Q:

Taylor Polynomial And Reminder In Lagrange Form

What is the error of Taylor polynomial of $\sin(x)$ of order 4 around $x_0=0$ at the point $-1$?

So the Taylor polynomial is $x-\frac{x^3}{3!}$ and the remainder in  Lagrange form is $$\frac{\cos(c)}{5!}\cdot x^5$$
So $-1\leq c\leq 1$ and therefore it is $\frac{\cos(c)}{5!}\cdot x^5\leq \frac{1}{120}$ so the error can be at most $+/-\frac{1}{120}$?

A:

If it just ask you to find the error, it should be
$$|\sin{(-1)}-((-1)-\frac{(-1)^3}{3!})|$$
If it asks you to find the error bound
$$\left|\frac{\cos{c}}{5!}(-1)^5\right|\leq \frac{1}{120}$$
since $-1\leq c\leq 0$.
Edit: As pointed out by @Bernard, the error is defined by 
$$f-T_n$$
In this case
$$\sin{(-1)}-((-1)-\frac{(-1)^3}{3!})$$

