Finding for what $x$ values the error of $\sin x\approx x-\frac {x^3} 6$ is smaller than $10^{-5}$ 
Find for what $x$ values the error of $\sin x\approx x-\frac {x^3} 6$ is smaller than $10^{-5}$

I thought of two ways but got kinda stuck:


*

*Since we know that $R(x)=f(x)-P(x)$ then we could solve: $\sin x-x+\frac {x^3} 6<10^{-2}$ but I have no idea how to do it, deriving the expression to a more simple expression makes it loose the information on $10^{-5}$.

*Find for what $x,c$ we have $R_5(x)=\frac {\cos (c)x^5}{5!}<10^{-5}$ but can I simply choose $c=0$? then the answer would be: $x<\frac {(5!)^{1/5}}{10}$ which looks like it's right from the graph of $R(x)$ in 1. But why can I choose $c$? what if it was a less convenient function?
Note: no integrals.
 A: Since
$$\sin x =\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)!} $$
for any $x$ such that $|x|\leq 1$ we have:
$$\left|\sin x-x+\frac{x^3}{6}\right|\leq \frac{|x|^5}{5!}.$$
Provided that $\frac{|x|^5}{5!}\leq 10^{-5}$ holds, we have a good approximation. So a sufficient condition (but not a necessary one) to have $\left|\sin x-x+\frac{x^3}{6}\right|\leq 10^{-5}$ is $ |x|\leq 0.26$ or just:
$$ |x|\leq\frac{1}{4}.$$
A: I like your 2. You can't "simply" choose $c=0$, but whatever $c$ is, you know that $|\cos c|\leq 1$, and so 
$$\left| R_5(x)\right|= \left|\cos(c)\frac{x^5}{5!}\right|=\left|\cos(c)\right|\left|\frac{x^5}{5!}\right|\leq \frac{|x|^5}{5!}$$
A: Consider
$$
\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots.
$$
This series is alternating, and the terms are strictly decreasing in magnitude if $|x| < 1$. So we get 
\begin{align*}
\text{estimate} - \text{ reality} &= x - \frac{x^3}{6} - \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots\right) \\
&=\frac{x^5}{120} - \ldots \\
&\leq \frac{x^5}{120} 
\end{align*}
The last estimate works only because the terms are strictly decreasing in magnitude. So it suffices to take $x$ satisfying
$$
-10^{-5} < \frac{x^5}{120} < 10^{-5}
$$
But this is not a necessary condition. 
