Determine the values of x for which the approximation cos

Determine the values of $x$ for which the approximation $\cos x\approx1-\dfrac{x^2}{2!}=1-\dfrac{x^2}2$ holds with an absolute accuracy of $0.0001$.

Step 1: Identify the remainder

From $\cos x=1-\tfrac{x^2}2+\tfrac{x^4}{24}-\cdots$, the remainder after the quadratic term is, by the Lagrange form (the $x^3$ coefficient vanishes, so the effective order is 4):

\[ R_2(x) = \frac{\cos\xi}{24}x^4, \qquad |R_2(x)| \le \frac{x^4}{24}. \tag{1} \]

Step 2: Impose the accuracy condition

\[ \frac{x^4}{24} \lt 10^{-4} \;\Longrightarrow\; x^4 \lt 2.4\times10^{-3} \;\Longrightarrow\; |x| \lt (2.4\times10^{-3})^{1/4} \approx 0.2213. \tag{2} \]

Step 3: Verification

\[x=0.22\text{: }(0.22)^4/24\approx9.74\times10^{-5}\lt 10^{-4}\ \checkmark;\qquad x=0.23\text{: }(0.23)^4/24\approx1.17\times10^{-4}\gt 10^{-4}\text{ (too large)}\]