Transcript The Language of Science I - Biblical Christian World View

Derivative of the Sine Function
By James Nickel, B.A., B.Th., B.Miss., M.A.
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Basic Trigonometry
• Given the unit circle
(radius = 1) with angle
 (theta), then, by
definition, PA = sin 
and OA = cos .
P
1
sin 
O

cos 
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A
Basic Trigonometry
• As  changes from 0 to
/2 radians (90), sin
also changes from 0 to 1.
• As  changes from 0 to
/2 radians, cos also
changes from 1 to 0.
P
1
sin 
O

cos 
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A
Question
• How does sin vary as  varies?
• To find out, we must calculate the
derivative of y = sin.
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The Method of Increments
• We add  to  and
determine what
happens.
• We need to find the
change in y or y.
y
1

P
sin 

O
cos 
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The Method of Increments
• PA has increased to QB
where QB = sin( + ).
• OA has decreased to OB
where OB = cos( + ).
•  also represents the
radian measure of the arc
from P to Q.

Q
P
1
1


O
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B
A
The Method of Increments
• Also remember that 
is just a “little bit” or
infinitesimally small.

Q
P
1
1


O
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B
A
Our Goal
• Find QD = y
• To do this, we must
show QDP ~ PAO
• We then let   0
• Hence,
y DQ OA


  PO

Q
P
D
1
1


O
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B
A
Some Geometry
• Let’s now draw the
tangent line l to the circle
at point P.
• We know that l  PO (the
tangent line intersects the
circle’s radius at a right
angle).
• Hence, QPO = 90

Q
P
D
1
1
l


O
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A
Some Geometry
Construct DP  OA
Note the transversal OP
Hence,  = DPO
QPO = 90  DPO
and DPQ are
complementary.
• Hence, DPQ = 90 – 
•
•
•
•

Q
P
D
1
1
sin


O
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B
A
l
Some Geometry
• DPQ = 90 –   DQP
=
• Hence, QDP ~ PAO
because both are right
triangles and DQP =
POA = 

Q
P
D
1
1
l


O
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A
Some Geometry
• Consider DQ
• It represents the change
in y (i.e., y) resulting
from the change in  (i.e.,
).
y DQ sin       sin 


 


Q
P
D
1
1
sin


O
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A
l
Derivative Formula
y DQ sin       sin 


 

DQ
In QDP, cos  =

Note that as  gets
infinitesimally small, QP
(the hypotenuse)
converges to .

Q
P
D
1
1
sin


O
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A
l
Derivative Formula
• Because QDP ~ PAO,
then:
DQ OA

 PO

Q
P
D
1
1
sin


O
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B
A
l
Derivative Formula
• Since PO = 1 and OA =
cos , then:
y DQ


 
sin       sin 


OA cos 

 cos 
PO
1

Q
P
D
1
1
sin


O
B
OA=cos
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A
l
Derivative Formula
• As we let  approach 0 as a limit, then:
y (sin) = cos
• This means as  increases, sin increases at a
instantaneous rate of cos.
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Derivative Graph
y-axis




-axis
y = sin 
• The solid line graph represents y = sin (the
sine curve).
• The dotted line graph represents y = cos (the
cosine curve).
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Derivative Graph
y-axis



y = sin 

-axis
• Note that when  = 0, then cos = 1.
• This means that the slope of the line tangent to
the sine curve at 0 is 1.
• The cosine curve plots that derivative.
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Derivative Graph
y-axis




-axis
y = sin 
• When  = /2, then cos = 0.
• This means that the slope of the line tangent to
the sine curve at /2 is 0 (the slope is parallel
to the -axis).
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Derivative Graph
y-axis




-axis
y = sin 
• The cosine curve traces the derivative of the
sine curve at every point on the sine curve.
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