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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
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  • Thales of Miletus Sources and Interpretations Miletli Thales Kaynaklar Ve Yorumlar

    Thales of Miletus Sources and Interpretations Miletli Thales Kaynaklar Ve Yorumlar

    Thales of Miletus Sources and Interpretations Miletli Thales Kaynaklar ve Yorumlar David Pierce October , Matematics Department Mimar Sinan Fine Arts University Istanbul http://mat.msgsu.edu.tr/~dpierce/ Preface Here are notes of what I have been able to find or figure out about Thales of Miletus. They may be useful for anybody interested in Thales. They are not an essay, though they may lead to one. I focus mainly on the ancient sources that we have, and on the mathematics of Thales. I began this work in preparation to give one of several - minute talks at the Thales Meeting (Thales Buluşması) at the ruins of Miletus, now Milet, September , . The talks were in Turkish; the audience were from the general popu- lation. I chose for my title “Thales as the originator of the concept of proof” (Kanıt kavramının öncüsü olarak Thales). An English draft is in an appendix. The Thales Meeting was arranged by the Tourism Research Society (Turizm Araştırmaları Derneği, TURAD) and the office of the mayor of Didim. Part of Aydın province, the district of Didim encompasses the ancient cities of Priene and Miletus, along with the temple of Didyma. The temple was linked to Miletus, and Herodotus refers to it under the name of the family of priests, the Branchidae. I first visited Priene, Didyma, and Miletus in , when teaching at the Nesin Mathematics Village in Şirince, Selçuk, İzmir. The district of Selçuk contains also the ruins of Eph- esus, home town of Heraclitus. In , I drafted my Miletus talk in the Math Village. Since then, I have edited and added to these notes.