<s>[INST] 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?
 [/INST] One may recall the definition of the generalized hypergeometric function
$$
{}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z): = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}
$$ which gives
$$
{}_0F_1\left(-;\frac{3}{2};-\frac{x^{2}}{4}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{\left(\frac32\right)_n} \, \frac {x^{2n}} {4^n \:n!}.
$$ Then one may check that
$$
\begin{align}
\frac1{\left(\frac32\right)_n\:4^n \:n!}&=\frac1{\left(\frac32\right)\left(\frac32+1\right)\left(\frac32+2\right)\cdots\left(\frac32+n-1\right)\:4^n \:n! }
\\\\&=\frac{2^n}{3\times 5\times 7 \cdots \times (2n+1)\:4^n \:n! }
\\\\&=\frac{2^n\times \color{blue}{2\times 4\times 6 \cdots \times (2n)} }{\color{blue}{2}\times3\times \color{blue}{4} \times5\times \color{blue}{6}\times7 \cdots \times \color{blue}{2n}\times (2n+1)\:4^n \:n! }
\\\\&=\frac{2^n\times 2^n\times n! }{(2n+1)!\:4^n \:n! }
\\\\&=\frac1{(2n+1)!}
\end{align}
$$ giving 

$$
x\:{}_0F_1\left(-;\frac{3}{2};-\frac{x^{2}}{4}\right) = \color{blue}{\sum_{n=0}^\infty (-1)^n\frac {x^{2n+1}} {(2n+1)!}}=\color{red}{\sin x}
$$

as announced.
</s>