How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$? How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$?
where $_{0}F_{1}$ is the hypergeometric series?

The function $_0F_1(;a;z)$ is a solution to the differential equation
\begin{equation}\tag{1}
  z \frac{d^2}{dz^2}y(z)+a\frac{d}{dz}y(z)= y(z).
\end{equation}
Knowing this, consider the function $u(x)=x y(-x^2/4)$ and take its second derivative. We obtain
$$\tag{2} \frac{d^2}{dx^2}u(x)=x\left[\frac{x^2}{4}y''\left(\frac{x^2}{4}\right)-\frac{3}{2}y'\left(\frac{x^2}{4}\right)\right] $$
Now choose $y(z)=$$_0F_1(;3/2;z)$, so that it satisfies eq.(1) with $a=3/2$ and $z=-x^2/4$. Using it we obtain
$$ \frac{d^2}{dx^2}u(x)=-xy\left(\frac{x^2}{4}\right)=-u(x) \tag{3}.$$
The general solution to (3) is $u(x)=\alpha \sin(x)+\beta \cos(x).$
Since $u$ is odd $u(x)=-u(x)$, $\beta=0$. Finally,
$$ \alpha  = \lim_{x\to0} \frac{u(x)}{x} =\; _0F_1\left(;\frac{3}{2};0\right)=1$$
so
$$ u(x)= \sin(x) = x  \; _0F_1\left(;\frac{3}{2};-\frac{x^2}{4}\right)$$
