Prove $ \left |\sin(x) - x + \frac{x^3}{3!} \right | < \frac{4}{15}$ Prove $ \left |\sin(x) - x + \dfrac{x^3}{3!} \right | < \dfrac{4}{15}$ $\forall x \in [-2,2]$
By Maclaurin's formula and Lagrange's remainder we have $\sin(x)  = x - \dfrac{x^3}{3!} + \dfrac{\sin(\xi)}{5!}x^5$ for some $0<\xi<2$
subbing this in we get $\left|\dfrac{\sin(\xi)}{5!}x^5 \right| \leq \left |\dfrac{x^5}{5!} \right| \leq \dfrac{2^5}{5!} = \dfrac{4}{15}$, but the question has $<$ rather than $\leq$ - where have I done wrong?
edit: thinking the $\cos(\xi)$ should be there rather than $\sin(\xi)$
 A: Use the exact form of the Taylor formula:
$$
\sin x - x  + \frac{x^3}3 = \int_0^x \frac{(x-t)^4}{4!}\cos(t) dt
\\
\left| \sin x - x  + \frac{x^3}3 \right| = 
\left| \int_0^x \frac{(x-t)^4}{4!}\cos(t) dt \right|
\le\int_0^x  \left| \frac{(x-t)^4}{4!}\cos(t) \right|dt \\
\le\int_0^x \left| \frac{(x-t)^4}{4!} \right|dt 
= \int_0^x \frac{t^4}{4!} dt = \frac{2^5}{5!} 
$$
Now if there is equality anywhere, every inequality becomes an equality, but considering the first implies that $x=0$, and the last implies that $x=2$.
A: From the Leibniz rule it is known that, if the sequence of $\frac{x^4}{5!},\frac{x^6}{7!},\frac{x^8}{9!},...$ is decreasing, which is the case for $x^2<6\cdot7=42$, $|x|\le6$ to get a round number, then
$$
0\le\frac{x^4}{5!}-\frac{x^6}{7!}
\le \frac{\sin x}{x}-1+\frac{x^2}{3!}
\le\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}
=\frac{x^4}{5!}-\frac{x^6}{7!}\left(1-\frac{x^2}{72}\right).
$$
Since under the assumed restrictions $1-\frac{x^2}{72}\ge\frac12$, we get
$$
\left|\sin x-x+\frac{x^3}{3!}\right|\le\frac{|x|^5}{5!}\left(1-\frac{x^2}{84}\right)<\frac{|x|^5}{5!}
$$
for $0<|x|<6$.
