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

[Lightf]

Homework Statement

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

Show

\dot{\phi} = -2m\gamma v \sin{\frac{\phi}{2}}

Homework Equations

The Attempt at a Solution

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

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

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

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 \dot{\phi} = -2m\gamma v \sin{\frac{\phi}{2}}

Can someone explain to me how: -m\gamma v \exp^{m\gamma(x-vt)} = -m\gamma v \tan{\frac{\phi}{2}}?

 

[haruspex]

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