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Differential equations proof

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that Acos(ωt) + Bsin(ωt) can be written in the form r*sin(ωt - θ). Determine r and θ in terms of A and B. If Rcos(ωt - ε) = r*sin(ωt - θ), deermine the relationship among R, r, ε and θ.

    2. Relevant equations

    Trig identities.

    3. The attempt at a solution

    I'm really confused on this question. What I first tried to do is to use the sum-difference forumla on r*sin(ωt - θ) = r*sin(ωt)cos(θ) - r*cos(ωt)sin(θ).
     
  2. jcsd
  3. Nov 8, 2012 #2

    LCKurtz

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    Start by writing your expression like this$$
    \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}}
    \sin(\omega t) \right)$$That quantity in the large parentheses looks like an addition formula. Think about a right triangle with legs ##A## and ##B##.
     
  4. Nov 8, 2012 #3
    Is that from the Pythagorean theorem?
    I'm still unsure how to set this problem up :(
     
  5. Nov 8, 2012 #4

    LCKurtz

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    Did you draw the triangle I suggested? Do those fractions look like sines and cosines of some angle? Can you make the expression in large parentheses look like an addition formula?
     
  6. Nov 8, 2012 #5
    It would be cos(theta) and sin(theta)
     
  7. Nov 8, 2012 #6

    LCKurtz

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    So.......?
     
  8. Nov 8, 2012 #7
    If I use the sum-difference identity I would get cosine term not the sine :(
     
  9. Nov 8, 2012 #8

    LCKurtz

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    Isn't ##\displaystyle{\frac A {\sqrt{A^2+B^2}}}## the cosine of some angle in your triangle?
     
  10. Nov 8, 2012 #9
    yeah so you get cos(θ)cos(ωt) + sin(θ)(sin(ωt))
     
  11. Nov 8, 2012 #10

    LCKurtz

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    This gives you the form ##r\cos(\omega t -\theta)## (don't forget there is that square root out front). Now getting it in a sine form is what I think last sentence in your problem is about.
     
  12. Nov 8, 2012 #11
    But I did not show the first part. I'm quite sure though that if you have rcos(θ) and rsin(θ) they're constants.
     
  13. Nov 8, 2012 #12
    Here is something else I did. I got r*sin(wt)cos(theta) - r*cos(wt)sin(theta)
    Then I set rewrote as r*cos(theta) + (-rsin(theta)cos(wt)

    I can then get A = r*cos(theta) and B = (-r*sin(theta))
     
  14. Nov 8, 2012 #13

    LCKurtz

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    Haven't we shown$$
    A\cos(\omega t)+B\sin(\omega t) = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}}
    \sin(\omega t) \right) = \sqrt{A^2+B^2}(\cos\theta\cos\omega t + \sin\theta \sin(\omega t) = \sqrt{A^2+B^2}\cos(\omega t - \theta)$$To get it to a sine function I think you have to work the last sentence in the original problem.
     
  15. Nov 8, 2012 #14
    What can I do though? Use the identity on both sides? The first line states to write it in the sin form. We did it in cos :(.
     
  16. Nov 8, 2012 #15

    LCKurtz

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    Yes, we have the cosine. Did you read my previous post, copied above? That is part of your problem, to figure out how to change the cosine to a sine. I have helped you with half of your problem. Now you work on finishing it.
     
  17. Nov 8, 2012 #16
    Is A = r*cos(theta) and B = (-r*sin(theta)) correct for the first part then? To get theta and r in terms of A and B?
     
  18. Nov 8, 2012 #17
    Hold on, I have a question. How can you assume that will give you cosine instead of sine? If I switch, A and B I would get sine.
     
  19. Nov 8, 2012 #18

    LCKurtz

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    You have A and B in terms of r and theta. You want r and theta in terms of A and B.
     
  20. Nov 8, 2012 #19

    LCKurtz

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    You can get either a sine or cosine form. That shouldn't surprise you because anything that can be expressed as one can be expressed in terms of the other since ##\sin\theta = \cos(\frac \pi 2 -\theta)## and similarly for the cosine.
     
  21. Nov 9, 2012 #20
    woud ε = be pi/2 - wt - theta?
     
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