The scattering angle advance $\mathrm{\Delta }$ and the deflection relation: the total polar advance is $2\mathrm{\Delta }$ and the velocity directions make the cosine $\cos (\pi -2\mathrm{\Delta })$

Outgoing half-angle: on $[t_{\ast },\mathrm{\infty })$, $\theta$ is $C^1$ and strictly increasing with ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)>0$, and $\rho$ is a $C^1$ strictly increasing bijection of $[t_{\ast },\mathrm{\infty })$ onto $[r_{\ast },\mathrm{\infty })$; substituting $\rho$ for $t$ first on compact subintervals of $(t_{\ast },\mathrm{\infty })$, where $\rho \mathrm{\prime }>0$, and then taking endpoint limits [L3] gives $\theta (\mathrm{\infty })-\theta (t_{\ast })={\int }_{t_{\ast }}^{\mathrm{\infty }}{\theta }^{\mathrm{\prime }}(t)\mathrm{d}t={\int }_{r_{\ast }}^{\mathrm{\infty }}\frac{J}{\mu s^2}\frac{\mathrm{d}s}{{\rho }^{\mathrm{\prime }}({\rho }^{-1}(s))}={\int }_{r_{\ast }}^{\mathrm{\infty }}\frac{J\mathrm{d}s}{s^2\sqrt{2\mu (E-V(s))}}=\mathrm{\Delta },$ using ${\rho }^{\mathrm{\prime }}=\sqrt{(2\mathrm{/}\mu )(E-V)}$ from [L1]; the integral converges at $r_{\ast }$ because $E-V(s)$ is comparable to $s-r_{\ast }$ there and diverges nowhere else, and it is positive.

Incoming half-angle: the same computation on the decreasing branch $(-\mathrm{\infty },t_{\ast }]$, where ${\rho }^{\mathrm{\prime }}=-\sqrt{(2\mathrm{/}\mu )(E-V(\rho ))}$ and the substitution reverses the sign, gives $\theta (t_{\ast })-\theta (-\mathrm{\infty })=\mathrm{\Delta }$; adding the two halves gives $\theta (\mathrm{\infty })-\theta (-\mathrm{\infty })=2\mathrm{\Delta }$, which is clause 1.

Direction limits: on the outgoing tail [L5], $r(t)\mathrm{/}\parallel r(t){\parallel }_2\to v_{+}\mathrm{/}\mathrm{\mid }{v}_{+}\mathrm{\mid }$ as $t\to \mathrm{\infty }$ (an affine curve with nonzero velocity direction tends to its velocity direction), and $r(t)\mathrm{/}\rho (t)=n(\theta (t))\to n(\theta (\mathrm{\infty }))$ because $\theta$ has a finite limit and $n$ is continuous; hence $n(\theta (\mathrm{\infty }))=v_{+}\mathrm{/}\mathrm{\mid }{v}_{+}\mathrm{\mid }$. On the incoming tail, $t-t_2<0$ tends to $-\mathrm{\infty }$, so $r(t)\mathrm{/}\parallel r(t){\parallel }_2\to -v_{-}\mathrm{/}\mathrm{\mid }{v}_{-}\mathrm{\mid }$, while $r(t)\mathrm{/}\rho (t)\to n(\theta (-\mathrm{\infty }))$; hence $n(\theta (-\mathrm{\infty }))=-v_{-}\mathrm{/}\mathrm{\mid }{v}_{-}\mathrm{\mid }$. This is clause 2.

Rotation form: by [L4], $v_{+}\mathrm{/}\mathrm{\mid }{v}_{+}\mathrm{\mid }=n(\theta (\mathrm{\infty }))=n(\theta (-\mathrm{\infty })+2\mathrm{\Delta })=R_{2\mathrm{\Delta }}(n(\theta (-\mathrm{\infty })))=-R_{2\mathrm{\Delta }}(v_{-}\mathrm{/}\mathrm{\mid }{v}_{-}\mathrm{\mid })$, the first equality of clause 3.

Cosine: writing $u:=v_{-}\mathrm{/}\mathrm{\mid }{v}_{-}\mathrm{\mid }$, the last identity gives $⟨v_{-},v_{+}⟩\mathrm{/}(\mathrm{\mid }{v}_{-}\mathrm{\mid }\mathrm{\mid }{v}_{+}\mathrm{\mid })=⟨u,-R_{2\mathrm{\Delta }}u⟩=-\cos 2\mathrm{\Delta }$, because $⟨u,R_{2\mathrm{\Delta }}u⟩=\cos 2\mathrm{\Delta }\mathrm{\mid }u{\mathrm{\mid }}^2+\sin 2\mathrm{\Delta }⟨u,e_3\times u⟩=\cos 2\mathrm{\Delta }$ by [L4]. Since $\cos (\pi -2\mathrm{\Delta })=-\cos 2\mathrm{\Delta }$, this gives the displayed cosine; if $0<\mathrm{\Delta }<\pi \mathrm{/}2$ then $\pi -2\mathrm{\Delta }\in (0,\pi )$ and the identity says that the angle between the two nonzero velocity vectors (the unique angle in $[0,\pi ]$ whose cosine is the displayed value, cosine being strictly decreasing there) is $\pi -2\mathrm{\Delta }$.