Where is the factor of $-i$ in rotation gates coming from?

As I understand it the Pauli-X, Y and Z gates are the same as their rotational gates with a rotation of $$\pi$$. But given the expression for those gates, I find that there is a factor of $$-i$$ in each of them. Can someone explain why?

For example the X-gate:

R_X(\theta)=\begin{align}\begin{bmatrix} \cos\frac{\theta}{2}&-i\;\sin\frac{\theta}{2}\\ -i\;\sin\frac{\theta}{2}&\cos\frac{\theta}{2} \end{bmatrix}\end{align}\\ R_X(\pi)=\begin{align}\begin{bmatrix} 0&-i\\ -i&0 \end{bmatrix}\end{align}\\ X=\begin{align}\begin{bmatrix} 0&1\\ 1&0 \end{bmatrix}\end{align}\\

Let's start from the basics. Any arbitrary single qubit state can be written as

$$|\Psi\rangle = e^{i\gamma} \left(\cos \frac {\theta}{2} |0\rangle+ e^{i\phi} \sin \frac{\theta}{2}|1\rangle\right),$$ where $$\theta, \phi, \gamma \in \Bbb R$$. $$0\leq \theta \leq \pi$$ and $$0\leq \phi\leq 2\pi$$. $$e^{i\gamma}$$ is a global phase here. Qubit states with arbitrary values of $$\gamma$$ represent the same quantum state as global phases don't have any observable effects. So the standard notation for qubit states is

$$|\Psi\rangle = \cos \frac {\theta}{2} |0\rangle+ e^{i\phi} \sin \frac{\theta}{2}|1\rangle$$

where we neglect the global phase factor $$e^{i\gamma}$$.

Remember that you are working on a vector space structure where $$|0\rangle$$ and $$|1\rangle$$ are your basis elements. The rotation operators $$R_X, R_Y$$ and $$R_Z$$ are simply linear maps in this context (in the standard $$\{|0\rangle, |1\rangle\}$$ basis). And so are the Pauli $$X, Y$$ and $$Z$$.

Now, note that you can write $$R_X(\pi)$$ as $$-iX$$. When $$X$$ acts on an arbitrary state $$|\Psi\rangle$$ it will produce $$X|\Psi\rangle$$. If $$-iX$$ acts, it will produce $$-iX|\Psi\rangle$$ instead. However, the states $$-iX|\Psi\rangle$$ and $$X|\Psi\rangle$$ simply differ by a global phase of $$-i$$, or in other words by a factor of $$e^{i\gamma}$$ where $$\gamma = (2n+1)\frac{\pi}{2}$$ and $$n\in \Bbb N$$. So, as you can see, $$R_X(\pi)$$ and $$X$$ are equivalent mappings as global phases don't matter in quantum mechanics!

As for why the Pauli operators and rotation operators differ by a phase factor, you'll have to check out the derivations for the following. You'll find an outline here.

$$R_X(\theta) = e^{-i \frac{\theta}{2}X},$$

$$R_Y(\theta) = e^{-i \frac{\theta}{2}Y},$$

$$R_Z(\theta) = e^{-i \frac{\theta}{2}Z}.$$