Pythagoras theorem trigonometry
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Pythagoras’ Theorem & Trigonometry. 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.

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Pythagoras’ Theorem & Trigonometry

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Pythagoras’ Theorem & Trigonometry


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


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


Getting to the “Point”

“Something Interesting”

Dad & Son


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


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


A Geometrical Proof

Most geometrical proofs revolve around the concept of

“Preservation of Area”


Class Activity

How many similar triangles can you see in the above triangle???

Use them to prove the Pythagoras’ Theorem again!


How to interest students with Pythagoras Theorem


8 x 8 squares

= 64 squares

Challenge Their Minds


2

2

3

1

4

1

4

3

Challenge Their Minds

13 x 5 squares

= 65 squares ?


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


h

5

2

3

4

1

13

Using Pythagoras Theorem

3

h= (52 + 132)

= (25+ 169)

= (194)

= 13.9283 units


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


h

y

x

Pythagorean Triplets

  • 3 special integers

  • Form the sides of right-angled triangle

  • Example: 3, 4 & 5

  • Non-example: 5, 6 & √61


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


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


Trick for Teachers

  • Why share this trick?

  • Can use this to set questions on Pythagoras Theorem with ease


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


Meaning of “Sine”, “Cosine” & “Tangent”

  • Sine – From half chord to bosom/bay/curve

  • Cosine – Co-Sine, sine of the complementary

  • angle

  • Tangent – to touch


Sine

Tangent

Cosine

The Story of 3 Friends


sin 

A (1,0)

cos 

Formal Definition of Sine and Cosine

1

Unit circle


Some Results from Definition

  • Definition of tan :

    sin 

    cos 

  • Pythagorean Identity:

  • sin2 + cos2 = 1


`

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


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


Side

Tide

Coside

Invasion by King Square!


side 

coside 

Extension to Non-Circular Functions

A (1,0)

Unit Square


Some Results from definition

  • Tide  = side  /coside 

  • BUT is side2 + coside2  = 1 ?


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


Comparison of other theorems

Circular FunctionSquare Function

Complementary Thm

Supplementary Thm

Half Turn Thm

Opposites Thm

AGREES !!


Comparison of Sine and Side Functions


Comparison of Cosine and Coside Functions


Comparison of Tan and Tide Functions


Further Extensions…

(0,1)

(0,1)

(1,0)

(1,0)

Hexagon

Diamond


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


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


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


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


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


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


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


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


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


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