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 $\sin(\xi)$

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 $x=2$.
