sin x over_x
TRANSCRIPT
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By Ron English
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EASY PROOF BUT
Only after you learn Derivatives and l’Hopital’s Rule
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l'Hôpital's Rule
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PROOF using
MATHEMATICS and
LIMITS!!!
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You will need to know …⋆Squeeze theorem⋆Also known as the pinching theorem
⋆Area of arcs ⋆Area of an Arc of radius 1
⋆Area of triangles
⋆Trig functions
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Squeeze Theorem
A B C≤ ≤if A C: =
∴ = =B A B C,
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Also
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TRIG of TRIANGLE
Adjacent
Opposite
Hypotenuse
x
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MORE trig of a triangle
Adjacent
Opposite
Hypotenuse
x
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AREA of a TRIANGLE RIGHT ANGLE TRIANGLE
HEIGHT
BASE
AREA =1/2 BASE * HEIGHT
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AREA of an ARC
θ
RadiusArea = ½ x radius x θ
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Start the PROOF
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An Arc Radius 1
X
1
1
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Draw lines from two intercepts
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Add labels
XO A
B
C
D
Radius of Circle = 1 = OC = OB
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THREE AREAS
XO A
B
C
D
Radius of Circle = 1 = OC = OB
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SMALLEST AREA
XO A
B
C
D
Area = ½ base * height = ½ OA * AB
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SMALLEST AreaArea = ½ base * height = ½ OA * AB
OA
BSin x = Opp / Hyp
= AB / OB
but: OB = 1
(radius of circle)
∴ Sin x = AB / 1
OR AB = Sin x
X
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SMALLEST AreaArea = ½ base * height = ½ OA * AB
OA
BCos x = Adj / Hyp
= OA / OB
but: OB = 1
(radius of circle)
∴ Cos x = OA / 1
∴ OA = Cos x
X
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SMALLEST AreaArea = ½ base * height = ½ AB * OA
OA
BAB = Sin x
OA = Cos x
X
Area = ½ Sin x * Cos x
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½ Sinx*Cosx
Start the SQUEEZE Equation
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MIDDLE Area
XO
B
CArea = ½ Radius * x
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MIDDLE Area
XO
B
C
Area = ½ Radius * xRadius = 1
Area = ½ x
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½ Sinx*Cosx <=½ x
Middle of SQUEZE Equation
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LARGEST Area
XO
C
D
Area = ½ base * height = ½ OC * CD
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LARGEST Area
XO C
DArea = ½ base * height = ½ OC * CD
OC = 1 (Radius of Circle)
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LARGEST Area
XO
C
DArea = ½ base * height = ½ OC * CD
TAN x = DC / OC
since OC = 1
∴ TAN x = DC
but TAN x =Sin x / Cos x
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LARGEST Area
XO
C
DArea = ½ base * height = ½ OC * CD
DC = TAN x
DC = Sin x / Cos x
Area = ½ Sin x / Cos x
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Complete SQUEEZE Equation
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Some Arithmetic
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Some more Arithmetic
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Even MORE Arithmetic
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a LIMIT
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Take LIMITS a X approaches Zero
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QED
Latin for
quod erat demonstrandum
(which was to be demonstrated“)