Lagrange Error Bound BC only

The Lagrange bound is $\left|f(x) - P_n(x)\right| \le \frac{M}{(n+1)!}\left|x-a\right|^{n+1}$, where the derivative order, the factorial and the power are all $n+1$ and move together, and $M$ bounds $\left|f^{(n+1)}\right|$ everywhere between the centre and the point. That interval requirement follows from the remainder theorem, which guarantees some $c$ between $a$ and $x$ with $f(x) - P_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ without saying which: not knowing $c$ is exactly why a bound is needed, so evaluating the derivative at either endpoint can produce a number that is not a bound at all. $M$ can come from a universal bound as for $\sin$ and $\cos$, from monotonicity as for $e^{x}$, or from any convenient over-estimate, since a larger $M$ stays valid.

Bounding the error in approximating $\cos 2$ by $P_2$ gives $\frac{1}{3!}\left|2\right|^{3} = \frac{4}{3} \approx 1.3333$ with $M = 1$, against a true error of $\left|\cos 2 + 1\right| \approx 0.5839$. The factor $\left|x-a\right|^{n+1}$ is where both halves of "good near its centre" live, since it grows quickly with distance and shrinks quickly with degree. Choosing between this and the alternating bound is a matter of what is available: an alternating series with decreasing terms gives a bound that is a single already-written term, while a Taylor approximation without useful alternation needs Lagrange. Being handed a bound on a derivative means Lagrange, since the alternating bound contains no derivative; being asked about a numerical series rather than a function approximation means the alternating one, since there is no function to differentiate.