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# trigonometric functions

Extracts from this document...

Introduction

Portfolio

Type 1 investigation

Transformation of Trigonometric Functions

Investigate the function:

y = a sin b(x-c) + d   in respect to the transformation of the base curve of y = sin x, depending on the values of a, b, c, and d. Be sure to consider all possible values for a, b, c, and d.

Describe the base curve

Then try values in b (sin2x, sin -3x, sin1/2x)

Does your hypothesis hold true for y = cos(x) and y = a cos b (x-c) +d? How about tan(x)?

Transformation of Trigonometric Functions

Introduction

The purpose of this study is to examine the transformation of trigonometric functions of y=A sin B (x-C) + D and determine the effect on the base curve y=sin(x). I am going to be systematically changing the values of A, B, C and D in the equation y= A sin B (x-C) +D. First I am going to examine different numbers for the value of A. I am going to use whole numbers, negative whole numbers, positive rational numbers and negative rational numbers for the value of A and see how this affects the Sine curve. Then I will examine different numbers for B, then C then D.

Middle

Figure 9                                                Figure 10

These graphs show that my prediction was correct, as figure 9 shows that the minimum value is -0.4 and figure 10 shows that the maximum value is 0.4, and both figures show that the period is 2 π.

Changing the value of constant B

I am now going to change the values of B in the equation y= sin(Bx) to see what affect it has on the base curve.

Y=sin (3x)

Figure 11

Figure 11 shows that the amplitude, maximum and minimum values have the same value as the curve y=sinx, i.e  1 and -1. The period of y=sin3(x) has decreased from 2 π to π/3. It is also noticed that increasing the value of ‘B’ to 3, the curve repeats itself three times within the period f 2 π. So this shows that large values for ‘B’ have a smaller period but a larger frequency. So this shows that ‘B’ affects the period which affects the frequency of the equation.

Y=sin (-3x)

Figure 12

Figure 12 shows that the amplitude and maximum value remain 1 while the minimum value remains -1. And that the period also decreased from 2 π to π /3. Once again using the trace option on the graphing calculator, we notice that using a negative value inverts the curve.

Y= sin (0.25x)

Figure 13                                                Figure 14

Conclusion

Y=sin (x)+ 2/5

Figure 26

Figure 26 shows my prediction is correct.

After examining what happens when changing all values in the equation, y= A sin B (x-C) +D, I am now going to predict what changes will occur is I used the equation y= 5 sin -2 (x-1) + ¾. I predict that the maximum value will change to 5, the minimum value will change to -5 and the amplitude will change to 5. The base curve is inverted because the ‘B’ value is a negative value and the period of the curve is decreased to π which increases the frequency to 2. The base curve also shifts to right by 1 unit and shifts vertically upwards by 0.75 (changing the maximum value to 5.75 and the minimum value to -4.25).

Figure 27

Figure 28                                                        Figure 29

Figure 27 above shows that my prediction is correct. Figure 28 proves shows that my prediction about the maximum value being 5.75 is correct. And Figure 29 using the trace button shows that my prediction about the curve having a phase shift to the right by one unit.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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