I think that the following is true. It arose while trying to prove the injectivity of a function:
If $0<a<R$, $v,w\in]-a,a[$ and $\phi,\theta\in]0,\pi[$, then
$$ 2R(v\cos\frac{\phi}{2}-w\cos\frac{\theta}{2})+v^2-w^2=0 \\ \text{iff} \\ v=w \text{ and } \phi=\theta $$
But I am clueless on how to prove this. Any hints?
EDIT: Also, assume that $v\sin(\frac{\phi}{2})=w\sin(\frac{\theta}{2})$
Assume WLOG $R=1\,$, which is equivalent to dividing by $R^2$ and redefining $\frac{v}{R} \mapsto v, \frac{w}{R} \mapsto w\,$.
Let $\,\lambda=v\sin(\frac{\phi}{2})=w\sin(\frac{\theta}{2}) \ne 0\,$ so that $\,v = \lambda / \sin(\frac{\phi}{2})\,$ and $\,w = \lambda / \sin(\frac{\theta}{2})\,$, then:
$$ \begin{align} 2\left(v\cos\frac{\phi}{2}-w\cos\frac{\theta}{2}\right)+v^2-w^2 &= 2 \lambda\left(\cot \frac{\phi}{2} - \cot \frac{\theta}{2}\right) + \lambda^2\left(\frac{1}{\sin^2 \frac{\phi}{2}} - \frac{1}{\sin^2 \frac{\theta}{2}}\right) = \\[5px] &= -\, \frac{2\lambda \sin \frac{\phi-\theta}{2}}{\sin \frac{\phi}{2} \, \sin \frac{\theta}{2}} -\, \frac{\lambda^2(\cos \phi - \cos \theta)}{2 \sin^2 \frac{\phi}{2} \, \sin^2 \frac{\theta}{2}} \\[5px] &= -\, \frac{\lambda \sin \frac{\phi - \theta}{2}}{\sin^2 \frac{\phi}{2} \, \sin^2 \frac{\theta}{2}} \left( 2\sin \frac{\phi}{2} \, \sin \frac{\theta}{2} + \lambda \sin \frac{\phi+\theta}{2}\right) \end{align} $$
The equation is therefore satisfied by $\phi=\theta\,$, and by any solutions to $2\sin \frac{\phi}{2} \, \sin \frac{\theta}{2} + \lambda \sin \frac{\phi+\theta}{2} = 0$.
For $\phi=\theta\,$ to be the only solution, it remains to show that the latter factor has no eligible solutions within the prescribed range $v,w \in (-a,a) \subset (-1,1)\,$ (remember that we normalized to $R=1\,$). Substituting back $\,\lambda = v \sin \frac{\phi}{2}\,$ into $\,2\sin \frac{\phi}{2} \, \sin \frac{\theta}{2} + \lambda \sin \frac{\phi+\theta}{2} = 0\,$ gives:
$$ v = -\, \frac{2 \sin \frac{\theta}{2}}{\sin \frac{\phi+\theta}{2}} \quad\quad \text{and} \quad\quad w = -\, \frac{2 \sin \frac{\phi}{2}}{\sin \frac{\phi+\theta}{2}} $$
But, since $\,\phi, \theta \in (0, \pi)\,$ the strict inequalities hold $\cos \frac{\phi}{2},\,\cos \frac{\theta}{2} \lt 1\,$, so:
$$ \left| \sin \frac{\phi+\theta}{2} \right| = \left| \sin \frac{\phi}{2} \cos \frac{\theta}{2} + \cos \frac{\phi}{2} \sin \frac{\theta}{2}\right| \lt \left| \sin \frac{\phi}{2} \right| + \left| \sin \frac{\theta}{2} \right| \quad\implies\quad |v|+|w| \gt 2 $$
Therefore at least one of $|v|,|w|$ would have to be greater than $1\,$, thus out of the allowed range, which completes the proof that the only eligible solution is $\phi = \theta\,$.