Trig/constructing sine function from graph

[Asphyxiated]

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!

 

[HallsofIvy]

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 2\pi and has period 2\pi - 0= 2\pi. Basically, whatever your "Bx+ C" is you must have B(1)+ C= B+ C= 0 and B(6)+ C= 6B+ C= 2\pi. Subtracting the first equation from the second, 6B+ C- (B+ C)= 5B= 2\pi so B= 2\pi/5. Then B+ C= 0 becomes 2\pi/5+ C= 0 so C= -2\pi/5. Your equation should look like A sin((2\pi/5)x- 2\pi/5)+ D or, equivalently, A sin(2\pi/5(x- 1))+ D

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!

 

[HallsofIvy]

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



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 x_0 to x_1, Solve Bx_0+ C= 0 and Bx_1+ C= 2\pi.

 

[Asphyxiated]

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?

 

[HallsofIvy]

Since you don't say what you got or what the answer in the book was, I can't answer that.

 

[Asphyxiated]

@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