R = 34380 (equilateral triangle) from right-angle triangle 2 2 crd(180◦ α) = 4R crd (α) • − − p

Development of trigonometryHipparchus & Ptolemy – p.7/31 How did he compute chords? Note

• crd(60◦) = R = 34380 (equilateral triangle) from right-angle triangle 2 2 crd(180◦ α) = 4R crd (α) • − − and (proof in book) p 2 crd (α/2) = R 2R crd(180◦ α) • − −

Development of trigonometryHipparchus & Ptolemy – p.7/31 • From crd(120◦) we get crd(30◦) and crd(180◦ 30◦) − • From crd(150◦) we get crd(15◦) and crd(180◦ 15◦) − • From crd(165◦) we get crd(7.5◦) and crd(180◦ 7.5◦) − • Smallest angle is 7.5◦

Table constructon So, get crd(60◦), crd(120◦), then

Development of trigonometryHipparchus & Ptolemy – p.8/31 • From crd(150◦) we get crd(15◦) and crd(180◦ 15◦) − • From crd(165◦) we get crd(7.5◦) and crd(180◦ 7.5◦) − • Smallest angle is 7.5◦

Table constructon So, get crd(60◦), crd(120◦), then

• From crd(120◦) we get crd(30◦) and crd(180◦ 30◦) −

Development of trigonometryHipparchus & Ptolemy – p.8/31 • From crd(165◦) we get crd(7.5◦) and crd(180◦ 7.5◦) − • Smallest angle is 7.5◦

Table constructon So, get crd(60◦), crd(120◦), then

• From crd(120◦) we get crd(30◦) and crd(180◦ 30◦) − • From crd(150◦) we get crd(15◦) and crd(180◦ 15◦) −

Development of trigonometryHipparchus & Ptolemy – p.8/31 Table constructon So, get crd(60◦), crd(120◦), then

• From crd(120◦) we get crd(30◦) and crd(180◦ 30◦) − • From crd(150◦) we get crd(15◦) and crd(180◦ 15◦) − • From crd(165◦) we get crd(7.5◦) and crd(180◦ 7.5◦) − • Smallest angle is 7.5◦

Development of trigonometryHipparchus & Ptolemy – p.8/31 Ptolemy 100–178 CE

• His Almagest, like Euclid’s Elements, made all other texts obsolete

• Influenced astronomy for a thousand years

• Starts with plane and spherical trigonometry

• Chord tables based on circle of radius 60 . . . 1◦ • . . . in increments of 2

Development of trigonometryHipparchus & Ptolemy – p.9/31 1◦ How to get to 2

• Establish crd(36◦) from geometry Establish 120 crd(α β) = • − crd(α) crd(180◦ β) crd(β) crd(180◦ α) − − − • Repeated applications of the above to get 1◦ crd(12 )

• Interpolate to get crd(1◦)

Key tools: Basic geometry and square root extractions

Development of trigonometryHipparchus & Ptolemy – p.10/31 Plane triangle problems “Calculate the length of the shadow of a pole 60 units high at Rhodes (latitude 36◦) at the vernal equinox”

Development of trigonometryHipparchus & Ptolemy – p.11/31 Preamble Angle from the center and from the circumference.

α

2α

Development of trigonometryHipparchus & Ptolemy – p.12/31 Shadow

B A

36

E

36

72 54 C F

Development of trigonometryHipparchus & Ptolemy – p.13/31 • Angle at center is twice 36◦, hence CF = crd(72◦) = 70;32,2

Then CE = crd(180◦ 72◦) = 97;4,56 • − • Now, since problem requires CE = 60 CF 70;32,3 shadow = = CE 97;4,56 60

So shadow is 60 70;32,3 = 43;36 97;4,56 · Can you solve this by modern methods?

Shadow

• Consider CE as the chord of a circle

Development of trigonometryHipparchus & Ptolemy – p.14/31 Then CE = crd(180◦ 72◦) = 97;4,56 • − • Now, since problem requires CE = 60 CF 70;32,3 shadow = = CE 97;4,56 60

So shadow is 60 70;32,3 = 43;36 97;4,56 · Can you solve this by modern methods?

