• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15

trigonometric functions

Extracts from this document...



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

Start with a (try some values)

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


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.

...read more.


              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

...read more.


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.

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. LACSAP's Functions

    - 0(7 - 0) = 28 1 28 ( (7+1)C2) - 1(7 - 1) = 22 2 28 ( (7+1)C2) - 2(7 - 2) = 18 3 28 ( (7+1)C2) - 3(7 - 3) = 16 4 28 ( (7+1)C2) - 4(7 - 4) = 16 5 28 ( (7+1)C2)

  2. Derivative of Sine Functions

    At every /4 unit is shown in the table below: X -2 - - - - - - - 0 2.00 1.41 0 -1.41 -2.00 -1.41 0 1.41 2 X 2 1.41 0 -1.41 -2.00 -1.41 0 1.41 2.00 The scatter plots of values in the table is sketched below: (joined by a smooth curve)

  1. In this essay, I am going to investigate the maximum number of pieces obtained ...

    2 4 7 11 16 n 1 2 3 4 5 P 2 4 8 15 26 Compare these three tables, we can find out that P2=R1+P1=2+2=4 P3=R2+P2=4+4=8 P4=R3+P3=7+8=15 P5=R4+P4=11+15=26 P=Y+X+S=Y+ (n^2-n)/2+ (n+1) Y=P-(n^2-n)/2+ (n+1) = (n^3+5n+6)/6-[(n^2-n)/2+ (n+1)] = (n^3+5n+6)/6-(n^2+n+2)/2 = n^3-3n^2+2n Therefore, P=Y+X+S= (n^3-3n^2+2n)

  2. Mathematics Higher Level Internal Assessment Investigating the Sin Curve

    In this equation the variables are as following, and . This means that the amplitude of this graph would be 3, an outwards stretch from the original graph. The negative means that the graph would be flipped over with respect to the - axis.

  1. Population trends. The aim of this investigation is to find out more about different ...

    In most cases the points are very close to the graphed line. The model shows how the population increased correctly although it should be more curved order to be able to represent what actually happened. The model shows a reasonable past where population was less than in 1950.

  2. Analysis of Functions. The factors of decreasing and decreasing intervals (in the y ...

    In power functions when the degree is positive then there would be always increasing values doesn't matter if it's odd or even, but when the degree is negative then it depends if it's odd or even, to have increasing or decreasing intervals.

  1. derivitaive of sine functions

    Later on starting at 0, the gradient begins to once again moves upwards positively until it reaches a maximum amplitude of 1 at =/2 or at positive /2. The graph then continues from /2 by moving negatively downwards until it reach minimum amplitude of -1 once again at 2.

  2. The investigation given asks for the attempt in finding a rule which allows us ...

    follows: A=12hfa+2fx1+2fx2+?+2f(xn-1+f(b)] Unlike previous time whereh=1n, the difference between these expressions is that the value of h is changed. Since a and b are defined as the first and last bases, therefore, they would be the most extreme values on the x-axis.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work