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Some trigonometry

  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data

    [tex]\phi = 4\arctan{\exp^{m\gamma(x-vt)}}[/tex]


    [tex]\dot{\phi} = -2m\gamma v \sin{\frac{\phi}{2}}[/tex]

    2. Relevant equations
    3. The attempt at a solution

    [tex]\phi = 4\arctan{\exp^{m\gamma(x-vt)}}[/tex]

    [tex]\tan{\phi/4} = \exp^{m\gamma(x-vt)}[/tex]

    [tex]\frac{\dot{\phi}}{4\cos^{2}{\frac{\phi}{4}}} = -m\gamma v \exp^{m\gamma(x-vt)}[/tex]

    From an example question, They say that

    $$ -m\gamma v \exp^{m\gamma(x-vt)} = -m\gamma v \tan{\frac{\phi}{2}}$$

    Which implies that [tex]\dot{\phi} = -2m\gamma v \sin{\frac{\phi}{2}}[/tex]

    Can someone explain to me how: [itex]-m\gamma v \exp^{m\gamma(x-vt)} = -m\gamma v \tan{\frac{\phi}{2}}[/itex]?
  2. jcsd
  3. Aug 21, 2013 #2


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    No, here it should be ## -m\gamma v \exp^{m\gamma(x-vt)} = -m\gamma v \tan{\frac{\phi}{4}}##. This follows from (1)
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