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Tricky Quadratic formula / trig identities

  1. May 25, 2015 #1
    • Member warned about posting without the homework template
    Im to solve ##(k+l)^{2}e^{-ila}-(k-l)^{2}e^{ila}=0##, for ## k##,

    The solution is ##k=l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##

    FIRST QUESTION

    So it's a quadratic in k, should be simple enough, my working so far using the quad. formula is ##k= (4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{4l^{2}(e^{-ila}+e^{ila})^{2}-4l^{2}(e^{-ila}-e^{ila})^{2}})/2(e^{-ila}-e^{ila})##
    =##(4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{16l^{2}})/2(e^{-ila}-e^{ila})##
    ... which doesnt give the correct answer.

    SECOND QUESTION.

    Even after attaining the correct solution, I'm struggling to show that ##l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##
    LHS: ##cos (al) + isin (al) -1 / cos (al) + isin (al) +1=(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)-1)/(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)+1)##

    using## cos 2x= cos^{2}x-sin^{2}x##
    and ##sin 2x=2sinxcosx,##

    but this doesnt seem to be going anywhere...

    Any help on either question really appreciated,
    Thanks in advance.
     
    Last edited: May 25, 2015
  2. jcsd
  3. May 25, 2015 #2

    SammyS

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    The given solution is consistent with there being a typo in your stated problem statement.

    It seems there should be no exponent on the k inside the first parenthesis .
     
  4. May 25, 2015 #3
    apologies, yes that was a typo,
    edited ta.
     
  5. May 25, 2015 #4

    SammyS

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    Does that help you in solving the problem, or are you still stuck?
     
  6. May 25, 2015 #5
    The original question remains.
    it was a typo.
     
  7. May 25, 2015 #6

    epenguin

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    For the first part, less tedious and error-prone than using your quadratic solution maybe, you have got there a slightly disguised difference of two squares.

    If you can't see that, you can also take a term from right to the left hand side and see how you could then best simplify.
     
    Last edited: May 25, 2015
  8. May 25, 2015 #7
    I think what you are saying is equivalent to the step I took in simpifying the part inside the square root of the quadratic formula, the last line where I get it simplifies to square rt of 16l^{2} ?
     
  9. May 25, 2015 #8

    SammyS

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    Second Question:
    Multiply the numerator & denominator by ## \ e^{-i\, l a/2}\ .##
     
  10. May 25, 2015 #9

    SammyS

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    First Question:

    Multiply the original equation by ##\ e^{ila} \ ## .

    Then it's pretty quick to get difference of squares else, add ##\displaystyle \ (k-l)^{2}e^{2ila} \ ## to both sides & take square root.

    The given solution only uses one of the two possible sign choices.
     
  11. May 26, 2015 #10
    how many variables have your equation?
     
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