To find an expression for $\sin A \sin B$ note first $\sin A \sim \pm e^{iA} \pm e^{-iA}$ so that products $\sin A \sin B \sim \pm e^{ i (A \pm B)}$, and we must have
\begin{align}
\sin A \sin B = \sin B \sin A
\end{align}
which means the form of the formula should not change at all under interchanges, so the right-hand side must be expressible in terms of $\cos$ alone. Thus we immediately have
\begin{align}
\sin A \sin B = \frac{1}{2} [ \pm \cos (A + B) \pm \cos(A - B)].
\end{align}
For $A = B = \frac{\pi}{2}$ we have