<s>[INST] 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:
$$
\displaystyle f(r)=\sin r = r - \frac{r^3}{6} + E\\
|\sin r - r^2| = |\sin r-(r-\frac{r^3}{6})|=|E|=E\\
E>0\text{ because } r>a=0\text{ and } f^4(x)=\sin x > 0 \text{ on }(0,\pi)\text{ containing }r\\
\text{Because }f^4(r)=\sin r <\sin 0.9\\
\therefore E<\sin0.9\frac{r^4}{4!}\approx 0.01895639019\\
\frac{1}{200}=0.005\\
$$
 [/INST] The series for $\sin x$ is an alternating series. So for numbers $x$ near $r$, we have $\sin x\lt x-\frac{x^3}{3!}+\frac{x^5}{5!}$. 
Since $r^2=r-\frac{r^3}{3!}$, the absolute value of the difference is $\lt \frac{(0.9)^5}{5!}$. This brings is a little below $\frac{1}{200}$. Using $\sqrt{15}-3$ instead of $0.9$ brings us more cleanly under $\frac{1}{200}$.  
The Lagrange remainder term when we cut off the expansion of $\sin x$ after the (zero) term in $x^4$ yields the same result.
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