 Commuting-involution unitary combination

If A and B are commuting self-adjoint operators with $A^2=B^2=I$, then the displayed combination is a unitary operator. Indeed,and the reverse product is also I. The factors are involutions, and their commutation is essential to cancel the cross terms. Operators $A=U_A\otimes I$ and $B=I\otimes U_B$ on separate tensor factors provide a natural example. More generally, $\cos \theta A+i\sin \theta B$ is unitary for real $\theta$ by the same proof. Replacing i by a real coefficient usually loses this property.