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)