This question refers to the Compton Scattering.
We have an elastic impact between a photon and an electron, so conservation of $E$ and $\vec{p}$ in a 2D plane:
$$\begin{cases}E^i_p+E^i_e=E^f_p+E^f_e \\ \vec{p}^i_p+\vec{p}^i_e=\vec{p}^f_p+\vec{p}^f_e\end{cases}. \tag{1}$$
this system of two equation can be explicited, keeping in mind that we want to use relativistic mechanics:
$$\begin{cases}h\nu+mc^2=h\nu '+m\gamma c^2 \\ \frac{h\nu}{c}=\frac{h\nu '}{c}\cos\phi+m\gamma v \cos \psi \\ 0=\frac{h \nu '}{c}\sin \phi - m\gamma v \sin \psi\end{cases}. \tag{2}$$ where $\phi , \psi$ are the angles of ejection of the photon and the electron. Easy enough. Solving this system we should get: $$\lambda ' -\lambda = 2 \lambda _c \sin^2\left(\frac{\phi}{2}\right) \tag{3}$$ where $\lambda _c = h/mc$.
My question is: I find the resolution of the system (2) difficult and laborious, seems to me a really tedious task to get (3) from (2). Is there some smart way of going about this? Some shortcut we can take to get easily (3) from (2)? Any help or tip would be appreciated.
Yes; it is a one-liner. (2) is gratuitous, if not sadistic.
You simply move $E_p^f$ and $\vec p_p^f$ to their respective l.h.sides in (1), square both resulting equations, and subtract the second from the first, to obtain (3), $$ (E^i_p+mc^2-E^f_p)^2- (\vec p_p^i-\vec p_e^f)^2=mc^2, \leadsto \\ mc p_p^i-p_p^ip_p^f -mc p_p^f -\vec p_p^i\cdot \vec p_p^f=0 ~~~\leadsto \\ p_p^i - p_p^f =\frac{p_p^i p_p^f (1-\cos\phi)}{mc} ~~\implies \\ \lambda ' -\lambda = \lambda_c (1-\cos\phi), $$ given $p=h/\lambda$, mutatis mutandis for the ', and $mc=h/\lambda_c$. Further recall $E_p=cp_p$ and $E_e= \sqrt{m^2 c^4+p_e^2 c^2}$.