Solution to Trig Homework: Show $\dot{\phi} = -2m\gamma v \sin{\frac{\phi}{2}}$

[Lightf]

Homework Statement

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

Show

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

Homework Equations

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]?

 

[haruspex]

[itex]\tan{\phi/4} = \exp^{m\gamma(x-vt)}[/itex] ... (1)

From an example question, They say that

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

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