Find a value of $h$ such that if $\left | x \right |< h$ then $\sin(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+R$ where $\left | R \right |< 10^{-4}$. I need to find a value of $h$ such that if $\left | x \right |< h$ then
$$\sin(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+R$$
where $\left | R \right |< 10^{-4}$.
Attempt: I tried to start with $\left | \sin(x) \right |\leq 1$, then using triangle inequality to $\left | x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+R \right |\leq 1$, but I'm not sure this is the correct way.
I would appreciate a hint or an answer. Thank you for your time.
 A: As $$x-\frac{x^3}{3!} + \frac{x^5}{5!}$$  is the taylor polynom of order 5, we know that 
$$\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} + R(x)$$ with
$R(x)=-\frac{\sin(\xi)}{6!} x^6 $ for a $\xi \in (0,x)$ (as the sixth derivative of $\sin(x)$ is $-\sin(x)$. Now we use use that $|\sin(x)|<1$ 
$$|R(x)| \leq \frac{x^6}{6!}=\frac{x^6}{720}$$ 
Now just find a $x$ for which this is smaller than $10^{-4}$.
As Maesumi mentioned you could interpret it as taylor polynom of order $6$ which gives you in nearly the same way 
$$|R(x)|\leq \frac{|x^7|}{7!} = \frac{|x^7|}{5040}$$
Here I plotted $\frac{x^6}{540}$ in green and the other one in red, as long as you are below zero, $|R(x)|<10^{-4}$
A: We will ultimately ask for a stronger condition than $|x|\le 1$. For $|x|\le 1$, our series is an alternating series. Thus the absolute value of the error when we truncate is less than the absolute value of the first neglected term. In this case, the first neglected term has absolute value $\dfrac{|x|^7}{7!}$. So choosing $h=(5040\times 10^{-4})^{1/7}$ will do. 
Remark: In this case, the alternating series criterion gives no stronger a result than the Lagrange form of the remainder. But fairly often, it is difficult to estimate the appropriate derivatives, and the Lagrange form of the estimate produces estimates that are excessively pessimistic. 
