Showing $ | 1 - \frac{x^2}{2} - \cos x | \leq \frac{1}{24}$ using Taylor Use Taylor's theorem to show that for $x \in [-1, 1] $ we have that $$ \left| 1 - \frac{x^2}{2} - \cos x \right| \leq \frac{1}{24}. $$
Attempt: Since $\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \ldots $, I see that $1 - \frac{x^2}{2} = P_2 (x)$ is the second order Taylor polynomial of $f(x) = \cos(x)$ around $x_0 = 0$. So we have that $\cos(x) =  P_2(x) + r(x)$, where $r(x)$ is the remainder term. But I'm not sure how to use this remainder term to get a proper upper bound for $ | 1 - \frac{x^2}{2} - \cos x | \leq \frac{1}{24}$.

Since the term in $x^3$ is $0$, your approximation $P_2(x)$ can also be thought of as $P_3(x)$.
By the Lagrange form of the remainder, the absolute value of the error in $P_3(x)$ is $\frac{1}{4!}|\cos(c_x)||x|^4$ for some $c_x$ between $0$ and $x$. We have $|\cos(c_x)|\le 1$, and $|x|\le 1$.
