1 / 43

# Pythagoras’ Theorem &amp; Trigonometry

Pythagoras’ Theorem &amp; Trigonometry. Our Presenters &amp; Objectives. Proving the theorem The Chinese Proof Preservation of Area – Applet Demo Class Activity – Proving the theorem using Similar Triangles. Boon Kah. Beng Huat. Applying the theorem Solving an Eye Trick Pythagorean Triplets. Download Presentation ## Pythagoras’ Theorem &amp; Trigonometry

E N D

### Presentation Transcript

1. Pythagoras’ Theorem & Trigonometry

2. Our Presenters & Objectives • Proving the theorem • The Chinese Proof • Preservation of Area – Applet Demo • Class Activity – Proving the theorem using Similar Triangles Boon Kah Beng Huat • Applying the theorem • Solving an Eye Trick • Pythagorean Triplets

3. Our Presenters & Objectives • Fundamentals of Trigonometry • Appreciate the definition of basic trigonometry functions from a circle • Apply the definition of basic trigonometry functions from a circle to a square. Lawrence Tang Keok Wen • The derivation of the double-angle formula

4. Getting to the “Point” “Something Interesting” Dad & Son

5. The Pythagoras Theorem The square described upon the hypotenuse of a right-angled triangle is equal to the sum of the squares described upon the other two sides. • Or algebraically speaking…… h2 = a2 + b2 h b a

6. The “Chinese” Proof b a b h a h 4(1/2 ab) + h2 = (a + b)2 2ab + h2 = a2 + 2ab + b2 h h2 = a2 + b2 a h b This proof appears in the Chou pei suan ching, a text dated anywhere from the time of Jesus to a thousand years earlier a b

7. A Geometrical Proof Most geometrical proofs revolve around the concept of “Preservation of Area”

8. Class Activity How many similar triangles can you see in the above triangle??? Use them to prove the Pythagoras’ Theorem again!

9. How to interest students with Pythagoras Theorem

10. 8 x 8 squares = 64 squares Challenge Their Minds

11. 2 2 3 1 4 1 4 3 Challenge Their Minds 13 x 5 squares = 65 squares ?

12. 8 h1 2 2 3 1 1 h2 4 4 3 3 5 2 Using Pythagoras Theorem h1 = (32 + 82) = (9+ 64) = (73) h2 = (22 + 52) = (4+ 25) = (29) h1 + h2 = (73 + 29) = 13.9292 units

13. h 5 2 3 4 1 13 Using Pythagoras Theorem 3 h= (52 + 132) = (25+ 169) = (194) = 13.9283 units

14. h h1 2 2 3 1 4 1 h2 4 3 h1 + h2 = 13.9292 units h = 13.9283 units Using Pythagoras Theorem h ≠h1 + h2

15. h y x Pythagorean Triplets • 3 special integers • Form the sides of right-angled triangle • Example: 3, 4 & 5 • Non-example: 5, 6 & √61

16. Trick for Teachers • Give me an odd number, except 1 (small value) • Form a Pythagorean Triplet • Form a right-angled triangle where sides are integers

17. Trick for Teachers • Shortest side = n • The other side = (n2 – 1)  2 • Hypotenuse = [(n2 – 1)  2] + 1 • For e.g., if n = 2 • Shortest side = 5 • The other side = 12 • Hypotenuse = 13

18. Trick for Teachers • Why share this trick? • Can use this to set questions on Pythagoras Theorem with ease

19. Trigonometry • Meaning of Sine,Cosine & Tangent • Formal Definition of Sine,Cosine and Tangent based on a unit circle • Extension to the unit square • Double Angle Formula

20. Meaning of “Sine”, “Cosine” & “Tangent” • Sine – From half chord to bosom/bay/curve • Cosine – Co-Sine, sine of the complementary • angle • Tangent – to touch

21. Sine Tangent Cosine The Story of 3 Friends

22. sin   A (1,0) cos  Formal Definition of Sine and Cosine 1 Unit circle

23. Some Results from Definition • Definition of tan : sin  cos  • Pythagorean Identity: • sin2 + cos2 = 1

24. ` slant length Opposite length 1 sin  cos  adjacent length Common Definition of Sine, Cosine & Tangent What happens if slant edge  1? By principal of similar triangles, (Sin )/ 1 = opposite/slant length (Cos )/1 = adjacent/slant length (Sin ) /(Cos ) = opposite/adjacent length For visual students

