quantum mechanics - Matrix exponentiation of Pauli matrix

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 $\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$.

Answer

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.

$e^{-i\pi/2\sigma_y}=1+(-i\pi/2\sigma_y)+\frac{(-i\pi/2\sigma_y)^2}{2!}+\frac{(-i\pi/2\sigma_y)^3}{3!}+\frac{(-i\pi/2\sigma_y)^4}{4!}+\frac{(-i\pi/2\sigma_y)^5}{5!}+...$

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

$$\begin{equation} \begin{aligned} e^{-i\pi/2\sigma_y}&=1-i\sigma_y(\pi/2)-\frac{(\pi/2)^2}{2!}+i\sigma_y\frac{(\pi/2)^3}{3!}+\frac{(\pi/2)^4}{4!}-i\sigma_y\frac{(\pi/2)^5}{5!}+...\\ &=\bigg\{1-\frac{(\pi/2)^2}{2!}+\frac{(\pi/2)^4}{4!}+...\bigg\}-i\sigma_y\bigg\{(\pi/2)-\frac{(\pi/2)^3}{3!}+...\bigg\}\\ &=\cos(\pi/2)-i\sigma_y\sin(\pi/2)\\ &=-i\sigma_y \end{aligned} \end{equation}$$

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 $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$