# Estimate Interval of Validity of $1-\frac{x^2}{2}$ for $\cos(x)$

I have been struggling with the following problem and was wondering if anyone could provide some insight or suggestions:

Use $1-\frac{x^2}{2}$ as an approximation to $\cos(x)$, with an error not greater than 0.0001. Estimate over what interval this will be valid.

My attempts:

I tried to use the Remainder term in the Taylor approximation as follows: $R_n(x)= \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{(n+1)}$, where $c$ is between $a$ and $x$. I used $n=2$ since we are given two terms in the Taylor expansion of $\cos(x)$. Taking the 3rd derivative and applying the formula, I obtain $\frac{\sin(c)}{3!}(x-0)^3$. Here is where I am stuck. I am not sure if I should use this result along with the desired tolerance or if I am still missing something to determine the interval. Also, I think that $| \cos(x)|<1$ would apply somewhere.

Thank you very much for any advice/suggestions. I am using the textbook Introduction to Analysis by Arthur Mattuck.

The Taylor Series for $\cos x$ is $\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$. You want the difference between this and $1-\frac{x^2}{2!}$ to be $<10^{-4}$. Then, for an estimate, just use the next term in the expansion: $$\frac{x^4}{4!}=10^{-4}\Rightarrow x=\pm\sqrt[4]{2.4\times 10^{-3}}\approx .221336$$ Note that the next term, $\frac{x^6}{6!}\approx 1.633\times 10^{-7}$, is negligible.
Since the Taylor polynomial of $\cos x$ has a null third term, you may as well use a fourth degree remainder. Then from $|\cos\xi|\leqslant 1$ you get that $|R_{4,0}(x)|\leqslant \dfrac{x^{4}}{4!}$. Then you want that $$\dfrac{x^{4}}{4!}\leqslant 10^{-4}$$ or $$x^4\leqslant 0.0024$$ Thus $|x|\leqslant 0.22(133638394...)$. Here's a plot to confirm the estimation:
$\hspace{0.5 cm}$