Answer
$0$
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)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
As we are given that $x=\pi$
Then, we have $\sin (x)=\sin \pi =0$