Shadow

• Consider CE as the chord of a circle

• Angle at center is twice 36◦, hence CF = crd(72◦) = 70;32,2

Development of trigonometryHipparchus & Ptolemy – p.14/31 • Now, since problem requires CE = 60 CF 70;32,3 shadow = = CE 97;4,56 60

So shadow is 60 70;32,3 = 43;36 97;4,56 · Can you solve this by modern methods?

Shadow

• Consider CE as the chord of a circle

• Angle at center is twice 36◦, hence CF = crd(72◦) = 70;32,2

Then CE = crd(180◦ 72◦) = 97;4,56 • −

Development of trigonometryHipparchus & Ptolemy – p.14/31 Can you solve this by modern methods?

Shadow

• Consider CE as the chord of a circle

• Angle at center is twice 36◦, hence CF = crd(72◦) = 70;32,2

Then CE = crd(180◦ 72◦) = 97;4,56 • − • Now, since problem requires CE = 60 CF 70;32,3 shadow = = CE 97;4,56 60

So shadow is 60 70;32,3 = 43;36 97;4,56 ·

Development of trigonometryHipparchus & Ptolemy – p.14/31 Shadow

• Consider CE as the chord of a circle

• Angle at center is twice 36◦, hence CF = crd(72◦) = 70;32,2

Then CE = crd(180◦ 72◦) = 97;4,56 • − • Now, since problem requires CE = 60 CF 70;32,3 shadow = = CE 97;4,56 60

So shadow is 60 70;32,3 = 43;36 97;4,56 · Can you solve this by modern methods?

Development of trigonometryHipparchus & Ptolemy – p.14/31 Abstractly

• Given a side b and angle α of a right triangle, . . .

• . . . find the other side, a

• crd(2α) 2R sin α a = b = b = b tan α crd(180 2α) 2R cos α ◦ −

Development of trigonometryHipparchus & Ptolemy – p.15/31 Length of seasons

• Observation: Seasons are of different length

• Model assumption: Earth is fixed at the center

• Model assumption: Everything on spheres

• Solution: Eccentric circles Summer Spring Sun

Earth

Fall Winter

Development of trigonometryHipparchus & Ptolemy – p.16/31 This is the bread and butter of applied mathematicians

Find parameters of model

• Distance of Earth to center of Sun’s circle

• Angles of the triangle Earth-Center-Sun

Development of trigonometryHipparchus & Ptolemy – p.17/31 Find parameters of model

• Distance of Earth to center of Sun’s circle

• Angles of the triangle Earth-Center-Sun

This is the bread and butter of applied mathematicians

Development of trigonometryHipparchus & Ptolemy – p.17/31 360◦ • Velocity ν = 365;14,48 = 0◦590080017000 per day

• Spring length is 94.5 days, summer is 92.5

• Then 90 + θ + τ = 94.5ν, and 90 + θ τ = 92.5ν − • Solve for θ = 2◦100 and τ = 0◦590

Find θ and τ

• Consider the angles τ and θ

τ

D θ

E

Development of trigonometryHipparchus & Ptolemy – p.18/31 • Spring length is 94.5 days, summer is 92.5

• Then 90 + θ + τ = 94.5ν, and 90 + θ τ = 92.5ν − • Solve for θ = 2◦100 and τ = 0◦590

Find θ and τ

• Consider the angles τ and θ 360◦ • Velocity ν = 365;14,48 = 0◦590080017000 per day

τ

D θ

E

Development of trigonometryHipparchus & Ptolemy – p.18/31 • Then 90 + θ + τ = 94.5ν, and 90 + θ τ = 92.5ν − • Solve for θ = 2◦100 and τ = 0◦590

Find θ and τ

• Consider the angles τ and θ 360◦ • Velocity ν = 365;14,48 = 0◦590080017000 per day

• Spring length is 94.5 days, summer is 92.5

τ

D θ

E

Development of trigonometryHipparchus & Ptolemy – p.18/31 • Solve for θ = 2◦100 and τ = 0◦590

Find θ and τ

• Consider the angles τ and θ 360◦ • Velocity ν = 365;14,48 = 0◦590080017000 per day

• Spring length is 94.5 days, summer is 92.5

• Then 90 + θ + τ = 94.5ν, and 90 + θ τ = 92.5ν −

τ

D θ

E

Development of trigonometryHipparchus & Ptolemy – p.18/31 Find θ and τ

• Consider the angles τ and θ 360◦ • Velocity ν = 365;14,48 = 0◦590080017000 per day

