= Solution
From $A^2=I$, induction gives $A^{2n}=I$ and $A^{2n+1}=A$. The <matrix exponential> converges absolutely in finite dimension, so its even and odd terms may be separated:
$$
e^{-i\theta A}
=\left[\sum_{n\geq0}\frac{(-1)^n\theta^{2n}}{(2n)!}\right]I
-i\left[\sum_{n\geq0}\frac{(-1)^n\theta^{2n+1}}{(2n+1)!}\right]A
=\boxed{\cos\theta\,I-i\sin\theta\,A}.
$$
This <involution exponential formula> requires no Hermiticity assumption on $A$. Hermiticity is needed if the exponential is additionally asserted to be unitary.
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