I was working through the operation of the time reversal operator on a spinor as was answered in this question, however, I cannot figure out how this step was done:
$$e^{-i \large \frac{\pi}{2} \sigma_y} = -i\sigma_y.$$
I suspect it has something to do with a taylor series expansion. Here $\sigma_y$ is the pauli matrix which has the form $\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$.
The relation is shown using a taylor series of the exponential: $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ so that $e^{-i\pi/2\sigma_y}$ can be expanded.
$e^{-\frac{i\pi\sigma_y}{2}}=1+\left(-\frac{i\pi\sigma_y}{2}\right)+\frac{\left(-\frac{i\pi\sigma_y}{2}\right)^2}{2!}+\frac{\left(-\frac{i\pi\sigma_y}{2}\right)^3}{3!}+\frac{\left(-\frac{i\pi\sigma_y}{2}\right)^4}{4!}+\frac{\left(-\frac{i\pi\sigma_y}{2}\right)^5}{5!}+...$
Noting that $\sigma_y^2=\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$ then
\begin{equation} \begin{aligned} e^{-\frac{i\pi\sigma_y}{2}}&=1-i\sigma_y(\pi/2)-\frac{(\pi/2)^2}{2!}+i\sigma_y\frac{(\pi/2)^3}{3!}+\frac{(\pi/2)^4}{4!}-i\sigma_y\frac{(\pi/2)^5}{5!}+...\\ &=\bigg\{1-\frac{(\pi/2)^2}{2!}+\frac{(\pi/2)^4}{4!}+...\bigg\}-i\sigma_y\bigg\{(\pi/2)-\frac{(\pi/2)^3}{3!}+...\bigg\}\\ &=\cos(\pi/2)-i\sigma_y\sin(\pi/2)\\ &=-i\sigma_y \end{aligned} \end{equation}
Here the taylor series for cos and sin were used to simplify the infinite sequence: $\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...$ and $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$
First unitarily diagonalize $\sigma_{y}$: $$ \sigma_{y} = U^{\dagger} D U $$ where $U$ is a unitary matrix satisfying $UU^{\dagger} = U^{\dagger} U = \mathbb{I}$. It's always true that $D = \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right]$, and I pick $U = \frac{1}{\sqrt{2}}\left[ \begin{matrix} 1 & 1 \\ i & - i \end{matrix}\right]$ here (notice that $D = \sigma_{z}$, but that doesn't really matter for this calculation).
Before moving on, notice the properties $$ \sigma_{y}^2 = (U^{\dagger} D U)(U^{\dagger} D U) = U^{\dagger}D^2 U $$ which you can generalize to $\sigma_{y}^{n} = U^{\dagger} D^{n} U$ for any $n$. Also notice that taking that $n^{\mathrm{th}}$ power of a diagonal matrix is simple: $$ \left[\begin{matrix} d_1 & 0 \\ 0 & d_2 \end{matrix} \right]^n \ = \ \left[\begin{matrix} d_1^n & 0 \\ 0 & d_2^n \end{matrix} \right] $$ We'll need this in a moment.
You are correct, to take the exponential of a matrix means to take the Taylor Series $$ e^{ - i \alpha \sigma_{y} } \ = \ \sum_{n=0}^{\infty} \frac{(- i \alpha \sigma_{y})^n}{n!} \ = \ \sum_{n=0}^{\infty} \frac{ U^{\dagger}( - i \alpha D)^{n} U }{n!} \ = \ U^{\dagger} \bigg( \sum_{n=0}^{\infty} \frac{( - i \alpha D)^{n} }{n!} \bigg) U $$ but we know how to take various powers of the diagonal matrix $- i \alpha D$: $$ (- i \alpha D)^n \ = \ \left[ \begin{matrix} ( - i \alpha )^{n} & 0 \\ 0 & ( + i \alpha )^{n} \end{matrix} \right] $$ which means that $$ \sum_{n=0}^{\infty} \frac{( - i \alpha D)^{n} }{n!} \ = \ \left[ \begin{matrix} e^{- i \alpha } & 0 \\ 0 & e^{ + i \alpha } \end{matrix} \right] $$
Now we simply have $$ e^{ - i \alpha \sigma_{y} } \ = \ U^{\dagger} \left[ \begin{matrix} e^{- i \alpha } & 0 \\ 0 & e^{ + i \alpha } \end{matrix} \right] U $$ Multiplying this out and simplifying gives $$ e^{ - i \alpha \sigma_{y} } \ = \ \left[ \begin{matrix} \cos(\alpha) & \sin(\alpha) \\ -\sin(\alpha) & \cos(\alpha) \end{matrix} \right] $$
I leave it up to you to plug in $\alpha = \pi/2$.
