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anonymous

  • one year ago

How to find the sin and cosine of a shifted sinusoidal function.

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  1. anonymous
    • one year ago
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  2. dan815
    • one year ago
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    looks like this is the sin graph shifted by pi/12 and the period is only pi/2 now

  3. dan815
    • one year ago
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    Let's figure out how to change the period

  4. dan815
    • one year ago
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    sin(x) has what period initially

  5. dan815
    • one year ago
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    Hi, are you there?

  6. anonymous
    • one year ago
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    i'm a little confused how you got pi/2

  7. dan815
    • one year ago
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    ah it's not pi/2

  8. dan815
    • one year ago
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    it's 2pi/3

  9. dan815
    • one year ago
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    http://i.imgur.com/KZfiTq6.png

  10. anonymous
    • one year ago
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    what is 2pi/3? Sorry i'm lost

  11. dan815
    • one year ago
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    take a look at that picture

  12. dan815
    • one year ago
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    the length of this full cycle is called the period

  13. anonymous
    • one year ago
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    where would the period start from?

  14. dan815
    • one year ago
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    you can start the period anywhere, as long as 1 full cycle takes place, however the period will always be 2pi/3

  15. dan815
    • one year ago
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    let me show you some examples

  16. anonymous
    • one year ago
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    Ok thank you..

  17. dan815
    • one year ago
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    http://i.imgur.com/KnoNRXN.png

  18. dan815
    • one year ago
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    here i have picked 3 sets of pairs of points

  19. anonymous
    • one year ago
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    Oh i see! Thank you :)

  20. dan815
    • one year ago
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    okay so lets start with the base graph y=sin(x), what period does this have initially

  21. anonymous
    • one year ago
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    2pi?

  22. dan815
    • one year ago
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    right

  23. dan815
    • one year ago
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    y=sin(x) looks like this |dw:1444540802308:dw|

  24. dan815
    • one year ago
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    now lets see how we can possibly make this period only 2pi/3

  25. dan815
    • one year ago
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    currently y=sin(x), a period of 2pi \[y=sin(2\pi x)\] this graph would have a period of only 1, when you multiply by 2pi, because now, if you put in values for x, its being sped up by that 2pi factor, when u put in x=1 you can already see it will be sin(2pi*1) = sin(2*pi) and we know by then this would have completed a full cycle

  26. dan815
    • one year ago
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    Tell me if that makes sense, this might be a little tricky when u first see it

  27. anonymous
    • one year ago
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    I get that, it's just I'm a little confused how we use 2p/3 to help us find out the equation

  28. dan815
    • one year ago
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    right now can u figure out how to make this period 2pi/3 instead of 1

  29. dan815
    • one year ago
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    we saw how multiplying by 2pi in the brackets managed to speed up this period to 1 instead of 2pi

  30. dan815
    • one year ago
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    sin(ax) , what should this a be so that when x=2pi/3, we have a complete period

  31. dan815
    • one year ago
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    in other words y=sin(ax) a*2pi/3= 2pi -----> when you input x=2pi/3 you want to make sure this 'ax' in the brackets turns into 2pi so we know a full period was done

  32. anonymous
    • one year ago
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    couldn't you just multiply it by 2pi?

  33. dan815
    • one year ago
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    2pi made the period go to 1

  34. dan815
    • one year ago
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    the answer is y=\[sin(\frac{2\pi ~x}{2\pi/3} )\]

  35. dan815
    • one year ago
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    see how when x=2pi/3 the stuff in the brackets turns to 2pi

  36. dan815
    • one year ago
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    now apply your shift and amplitude change

  37. dan815
    • one year ago
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    \[y=A*sin(p*(x-b))\] this is the form for sin graph where A is the amplitude,p determines the period and b is the shift \[y=3*sin(\frac{2\pi ~(x-\pi/12)}{2\pi/3} )\]

  38. anonymous
    • one year ago
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    i'm still confused how we got 2pi from the 2pix/2pi/3?

  39. dan815
    • one year ago
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    when x=2pi/3

  40. dan815
    • one year ago
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    |dw:1444541639553:dw|

  41. anonymous
    • one year ago
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    oh ok.. thank you

  42. carlyleukhardt
    • one year ago
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    alright time to investigate sana xD

  43. anonymous
    • one year ago
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    go ahead XD I have a bf

  44. dan815
    • one year ago
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    I think you need to understand better why y=sin(2pi x) in making the period 1 from 2pi previously

  45. anonymous
    • one year ago
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    sorry it's just so new for me it's hard to follow. So I understand how we got 2pi, are we following a formula or just looking for it to complete just a whole period?

  46. dan815
    • one year ago
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    just looking to complete a whole period

  47. dan815
    • one year ago
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    think about it like this y=sin(x), x has to be all the number from 0 to 2pi to complete a period

  48. dan815
    • one year ago
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    when you multiply y=sin(2pi * x), now x has to be all the number from x=0 to 1, because this will generate the whole set 0 to 2pi in the brackets anyway

  49. dan815
    • one year ago
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    the point is you have to make sure you have interval lengths of 2pi showing up in the brackets for a sin function so that it completes a period

  50. anonymous
    • one year ago
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    so now I must plug in x values in for y?