• Spring length is 94.5 days, summer is 92.5 Then 90 + θ + τ = 94.5ν, and 90 + θ τ = 92.5ν • − • Solve for θ = 2◦100 and τ = 0◦590

τ

D θ

E

Development of trigonometryHipparchus & Ptolemy – p.18/31 • Assume DX = 60 1 1 • LE = OV = 2 crd(2θ) = 2 crd(4◦200) = 2;16 1 1 • DL = 2 crd(2τ) = 2 crd(1◦580) = 1;2 And DE = √LE2 + DL2 = 2;29,30 21 • ≈ 2

Find DE

w

τ

X L O D θ

E v

Development of trigonometryHipparchus & Ptolemy – p.19/31 1 1 • LE = OV = 2 crd(2θ) = 2 crd(4◦200) = 2;16 1 1 • DL = 2 crd(2τ) = 2 crd(1◦580) = 1;2 And DE = √LE2 + DL2 = 2;29,30 21 • ≈ 2

Find DE

• Assume DX = 60

w

τ

X L O D θ

E v

Development of trigonometryHipparchus & Ptolemy – p.19/31 1 1 • DL = 2 crd(2τ) = 2 crd(1◦580) = 1;2 And DE = √LE2 + DL2 = 2;29,30 21 • ≈ 2

Find DE

• Assume DX = 60 1 1 • LE = OV = 2 crd(2θ) = 2 crd(4◦200) = 2;16

w

τ

X L O D θ

E v

Development of trigonometryHipparchus & Ptolemy – p.19/31 And DE = √LE2 + DL2 = 2;29,30 21 • ≈ 2

Find DE

• Assume DX = 60 1 1 • LE = OV = 2 crd(2θ) = 2 crd(4◦200) = 2;16 1 1 • DL = 2 crd(2τ) = 2 crd(1◦580) = 1;2

w

τ

X L O D θ

E v

Development of trigonometryHipparchus & Ptolemy – p.19/31 Find DE

• Assume DX = 60 1 1 • LE = OV = 2 crd(2θ) = 2 crd(4◦200) = 2;16 1 1 • DL = 2 crd(2τ) = 2 crd(1◦580) = 1;2 And DE = √LE2 + DL2 = 2;29,30 21 • ≈ 2

w

τ

X L O D θ

E v

Development of trigonometryHipparchus & Ptolemy – p.19/31 • Circumscribe LDE in a circle LD = 1;2 = ? ? = 49;46 • DE 2;29,30 120 → • From tables (in reverse), 1 crd− (49;46) = 49◦ := α

49◦ • Therefore LED = 2 = 24◦300

Find angle LED

L D

α

α/2

E

Development of trigonometryHipparchus & Ptolemy – p.20/31 LD = 1;2 = ? ? = 49;46 • DE 2;29,30 120 → • From tables (in reverse), 1 crd− (49;46) = 49◦ := α

49◦ • Therefore LED = 2 = 24◦300

Find angle LED

• Circumscribe LDE in a circle

L D

α

α/2

E

Development of trigonometryHipparchus & Ptolemy – p.20/31 • From tables (in reverse), 1 crd− (49;46) = 49◦ := α

49◦ • Therefore LED = 2 = 24◦300

Find angle LED

• Circumscribe LDE in a circle LD = 1;2 = ? ? = 49;46 • DE 2;29,30 120 →

L D

α

α/2

E

Development of trigonometryHipparchus & Ptolemy – p.20/31 49◦ • Therefore LED = 2 = 24◦300

Find angle LED

• Circumscribe LDE in a circle LD = 1;2 = ? ? = 49;46 • DE 2;29,30 120 → • From tables (in reverse), 1 crd− (49;46) = 49◦ := α

L D

α

α/2

E

Development of trigonometryHipparchus & Ptolemy – p.20/31 Find angle LED

• Circumscribe LDE in a circle LD = 1;2 = ? ? = 49;46 • DE 2;29,30 120 → • From tables (in reverse), 1 crd− (49;46) = 49◦ := α

49◦ • Therefore LED = 2 = 24◦300

L D

α

α/2

E

Development of trigonometryHipparchus & Ptolemy – p.20/31 Parameters are Earth–Center : 2;29,30 2 1 • ≈ 2 • Angle Center-Earth-Vertical : 24◦300

Summer Spring Sun

C

Earth

Fall Winter

Development of trigonometryHipparchus & Ptolemy – p.21/31 Where will the Sun be at time T?

