Answer
$(\dfrac{x}{2})-\dfrac{1}{2^3 \cdot 3!} x^3+\dfrac{1}{2^{5} \cdot 5!}x^5-...$
Work Step by Step
Consider the Maclaurin Series for $\sin x$ as follows:
$ \sin (x)=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$
Plug $\dfrac{x}{2}$ instead of $x$ in the above series.
This implies that $ \sin (\dfrac{x}{2})=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\dfrac{x}{2})^{2n+1}}{(2n+1)!}$
and $ \sin (\dfrac{x}{2})=(\dfrac{x}{2})-\dfrac{1}{2^3 \cdot 3!} x^3+\dfrac{1}{2^{5} \cdot 5!}x^5-...$
