# Pythagoras’ Theorem & Trigonometry - PowerPoint PPT Presentation

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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”

### 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!

8 x 8 squares

= 64 squares

2

2

3

1

4

1

4

3

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

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

• 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

sin 

A (1,0)

cos 

1

Unit circle

### Some Results from Definition

• Definition of tan :

sin 

cos 

• Pythagorean Identity:

• sin2 + cos2 = 1

`

slant length

Opposite

length

1

sin 

cos 

### Common Definition of Sine, Cosine & Tangent

What happens if slant edge  1?

By principal of similar triangles,

(Sin )/ 1 = opposite/slant length

(Sin ) /(Cos ) = opposite/adjacent length

For visual students

hypotenuse

opposite

Therefore for a given angle  in ANY right

angled triangle,

Opposite Length

• sin = Hypotenuse

• cos  = Hypotenuse

Opposite Length

Side

Tide

Coside

side 

coside 

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 

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 !!

(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

Slant length

Opposite 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