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

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Presentation on theme: "Derivative of the Sine Function By James Nickel, B.A., B.Th., B.Miss., M.A. www.biblicalchristianworldview.net."— Presentation transcript:

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

2 Basic Trigonometry Given the unit circle (radius = 1) with angle  (theta), then, by definition, PA = sin  and OA = cos . Copyright  2009

3 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. Copyright  2009

4 Question How does sin  vary as  varies? To find out, we must calculate the derivative of y = sin . Copyright  2009

5 The Method of Increments We add  to  and determine what happens. We need to find the change in y or  y. Copyright  2009

6 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. Copyright  2009

7 The Method of Increments Also remember that  is just a “little bit” or infinitesimally small. Copyright  2009

8 Our Goal Find QD =  y To do this, we must show  QDP ~  PAO We then let   0 Hence, Copyright  2009

9 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  Copyright  2009

10 Some Geometry Construct DP  OA Note the transversal OP Hence,  =  DPO  QPO = 90    DPO and  DPQ are complementary. Hence,  DPQ = 90  –  Copyright  2009

11 Some Geometry  DPQ = 90  –    DQP =  Hence,  QDP ~  PAO because both are right triangles and  DQP =  POA =  Copyright  2009

12 Some Geometry Consider DQ It represents the change in y (i.e.,  y) resulting from the change in  (i.e.,  ). Copyright  2009

13 Derivative Formula In  QDP, cos  = Note that as  gets infinitesimally small, QP (the hypotenuse) converges to . Copyright  2009

14 Derivative Formula Because  QDP ~  PAO, then: Copyright  2009

15 Derivative Formula Since PO = 1 and OA = cos , then: Copyright  2009

16 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 . Copyright  2009

17 Derivative Graph The solid line graph represents y = sin  (the sine curve). The dotted line graph represents y = cos  (the cosine curve). Copyright  2009

18 Derivative Graph 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. Copyright  2009

19 Derivative Graph 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). Copyright  2009

20 Derivative Graph The cosine curve traces the derivative of the sine curve at every point on the sine curve. Copyright  2009


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