Answer
$$\dfrac{\sqrt 3}{2}$$
Work Step by Step
Recall the Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....; |x| \lt \infty $
or,$=\dfrac{\pi}{3} - \dfrac{\pi^3}{3^3 \cdot 3!} + \dfrac{\pi^5}{3^5 \cdot 5!}-....$
or, $=(\dfrac{\pi}{3})-( \dfrac{1}{3!}) (\dfrac{\pi}{3})^3+( \dfrac{1}{5!}) (\dfrac{\pi}{3})^5-...$
or, $=\sin (\dfrac{\pi}{3})$
or, $=\dfrac{\sqrt 3}{2}$
