Answer
$sin\frac{\pi}{2}=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(x-\frac{\pi}{2})^{2n}}{(2n)!}$
The radius of convergence is $R=\infty$.
Work Step by Step
Taylor series centered at $a= \frac{\pi}{2}$ is
$1-\frac{1}{2!}(x-\frac{\pi}{2})^{2}+\frac{1}{4!}(x-\frac{\pi}{2})^{4}+....$
$sin\frac{\pi}{2}=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(x-\frac{\pi}{2})^{2n}}{(2n)!}$
The radius of convergence is $R=\infty$.
