Error in approximation of nonzero root to $x^2=\sin x$ using Taylor's cubic polynomial. I have successfully obtained the root's approximation $r=\sqrt{15}-3$ as I'm supposed to as following:$$\begin{align}
\displaystyle f(x)=\sin x &= x - \frac{x^3}{6} + E\\
x^2 &\approx x - \frac{x^3}{6}\\
r^2 &= r - \frac{r^3}{6}\\
r &= \sqrt{15}-3
\end{align}$$
Given $r=\sqrt{15}-3<0.9$, show that $|\sin r-r^2|<\frac{1}{200}$. I have tried but of no avail like this:
$$
\displaystyle f(r)=\sin r = r - \frac{r^3}{6} + E\\
|\sin r - r^2| = |\sin r-(r-\frac{r^3}{6})|=|E|=E\\
E>0\text{ because } r>a=0\text{ and } f^4(x)=\sin x > 0 \text{ on }(0,\pi)\text{ containing }r\\
\text{Because }f^4(r)=\sin r <\sin 0.9\\
\therefore E<\sin0.9\frac{r^4}{4!}\approx 0.01895639019\\
\frac{1}{200}=0.005\\
$$

Hint: It's often worth noticing that for an odd function $f$, the third and fourth degree Taylor polynomials of $f$ at $0$ are the same.
