Problem
$f(x)=\cos(x)$, $a=\frac{\pi}{2}$. What is the smallest degree taylor polynomial needed so that the approximation for $\cos(\frac{4\pi}{9})$ is correct to five decimal places?
$\cos(x)=$$\sum_{n=0}^{\infty}\frac{f^n(\frac{\pi}{2})}{n!}(x-\frac{\pi}{2})^n$
If we know $|R_n(\frac{4\pi}{9})|<0.00001$ then $|R_n(x)|<0.00001$ for $\frac{4\pi}{9}\leq x\leq \frac{5\pi}{9}$
If we take the absolute value of all derivatives of cosine, the maximum value on $\frac{4\pi}{9}\leq x\leq \frac{5\pi}{9}$ is $1$
$\frac{1}{(n+1)!}|x-\frac{\pi}{2}|^{n+1}<0.00001$ for $\frac{4\pi}{9}\leq x\leq \frac{5\pi}{9}$. The textbook answer is $n=3$ and the approximation is $T_3(\frac{4\pi}{9})$. But when I plug in $1.4$ for $x$ and $3$ for $n$, the inequality $\frac{1}{(n+1)!}|x-\frac{\pi}{2}|^{n+1}<0.00001$ does not hold true. It should though because $1.4$ is on the interval $\frac{4\pi}{9}\leq x\leq \frac{5\pi}{9}$. What am I doing wrong?