  51. dan815
    • one year ago
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    What do you mean

  52. dan815
    • one year ago
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    http://i.imgur.com/CbqYn0Z.png

  53. dan815
    • one year ago
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    *side note* when you multiply by p the period becomes 2pi/p, so you can figure out what p has to be to get the period you want

  54. dan815
    • one year ago
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    if you set p=2pi/L where L is the length of the period you read on the graph, thats a simple way to force your period to be L

  55. anonymous
    • one year ago
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    is there a more simpler way to do this.. I feel like there is.

  56. dan815
    • one year ago
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    yeah you can memorize the formula

  57. dan815
    • one year ago
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    \[Y=A* Sin(\frac {2\pi}{L} (X-B))\] A=Amplitude L=Length B=Phase

  58. nincompoop
    • one year ago
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    grab a precal textbook

  59. nincompoop
    • one year ago
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    get the one by larson or stewart

  60. anonymous
    • one year ago
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    my professor doesn't use the book, he expects us to "automatically know this"

  61. anonymous
    • one year ago
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    and i've tried using the book

  62. dan815
    • one year ago
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    okay so can you plug into the formula?

  63. dan815
    • one year ago
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    You know the the length,the phase, the Amplitude

  64. anonymous
    • one year ago
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    the amplitude would be 3?

  65. dan815
    • one year ago
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    yes

  66. dan815
    • one year ago
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    I already showed you what the answer should look like in the screenshot a couple comments ago

  67. dan815
    • one year ago
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    just ask one of these people if you are still confused, i have to go now

  68. anonymous
    • one year ago
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    the 3sin(2pi(x-pi/12))/2pi/3)?

  69. anonymous
    • one year ago
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    thanks for your time

  70. carlyleukhardt
    • one year ago
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    i keep on getting notifications. oh nvm your done/

  71. imqwerty
    • one year ago
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    now u need to pay $50 for the help using paypal

  72. dan815
    • one year ago
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    \[Y=3* Sin(\frac {2\pi}{2pi/3} (X-\pi/12))\]

  73. anonymous
    • one year ago
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    ahaha @imqwerty

  74. anonymous
    • one year ago
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    sorry dan, but @imqwerty deserves that medal for that comment

  75. anonymous
    • one year ago
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    imqwerty, I already did pay with the plan.

  76. imqwerty
    • one year ago
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    ok then (:

  77. anonymous
    • one year ago
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    imqwerty, and it's sad to see that I'm paying so much money and not understanding these things. You guys should explain it step by step.

  78. ganeshie8
    • one year ago
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    just saying.. paying money wont guarantee you quick understanding, it only guarantees you quick assistance, understanding the concept depends on how much effort you put and engage with the helper

  79. imqwerty
    • one year ago
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    u can get only this much with $50

  80. anonymous
    • one year ago
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    ganeshie8, I put in a lot of effort thank you very much.

  81. ganeshie8
    • one year ago
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    Maybe lets start over ?

  82. anonymous
    • one year ago
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    I don't understand how to solve this problem. I know there is an equation that you follow. That's all that I got from it.

  83. imqwerty
    • one year ago
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    In which part are u having problem?

  84. anonymous
    • one year ago
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    all of it.

  85. Astrophysics
    • one year ago
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    Well, I think the difficult part is, that there's many ways to do this, so you have to sort of find your own way... A few steps that may help you: 1. Draw the Max and Min lines 2. Draw in the Mid line, i.) the mid line gives you the vertical displacement (k), ii.) the distance from the mid to the max gives you the amplitude (a) 3. Find the distance from crest to crest (or trough to trough) i.) This gives you the period (b), ii.) Period = 2pi/b 4. k, a, and b remain the same throughout the equation but note that a's sign can change 5. Determine the phase shift (h), which helps us determine whether you want sine or cosine i.) For cosine you need to start at a crest (MAX) or trough (MIN) ii.) For a sine you would have to start at the MID line 6. Determine whether your equation is positive or negative haha, there's many ways for this one I think that's like a small tutorial on graphing and finding these equations, also know what amplitude, period, range, domain represent, if you want to know I can tell you I think this post is long enough though.

  86. Astrophysics
    • one year ago
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    |dw:1444544503214:dw| MID = red, MAX = pink, MIN = green to give you an idea of what I mean

  87. anonymous
    • one year ago
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    then how would I find the equation.. i'm aware of the basics

  88. Astrophysics
    • one year ago
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    Ok so your equation is of the form \[y=a*f(b(x-h))+k\] so use what I've said above and see if you can figure it out

  89. Astrophysics
    • one year ago
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    Or I should put it as \[y = a*\sin(b(x-h \pi) ) +k \] similarly for cosine, just switch the trig function

  90. anonymous
    • one year ago
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    I plug in just the amplitude and midpoint?

  91. baru
    • one year ago
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    maybe it would help if you could tell exactly where you stand...have you understood what shifted sinusoidal function means?

  92. imqwerty
    • one year ago
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    maybe u shuld leave it (:

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