25. hypotenuse opposite adjacent Therefore for a given angle  in ANY right angled triangle, Opposite Length • sin = Hypotenuse Adjacent Length • cos  = Hypotenuse Opposite Length • tan = Adjacent Length 

26. Side Tide Coside Invasion by King Square!

27. side   coside  Extension to Non-Circular Functions A (1,0) Unit Square

28. Some Results from definition • Tide  = side  /coside  • BUT is side2 + coside2  = 1 ?

29. side   Corresponding Pythagorean Thm: side2+ coside2  = sec2  coside  Corresponding Pythagorean Thm: side2+ coside2  = cosec2  Pythagorean Theorem for Square Function For 0 <  < 45 coside  =1 side  = tan  tide  = tan For 45 <  < 90 side  = 1 coside  =cot  tide  = tan

30. Comparison of other theorems Circular FunctionSquare Function Complementary Thm Supplementary Thm Half Turn Thm Opposites Thm AGREES !!

31. Comparison of Sine and Side Functions

32. Comparison of Cosine and Coside Functions

33. Comparison of Tan and Tide Functions

34. Further Extensions… (0,1) (0,1) (1,0) (1,0) Hexagon Diamond

35. References • http://www.arcytech.org/java/pythagoras/history.html • http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Pythagoras.html • http://www.ies.co.jp/math/products/geo2/applets/pytha2/pytha2.html • The teaching of trigonometry in schools London G Bell & Sons, Ltd • Functions, Statistics & Trigonometry, Intergrated Mathematics 2nd Edition, University of Chicago School Math Project

36. 1 o a 1 = 2(o)/2 = o = sin  o = 2(o)/2(a) = o/a = tan  1 = 2(a)/2 = a = cos  o a a 1 = 3(o)/3 = o = sin  o = 3(a)/3 = a = cos  = 3(o)/3(a) = o/a = tan  1 o 1 o a a a = x(o)/x(1) = o = sin  = x(a)/x(1) = a = cos  = x(o)/x(a) = o/a = tan  x x(o) x(a) Sine, Cosine & Tangent Opposite Length Slant length Adjacent Length Slant length Opposite Length Adjacent length o defined as sin  a defined as sin  o/a defined as tan  For an angle , Return

37. side (90-) Unit Square coside (90-) Comparison of Complementary Theorems Square Function Circular Function For 0 <  < 90 For 0 <  < 45 sin(90 - ) = cos  side(90 - ) = coside  cos(90 - ) = sin  coside(90 - ) = side  tide(90 - ) = cotide  tan(90 - ) = cot  Return

38. side (90+) Unit Square coside (90+) Comparison of functions of (90 + ) Square Function Circular Function For 0 <  < 90 For 0 <  < 45 sin(90+ ) = cos  side(90 + ) = coside  cos(90+ ) = -sin  coside(90 + ) = -side  tan(90+ ) = -cot  tide(90 + ) = -cotide  Return

39. side (180-)  Unit Square coside (180-) Comparison of Supplement Theorems Square Function Circular Function For 0 <  < 90 For 0 <  < 45 side(180 - ) = side  sin(180 - ) = sin  coside(180 - ) = -coside  cos(180 - ) = -cos tide(180 - ) = -tide  tan(180 - ) = -tan  Return

40. side (180+) Unit Square coside (180+) Comparison of ½ Turn Theorems Square Function Circular Function For 0 <  < 90 For 0 <  < 45 side(180 + ) = - side  sin(180 + ) = - sin  coside(180 + ) = - coside  cos(180 + ) = - cos tide(180 + ) = tide  tan(180 + ) = tan  Return

41. coside (270-) side (270-)  Unit Square Comparison of Functions of (270 - ) Square Function Circular Function For 0 <  < 90 For 0 <  < 45 side(270 - ) = - coside  sin (270-) =-cos  cos(270-) = -side  coside(270 - ) = - side  tide(270 - ) = cotide  tan (270-) = cot  Return

42. side (180-) Unit Square coside (270+) Square Function Circular Function For 0 <  < 90 For 0 <  < 45 Comparison of Functions of (270 + ) side (270+ )= - coside  sin(270+ )= - cos  coside (270+ ) = side  cos(270+  ) = sin  tide (270+ )= - cotide  tan(270+) = - tan  Return

43. Square Function Circular Function For 0 <  < 90 For 0 <  < 45 Comparison of Opposite Theorems side(- ) = - side  sin(- ) = - sin  cos(- ) = cos  coside(- ) = coside  tan(- ) = - tan  tide(- ) = - tide  side (-)  Unit Square coside (-) Return

More Related