= Solution
Because $(\gamma_5)^2=\mathbf1$, its even and odd powers are $\gamma_5^{2r}=\mathbf1$ and $\gamma_5^{2r+1}=\gamma_5$. Separate the <matrix exponential> into these two <power series>:
$$
\begin{aligned}
e^{i\alpha\gamma_5}
&=\mathbf1\sum_{r=0}^\infty\frac{(i\alpha)^{2r}}{(2r)!}
+\gamma_5\sum_{r=0}^\infty\frac{(i\alpha)^{2r+1}}{(2r+1)!}\\
&=\boxed{\mathbf1\cos\alpha+i\gamma_5\sin\alpha.}
\end{aligned}
$$
No representation of the <gamma matrices> is needed. Hermiticity of $\gamma_5$ also makes this exponential unitary for real $\alpha$, while the projectors $(1\pm\gamma_5)/2$ acquire opposite phases.
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