Taylor series of $\sin(x)$ converges uniformly on $[-\pi,\pi]$? According to my notes, the Taylor series of $\sin(x)$ converges uniformly on $[-\pi,\pi]$.
I know that the remainder term needs to converge uniformly to $0$ for this to be the case.
But I really don't know how to begin showing that this series converges uniformly. I think it's the domain that really stumps me. I think I should start showing that the remainder term converges to $0$. So we have:
$$R_n= \frac{(x-x_0)^{N+1}}{N!}\int_0^1 (1-t)^Nf^{(N+1)}(x_0+t(x-x_0))dt$$
Where $R_n$ denotes the remainder term.
What should I do?
Thanks in advance.

Since all of the derivatives of $\sin(x)$ satisfy $$|f^{(N+1)}(x)| \le 1$$ for all $x$, we see that $$|R_n| \le \frac{|x-x_0|^{N+1}}{N!} \le \frac{(2\pi)^{N+1}}{N!}$$ and the term on the right converges to zero independently of $x$. Thus we can conclude that the Taylor series converges uniformly.
Here we used that the integrand is bounded in absolute value by 1.
