Answer
$\dfrac{x}{2}-\dfrac{x^3}{2^3 \cdot 3!}+\dfrac{x^5}{2^{5} \cdot 5!}-$
Work Step by Step
Since, we know that the Maclaurin Series for $\sin{x}$ is defined as:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$
We need to replace in the above series $x$ with $x/2$.
we have
$ \sin \dfrac{x}{2}=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\dfrac{x}{2})^{2n+1}}{(2n+1)!}=\dfrac{x}{2}-\dfrac{x^3}{2^3 \cdot 3!}+\dfrac{x^5}{2^{5} \cdot 5!}-$