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Trigonometric limit

  1. Feb 1, 2010 #1
    I met the following expression in a QM book:

    [itex]
    \frac{sin[(n+1/2)\pi+\epsilon]}{cos[(n+1/2)\pi+\epsilon]}=\frac{(-1)^n\cos(\epsilon)}{(-1)^{n+1}\sin(\epsilon)}
    [/itex]

    where [itex]
    \epsilon << 1
    [/itex]



    No matter how hard I try (sine of sum, etc.), I can't see the intermediate steps to this result.

    Please, help.
     
  2. jcsd
  3. Feb 1, 2010 #2

    tiny-tim

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    Hi intervoxel! :smile:

    (have a pi: π and an epsilon:ε :wink:)

    Learn your trigonometric identities …

    sin(A+B) = sinAcosB + cosAsinB

    cos(A+B) = cosAcosB - sinAsinB

    sin(n + 1/2)π = (-1)n

    cos(n + 1/2)π = (-1)n+1 :wink:
     
  4. Feb 1, 2010 #3
    Oh, come on, tiny-tim, I'm stuck in this problem.

    I arrive at the (wrong) answer: -(cos(ε)-sin(ε)) / (cos(ε)+sin(ε))=-(1-ε)/(1+ε) and not -1/ε, which is correct.
     
    Last edited: Feb 1, 2010
  5. Feb 2, 2010 #4

    tiny-tim

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    oops!

    oops! :redface:

    What was I thinking? :rolleyes:

    (lesson: check what people tell you, not only to see why it works but sometimes to see whether it works, or you'll never learn anything!)

    Try it again … this time with cos(n + 1/2)π = 0. :smile:
     
  6. Feb 2, 2010 #5
    Thank you for your help.
     
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