= Solution
Let the <photon> momenta be $\mathbf p_1,\mathbf p_2$, with $|\mathbf p_i|=E_i/c$. <Four-momentum conservation> gives total energy $E_1+E_2$ and total momentum $\mathbf p_1+\mathbf p_2$ for the <pion>. Its <mass shell> relation therefore gives
$$
m_\pi^2c^4=(E_1+E_2)^2-c^2|\mathbf p_1+\mathbf p_2|^2=2E_1E_2(1-\cos\theta)=4E_1E_2\sin^2(\theta/2).
$$
Since $0\le\theta\le\pi$, the <sine> is nonnegative, and taking the positive <square root> yields
$$
\boxed{\sin\frac\theta2=\frac{m_\pi c^2}{2\sqrt{E_1E_2}}.}
$$
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