Trig/constructing sine function from graph

In summary, the homework statement is telling you to work backwards from a sin graph to a sin formula. The amplitude, vertical shift, and arguments of the function are shown on the graph. The phase shift and period are found by solving the equation for x. The equation should look like A sin((2\pi/5)x- 2\pi/5)+ D. The standard graph goes from 0 to 2\pi and has a period of 2\pi - 0= 2\pi. The solution to this problem is in 0 <= Yx+-Z <= 2pi. If you need help solving the equation, let me know.
  • #1
Asphyxiated
264
0

Homework Statement



Construct or "work backwards" from a sine graph to a sine formula. The formula should be in:

y = A sin(Bx+C) +D

Where A is amplitude, D is Vertical Shift (on y access) and Bx+C are arguments of the function

The graph shows that the Phase Shift is 1 and Period is 6 (meaning the basic cycle begins at 1 and ends at 6)


Homework Equations



0 <= x <= 2pi (pi as in 3.14..) ("<=" is less than or equal too)

The Attempt at a Solution



1 <= x <= 6

0 <= x-1 <= 5 /** subtract 1 from all terms **/

0 <= pi(x)-1 <= 5pi /** multiply all terms by pi **/

0 <= (pi(x)-1)/2 <= 5pi/2 /** divide all terms by 2 **/

STUCK

The solution should be in 0 <= Yx+-Z <= 2pi (Where Y and Z can be any integer, the only thing that matters is that the 0 is at the beginning and that 2pi is the end)
________________________________

This is not the only solution I tried, just the last one, so unless I have missed something rather fundamental I can't seem to solve this one. Please note that I solved 14 other problems that were basically of the same thing using the same methods that I have used to try to solve this problem and they all worked out and check out in the answers section (I am not a student, just picking up where I left off in high school).

If I have done something wrong (in the formatting of this posting) or you need something that I neglected to mention just let me know as this is my first time posting anything on this forum.

Also just to clarify all i am looking for is a solution the equalities statement not the whole y = equation as I understand that entirely.

thanks immensely!
 
Last edited:
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  • #2
Asphyxiated said:

Homework Statement



Construct or "work backwards" from a sine graph to a sine formula. The formula should be in:

y = A sin(Bx+C) +D

Where A is amplitude, D is Vertical Shift (on y access) and Bx+C are arguments of the function

The graph shows that the Phase Shift is 1 and Period is 6 (meaning the basic cycle begins at 1 and ends at 6)
Then the period is 6-1= 5, not 6. The "standard" graph goes from 0 to [itex]2\pi[/itex] and has period [itex]2\pi - 0= 2\pi[/itex]. Basically, whatever your "Bx+ C" is you must have B(1)+ C= B+ C= 0 and [itex]B(6)+ C= 6B+ C= 2\pi[/itex]. Subtracting the first equation from the second, [itex]6B+ C- (B+ C)= 5B= 2\pi[/itex] so [itex]B= 2\pi/5[/itex]. Then B+ C= 0 becomes [itex]2\pi/5+ C= 0[/itex] so [itex]C= -2\pi/5[/itex]. Your equation should look like [itex]A sin((2\pi/5)x- 2\pi/5)+ D[/itex] or, equivalently, [itex]A sin(2\pi/5(x- 1))+ D[/itex]


Homework Equations



0 <= x <= 2pi (pi as in 3.14..) ("<=" is less than or equal too)

The Attempt at a Solution



1 <= x <= 6

0 <= x-1 <= 5 /** subtract 1 from all terms **/

0 <= pi(x)-1 <= 5pi /** multiply all terms by pi **/

0 <= (pi(x)-1)/2 <= 5pi/2 /** divide all terms by 2 **/
You are going the wrong way here!

STUCK

The solution should be in 0 <= Yx+-Z <= 2pi (Where Y and Z can be any integer, the only thing that matters is that the 0 is at the beginning and that 2pi is the end)
Why do you say they should be integers?

________________________________

This is not the only solution I tried, just the last one, so unless I have missed something rather fundamental I can't seem to solve this one. Please note that I solved 14 other problems that were basically of the same thing using the same methods that I have used to try to solve this problem and they all worked out and check out in the answers section (I am not a student, just picking up where I left off in high school).

