= Solution
Here the dimensionless <spin one-half> operator is $S_x=\sigma_x/2$. Since the <Pauli matrix> obeys $\sigma_x^2=I$, the even and odd terms of the <matrix exponential> give
$$
e^{-i\theta S_x}=\sum_{m=0}^{\infty}\frac{(-1)^m(\theta/2)^{2m}}{(2m)!}I
-i\sum_{m=0}^{\infty}\frac{(-1)^m(\theta/2)^{2m+1}}{(2m+1)!}\sigma_x
=I\cos\frac\theta2-i\sigma_x\sin\frac\theta2.
$$
This is the <exponential of a Pauli vector> for an $x$-axis <rotation>. With a physical <spin angular momentum> $S_x^{\rm phys}=\hbar\sigma_x/2$, the exponent is $-i\theta S_x^{\rm phys}/\hbar$.
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