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 = pLE2 + 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 = pLE2 + 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