[tex]f(x)=\sin x\implies f(\pi)=0[/tex]
[tex]f'(x)=\cos x\implies f'(\pi)=-1[/tex]
[tex]f''(x)=-\sin x\implies f''(\pi)=0[/tex]
[tex]f'''(x)=-\cos x\implies f'''(\pi)=1[/tex]
and so on, so that the Taylor series about [tex]x=\pi[/tex] takes the form
[tex]-(x-\pi)+\dfrac{(x-\pi)^3}{3!}-\dfrac{(x-\pi)^5}{5!}+\dfrac{(x-\pi)^7}{7!}-\cdots=\displaystyle\sum_{n=1}^\infty\frac{(-1)^n(x-\pi)^{2n-1}}{(2n-1)!}[/tex]
The ratio test gives
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}(x-\pi)^{2n+1}}{(2n+1)!}}{\frac{(-1)^n(x-\pi)^{2n-1}}{(2n-1)!}}\right|=(x-\pi)^2\lim_{n\to\infty}\frac{(2n-1)!}{(2n+1)!}=0[/tex]
which means the series converges everywhere, so the radius of convergence is infinite.
