= Solution
Each <photon> has energy $\varepsilon=\hbar\omega$. Their total <four-momentum> is
$$
P=\frac\varepsilon c(2,1+\cos\theta,\sin\theta,0).
$$
Its squared invariant is $P^2=2\varepsilon^2(1-\cos\theta)/c^2$. Therefore the system's <rest energy> is
$$
\boxed{Mc^2=\sqrt{E_{\rm tot}^2-c^2|\mathbf p_{\rm tot}|^2}
=2\hbar\omega\,\left|\sin\frac\theta2\right|.}
$$
Two individually massless <particles> can thus have positive total <invariant mass>. Back-to-back photons give rest energy $2\hbar\omega$, whereas exactly parallel photons have zero invariant mass and no massive-system rest frame.
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