# Some trigonometry

1. Aug 20, 2013

### Lightf

1. The problem statement, all variables and given/known data

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

Show

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

2. Relevant equations
3. 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: $-m\gamma v \exp^{m\gamma(x-vt)} = -m\gamma v \tan{\frac{\phi}{2}}$?

2. Aug 21, 2013

### haruspex

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