Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 550: 45

$$\dfrac{\sqrt 3}{2}$$

The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....; |x| \lt \infty $ We have $\\=\dfrac{\pi}{3} - \dfrac{\pi^3}{3^3 \cdot 3!} + \dfrac{\pi^5}{3^5 \cdot 5!}-....\\=\dfrac{\pi}{3}-( \dfrac{1}{3!}) (\dfrac{\pi}{3})^3+( \dfrac{1}{5!}) (\dfrac{\pi}{3})^5-....\\=\sin (\dfrac{\pi}{3})\\=\dfrac{\sqrt 3}{2}$