If I have done something wrong (in the formatting of this posting) or you need something that I neglected to mention just let me know as this is my first time posting anything on this forum.

Also just to clarify all i am looking for is a solution the equalities statement not the whole y = equation as I understand that entirely.

thanks immensely!
 
  • #3
HallsofIvy said:
Then the period is 6-1= 5, not 6. The "standard" graph goes from 0 to [itex]2\pi[/itex] and has period [itex]2\pi - 0= 2\pi[/itex]. Basically, whatever your "Bx+ C" is you must have B(1)+ C= B+ C= 0 and [itex]B(6)+ C= 6B+ C= 2\pi[/itex]. Subtracting the first equation from the second, [itex]6B+ C- (B+ C)= 5B= 2\pi[/itex] so [itex]B= 2\pi/5[/itex]. Then B+ C= 0 becomes [itex]2\pi/5+ C= 0[/itex] so [itex]C= -2\pi/5[/itex]. Your equation should look like [itex]A sin((2\pi/5)x- 2\pi/5)+ D[/itex] or, equivalently, [itex]A sin(2\pi/5(x- 1))+ D[/itex]



You are going the wrong way here!


Why do you say they should be integers?
To find the "Bx+ C" for any sine or cosine with a period from [itex]x_0[/itex] to [itex]x_1[/itex], Solve [itex]Bx_0+ C= 0[/itex] and [itex]Bx_1+ C= 2\pi[/itex].
 
  • #4
THANKS!

I looked over the other problems and noticed that they were all with phase shift 0, I don't know how I missed that...

Anyway I appreciate your replies

@HallsOfIvy

I tried it your way and it makes sense but when I solve the equation using period of 5 (6-1) I end with an equation of 0 <= ((pi)x-1)/2 <= 2pi which is markedly different from your procedure and end result. My equation checks out in the answers section, so I guess my question is why did you end up with something different for the same problem?
 
Last edited:
  • #5
Since you don't say what you got or what the answer in the book was, I can't answer that.
 
  • #6
@HallsOfIvy

I did say what my end result was with "0 <= ((pi)x-1)/2 <= 2pi" but if you want the sine equation it was y = 2 sin(((pi)x-1)/2)

EDIT sorry for all the parentheses, I really need to learn LaTeX
 

Related to Trig/constructing sine function from graph

1. How do I construct a sine function from a given graph?

To construct a sine function from a graph, you will need to identify the key points on the graph, including the amplitude, period, and phase shift. The amplitude is the distance from the middle of the graph to the highest or lowest point. The period is the length of one complete cycle of the graph, and the phase shift is any horizontal shift of the graph. Once you have identified these key points, you can use the general form of the sine function, y = a sin(bx + c), to create your specific function.

2. How do I determine the amplitude of a sine function from a graph?

The amplitude of a sine function can be found by measuring the distance from the middle of the graph to the highest or lowest point. If the highest point is above the middle of the graph, the amplitude is positive. If the lowest point is below the middle of the graph, the amplitude is negative. The amplitude is equal to half the distance between the highest and lowest points.

3. What is the period of a sine function and how is it determined from a graph?

The period of a sine function is the length of one complete cycle of the graph. It can be determined by measuring the distance between two consecutive peaks or two consecutive troughs on the graph. The period can also be calculated using the formula 2π/b, where b is the coefficient of x in the general form of the sine function, y = a sin(bx + c).

4. How does the phase shift affect a sine function on a graph?

The phase shift is any horizontal shift of the graph. It affects the starting point of the graph and determines the position of the graph on the x-axis. The phase shift can be positive, negative, or zero. A positive phase shift moves the graph to the left, while a negative phase shift moves the graph to the right. A phase shift of zero means there is no horizontal shift of the graph.

5. Can a sine function have a negative amplitude?

Yes, a sine function can have a negative amplitude. The amplitude of a sine function is determined by the distance between the middle of the graph and the highest or lowest point. If the highest point is below the middle of the graph, the amplitude will be negative. This means that the graph will be reflected over the x-axis, resulting in a negative amplitude.

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