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Why 2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2

  1. Jan 30, 2007 #1
    2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2
    Last edited by a moderator: Jan 30, 2007
  2. jcsd
  3. Jan 30, 2007 #2
    why not????
  4. Jan 30, 2007 #3
    i think you wrote it wrong. it should be
    [tex]\sin^{2}t = \left(\frac{1}{\cosh p}\right)^2[/tex]
  5. Jan 30, 2007 #4


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    What is the actual question you are asking here? Are you asking why, or if, the left implies the right. Have you done anything to attempt to show this? I doubt anyone will help unless you firstly specify your question, and secondly show some effort!
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