• We know the angular speed of the Sun. . .

• so we can reformulate time as angle P DS

• Given angle P DS what is angle DES?

S

P

θ

D E

Development of trigonometryHipparchus & Ptolemy – p.22/31 • If DE = 120, then DK = crd(120◦) = 103;55

• But DE = 2;30, so DE 2;30 120 = = ? = 2;10 DK ? 103;55 →

Specific problem: P DS = 30◦

S

D P 30 θ 120 D E

K

E K

Development of trigonometryHipparchus & Ptolemy – p.23/31 • But DE = 2;30, so DE 2;30 120 = = ? = 2;10 DK ? 103;55 →

Specific problem: P DS = 30◦

• If DE = 120, then DK = crd(120◦) = 103;55

S

D P 30 θ 120 D E

K

E K

Development of trigonometryHipparchus & Ptolemy – p.23/31 Specific problem: P DS = 30◦

• If DE = 120, then DK = crd(120◦) = 103;55

• But DE = 2;30, so DE 2;30 120 = = ? = 2;10 DK ? 103;55 →

S

D P 30 θ 120 D E

K

E K

Development of trigonometryHipparchus & Ptolemy – p.23/31 • SK = SD + DK = 60 + 2;10 = 62;10

• Since angle KDE = 30◦, then 1 2;10 EK = 2DE = 2 = 1;15 √ 2 2 • So SE = EK + SK = 62;11

Find SE

S

P

θ

D E

K

Development of trigonometryHipparchus & Ptolemy – p.24/31 • Since angle KDE = 30◦, then 1 2;10 EK = 2DE = 2 = 1;15 √ 2 2 • So SE = EK + SK = 62;11

Find SE

• SK = SD + DK = 60 + 2;10 = 62;10

S

P

θ

D E

K

Development of trigonometryHipparchus & Ptolemy – p.24/31 √ 2 2 • So SE = EK + SK = 62;11

Find SE

• SK = SD + DK = 60 + 2;10 = 62;10

• Since angle KDE = 30◦, then 1 2;10 EK = 2DE = 2 = 1;15

S

P

θ

D E

K

Development of trigonometryHipparchus & Ptolemy – p.24/31 Find SE

• SK = SD + DK = 60 + 2;10 = 62;10

• Since angle KDE = 30◦, then 1 2;10 EK = 2DE = 2 = 1;15 √ 2 2 • So SE = EK + SK = 62;11

S

P

θ

D E

K

Development of trigonometryHipparchus & Ptolemy – p.24/31 • KE 1;15 ? = = ? = 2;25 SE 62;11 120 →

1 • Use table in reverse to find crd− (2;25) = 2◦180

• Therefore KSE = 1◦190 and DES = 180◦ 150◦ 1◦90 = 28◦510 − −

Finally angle DES

S S

P D θ

D E

K

DeKvelopment of trigonometryHipparchus & Ptolemy – p.25/31 E 1 • Use table in reverse to find crd− (2;25) = 2◦180

• Therefore KSE = 1◦190 and DES = 180◦ 150◦ 1◦90 = 28◦510 − −

Finally angle DES

• KE 1;15 ? = = ? = 2;25 SE 62;11 120 →

S S

P D θ

D E

K

DeKvelopment of trigonometryHipparchus & Ptolemy – p.25/31 E • Therefore KSE = 1◦190 and DES = 180◦ 150◦ 1◦90 = 28◦510 − −

Finally angle DES

• KE 1;15 ? = = ? = 2;25 SE 62;11 120 → 1 • Use table in reverse to find crd− (2;25) = 2◦180

S S

P D θ

D E

K

K E Development of trigonometryHipparchus & Ptolemy – p.25/31 Summary

• Chords are as powerful as trig functions

• but they require more steps Hidden in these procedures

• Law of chordal cosines

• Law of chordal sines

Development of trigonometryHipparchus & Ptolemy – p.26/31 • Consider triangle with a, b, γ given

• Given a table of chords based on circle radius R 180◦ = 180◦ γ + 90◦ + γ hence γ = γ 90◦ • − 1 1 − γ = 180◦ 2γ = 360◦ 2γ • 2 − 1 −

