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Help derivation of this equation.

  1. Feb 9, 2013 #1
    Given a wave:
    [tex] e(t)=\hat x E_{x0}\cos \omega t+\hat y E_{y0}\cos( \omega t+ \delta)[/tex](1)
    The book claimed:
    [tex]\sin^2\delta\;=\; \left[\frac {e_x(t)}{E_{x0}}\right]^2-2\left[\frac {e_x(t)}{E_{x0}}\right]\left[\frac {e_y(t)}{E_{y0}}\right]\cos\delta+ \left[\frac {e_y(t)}{E_{y0}}\right]^2[/tex]
    I can not get this. Below is my work.

    From (1)
    [tex]e_x(t)=E_{x0}\cos\omega t\;\Rightarrow \;\cos \omega t=\frac{e_x(t)}{E_{x0}}[/tex]
    [tex]\Rightarrow\; \sin\omega t\;=\;\sqrt{1-\left(\frac{e_x(t)}{E_{x0}}\right)^2}[/tex]
    [tex]e_y(t)=E_{y0}\cos(\omega t+\delta) \;\Rightarrow \;\frac{e_y(t)}{E_{y0}}\;=\;\cos\omega t \cos\delta-\sin\omega t sin\delta\;=\; \frac{e_x(t)}{E_{x0}}\cos\delta-\sqrt{1-\left(\frac{e_x(t)}{E_{x0}}\right)^2}sin\delta[/tex]
    [tex]\Rightarrow\;\sqrt{1-\left(\frac{e_x(t)}{E_{x0}}\right)^2}sin\delta \;=\; \frac{e_x(t)}{E_{x0}}\cos\delta-\frac{e_y(t)}{E_{y0}}[/tex]
    [tex]\Rightarrow\;\left(1-(\frac{e_x(t)}{E_{x0}})^2\right) sin^2\delta \;=\;\left[\frac{e_x(t)}{E_{x0}}\right]^2 \cos^2\delta-2\frac{e_x(t)}{E_{x0}}\frac{e_y(t)}{E_{y0}}\cos \delta + \left[\frac{e_y(t)}{E_{y0}}\right]^2[/tex]
    It is obvious that I am not going to get what the book gave:
    [tex]\sin^2\delta\;=\; \left[\frac {e_x(t)}{E_{x0}}\right]^2-2\left[\frac {e_x(t)}{E_{x0}}\right]\left[\frac {e_y(t)}{E_{y0}}\right]\cos\delta+ \left[\frac {e_y(t)}{E_{y0}}\right]^2[/tex]

    Please help my derivation.

    Thanks
     
  2. jcsd
  3. Feb 9, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi yungman! :smile:

    (i can't see where you've gone wrong, but try it this way …)

    you are essentially asked to prove that cos2A + cos2B - 2cosAcosBcos(A-B) = sin2(A-B)

    just expand that! :wink:

    when you've done that, try drawing the diameter PQ of a unit circle, with two points A and B such that angleAPD = A, angle BPD = B, and using simple geometry to prove that AB = sin(A-B)
     
  4. Feb 9, 2013 #3
    Thanks Tiny Tim, I'll try it and see.

    You think you can look at the other questions I posted in page 3 of the "Classical Physics" on Balanis a few days ago, I have no luck at all.

    Many thanks
     
    Last edited: Feb 9, 2013
  5. Feb 9, 2013 #4
    Thanks Tiny Tim. I worked out using [itex]A=\omega t +\delta\;\hbox{ and }\; B=\omega t[/itex]. This is a little tricky and unobvious.
     
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