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Derivatives of Sine Fucntions

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Introduction

Math Portfolio Assignment:

Derivatives of Sine Functions

Bridget Belsher

          Derivatives of Sine Functions, Type I

 Date Set: January 14 2008

Date Submitted: February 5 2008

Father Lacombe High School

Mrs. Gabel

I certify this portfolio Assignment is entirely my own work

image00.pngimage01.png

The above graph, displays the unaltered sin function, in its original form. The functions domain on the graph is -2π ≤ 2π[1]. The result of this is can be seen in the two oscillations with the range. Its behavior can be determined by the behavior of the first oscillation (the oscillation starting at -2π), as the same behavior will be observed in the second oscillation (that starting at 0), due to the repetitive nature of sine functions. From the point (-2π, 0), the slope can be described as increasing with a value of one.  However, one can clearly see that at (image10.png the function has no slope, as it is a point of discontinuity, that is having a slope of zero, as the line is straight. Just after that point, the slope is -1, which causes a constant decrease in the functions y-value and increase in the functions x value, which continues until the point (image13.pngπ, -1), at which it reaches another point of discontinuity.

The derivative on sin(x) is Cos(x), which is seen in the graph below and further explained as well.

Cos(x)’s behavior describes the slope of sine(x).

...read more.

Middle

image00.pngimage03.png

Basically, when the Sin functions slope is defined as a number, then in the derived function would be directly affected. This can be most clearly explained in the following conjecture. In the derived function Cos(x), the maximum is equivalent to the highest positive slope of Sin(x), in this case being 1 and the minimum of the Cos(x) is equivalent to the greatest negative slope of Sin(x), being -1. The points at which Sin(x) has no slope, the x-value in cos(x) is 0.

This is highlighted in the graphs below, where in the Sin(x) function the points at which the slope is zero are colored red. The corresponding points in Cos(x) at which the actual x-value is zero is also highlighted, but in green. In the last graph below, to the right, the two graphs are compared, at the points where the slope of sin(x) =0 and the y value of Cos(x) actually is 0. The same could be done for the maximum and minimum values at which the slope is at its greatest negative and greatest positive slope; however I think at this point the explanation is clear enough for the concept to be understood without additional graphs.

image02.pngimage16.pngimage17.png

When the sin function is further complicated by being multiplied by a constant, its function changes from y=sin(x) to y=aSin(x).

...read more.

Conclusion

b value is 2.  image12.pngimage11.pngimage09.png

h(x) = sin2x

Lim h(x + h) – h(x)

h→ 0            h

Lim sin(2(x + h)) – sin(2x)

h→ 0                h                        

Lim sin(2x)cos(2h) + sin(2h)cos(2x) – sin(2x)

h→ 0                           h

Lim sin(2x)(cos(2h) – 1) + sin(2h)cos(2x)

h→ 0                                       h

sin(2x) Lim (cos(2h) – 1)  +  cos(2x) Lim sin(2h)

                       h→ 0             h                             h→ 0      h

sinx(0)(2) + cos(2x)(1)(2)

h’(x) = 2cos(2x)

The conjecture when the derivative of Sin(x+c) is Cos(x+c), is the same as in the original conjecture, in which the minimum and maximum slopes of Sinx determine the minimum and maximum values for Cosx. The only change is that the x0intercept change depending on the positive or negative value of c. An example of this can be seen below, in which sin(x + π/4) and Cos (x + π/4) are shown.

j(x) = sin(x + 2)

Lim j(x + h) – j(x)

h→ 0            h

Lim sin((x + 2)+ h) – sin(x + 2)

h→ 0                h                        

Lim sin(x + 2)cos(h) + sin(h)cos(x + 2) – sin(x + 2)

h→ 0                           h

Lim sin(x + 2)(cos(h) – 1) + sin(h)cos(x + 2)

h→ 0                             h

sin(x + 2) Lim (cos(h) – 1)  +  cos(x + 2) Lim sin(h)

                                  h→ 0             h                                   h→ 0      h

sin(x + 2)(0) + cos(x + 2)(1)

h’(x) = cos(x + 2)


[1] Note that this limit is present throughout all graphs in this paper, and that the said graphs increase and decrease in increments of  .

[2] Due to the fact that sin(x +h) = sin(x) multiplied by cos(h) + sin(h) multiplied by cos(x), I have taken the liberaty of substituting in the equation.

...read more.

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