Cosine chord law

α γ1

h γ2 b c

2γ1

p 180−γ γ β

a

Development of trigonometryHipparchus & Ptolemy – p.27/31 180◦ = 180◦ γ + 90◦ + γ hence γ = γ 90◦ • − 1 1 − γ = 180◦ 2γ = 360◦ 2γ • 2 − 1 −

Cosine chord law

• Consider triangle with a, b, γ given

• Given a table of chords based on circle radius R

α γ1

h γ2 b c

2γ1

p 180−γ γ β

a

Development of trigonometryHipparchus & Ptolemy – p.27/31 γ = 180◦ 2γ = 360◦ 2γ • 2 − 1 −

Cosine chord law

• Consider triangle with a, b, γ given

• Given a table of chords based on circle radius R 180◦ = 180◦ γ + 90◦ + γ hence γ = γ 90◦ • − 1 1 −

α γ1

h γ2 b c

2γ1

p 180−γ γ β

a

Development of trigonometryHipparchus & Ptolemy – p.27/31 Cosine chord law

• Consider triangle with a, b, γ given

• Given a table of chords based on circle radius R 180◦ = 180◦ γ + 90◦ + γ hence γ = γ 90◦ • − 1 1 − γ = 180◦ 2γ = 360◦ 2γ • 2 − 1 −

α γ1

h γ2 b c

2γ1

p 180−γ γ β

a

Development of trigonometryHipparchus & Ptolemy – p.27/31 b b p = crd(2γ ) = crd(2γ 180◦) • 2R 1 2R − b b h = crd(γ ) = crd(360◦ 2γ) • 2R 2 2R − 2 2 2 • c = h + (a + p)

Cosine chord law

α γ1

h γ2 b c

2γ1

p 180−γ γ β

a

Development of trigonometryHipparchus & Ptolemy – p.28/31 b b h = crd(γ ) = crd(360◦ 2γ) • 2R 2 2R − 2 2 2 • c = h + (a + p)

Cosine chord law b b p = crd(2γ ) = crd(2γ 180◦) • 2R 1 2R −

α γ1

h γ2 b c

2γ1

p 180−γ γ β

a

Development of trigonometryHipparchus & Ptolemy – p.28/31 2 2 2 • c = h + (a + p)

Cosine chord law b b p = crd(2γ ) = crd(2γ 180◦) • 2R 1 2R − b b h = crd(γ ) = crd(360◦ 2γ) • 2R 2 2R −

α γ1

h γ2 b c

2γ1

p 180−γ γ β

a

Development of trigonometryHipparchus & Ptolemy – p.28/31 Cosine chord law b b p = crd(2γ ) = crd(2γ 180◦) • 2R 1 2R − b b h = crd(γ ) = crd(360◦ 2γ) • 2R 2 2R − 2 2 2 • c = h + (a + p)

α γ1

h γ2 b c

2γ1

p 180−γ γ β

a

Development of trigonometryHipparchus & Ptolemy – p.28/31 a2 + b2 2ab cos γ −

Cosine chord law

c2 = h2 + (a + p)2 b 2 b 2 = crd(360◦ 2γ) + a + crd(2γ 180◦) 2R − 2R −     crd(2γ 180◦) = a2 + b2 + 2ab − 2R =

Development of trigonometryHipparchus & Ptolemy – p.29/31 Cosine chord law

c2 = h2 + (a + p)2 b 2 b 2 = crd(360◦ 2γ) + a + crd(2γ 180◦) 2R − 2R −     crd(2γ 180◦) = a2 + b2 + 2ab − 2R = a2 + b2 2ab cos γ −

Development of trigonometryHipparchus & Ptolemy – p.29/31 Stereographic Projections

• Points on sphere to points on plane of equator

• Great circles are mapped to?

• Circles are mapped to?

N

P

P’

S

Development of trigonometryHipparchus & Ptolemy – p.30/31 Menelaus’s theorem

sin n sin s sin r = 2 sin n1 sin s1 · sin r sin n sin s sin m 2 = 2 2 sin n1 sin s1 · sin m

A

m1 n1

D F E s1 r1 m2 n2

r2 s2

B C

Development of trigonometryHipparchus & Ptolemy – p.31/31