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Derivative of the Sine Function By James Nickel, B.A., B.Th., B.Miss., M.A. www.biblicalchristianworldview.net

Derivative of the Sine Function By James Nickel, B.A., B.Th., B.Miss., M.A

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Derivative of the Sine Function

By James Nickel, B.A., B.Th., B.Miss., M.A.www.biblicalchristianworldview.net

Basic Trigonometry

• Given the unit circle (radius = 1) with angle (theta), then, by definition, PA = sin and OA = cos . 1

P

O cos

sin

A

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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.

1

P

O cos

sin

A

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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.

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1

O

A

P

sin

cos

y

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.

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1

O

A

P

Q

B

1

The Method of Increments

• Also remember that is just a “little bit” or infinitesimally small.

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1

O

A

P

Q

B

1

Our Goal

• Find QD = y• To do this, we must

show QDP ~ PAO• We then let 0• Hence,

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y DQ OAPO

1

O

A

P

Q

B

1

D

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

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1

O

A

P

Q

B

1l

D

Some Geometry

• Construct DP OA• Note the transversal OP• Hence, = DPO QPO = 90 DPO and

DPQ are complementary.• Hence, DPQ = 90 –

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1

O

A

P

Q

B

1l

D

sin

Some Geometry

DPQ = 90 – DQP =

• Hence, QDP ~ PAO because both are right triangles and DQP = POA =

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1

O

A

P

Q

B

1l

D

Some Geometry

• Consider DQ• It represents the change

in y (i.e., y) resulting from the change in (i.e., ).

sin siny DQ

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1

O

A

P

Q

B

1l

D

sin

Derivative Formula

sin siny DQ

In QDP, cos = DQ

Note that as gets infinitesimally small, QP (the hypotenuse) converges to .

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1

O

A

P

Q

B

1l

D

sin

Derivative Formula

• Because QDP ~ PAO, then:

DQ OAPO

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1

O

A

P

Q

B

1l

D

sin

Derivative Formula• Since PO = 1 and OA =

cos , then:

y DQ

sin sin

OA coscos

PO 1

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1

O

A

P

Q

B

1

l

D

sin

OA=cos

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 Graphy-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 Graphy-axis

-axis

y = sin

• 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 Graphy-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 Graphy-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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