Answer
$\dfrac{\pi x}{2}+\dfrac{\pi^3}{2^3 \cdot 3!}x^3-\dfrac{\pi^5}{2^5 \cdot 5!}x^5-\dfrac{\pi^7}{2^7 \cdot 7!}x^7+...$
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)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Now, $ \sin(\dfrac{\pi x}{2})=(\dfrac{\pi x}{2})-\dfrac{(\dfrac{\pi x}{2})^3}{3!}+\dfrac{(\dfrac{\pi x}{2})^5}{5!}-\dfrac{(\dfrac{\pi x}{2})^7}{7!}+...]$
or, $=\dfrac{\pi x}{2}+\dfrac{\pi^3}{2^3 \cdot 3!}x^3-\dfrac{\pi^5}{2^5 \cdot 5!}x^5-\dfrac{\pi^7}{2^7 \cdot 7!}x^7+...$
