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

  • #1
why
2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2
:surprised
?????????????????????????????????????
?????????????????????????????????????
 
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Answers and Replies

  • #3
i think you wrote it wrong. it should be
[tex]\sin^{2}t = \left(\frac{1}{\cosh p}\right)^2[/tex]
 
  • #4
cristo
Staff Emeritus
Science Advisor
8,107
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why
2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2
:surprised
?????????????????????????????????????
?????????????????????????????????????
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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