Lagrange remainder and Taylor series Use the Lagrange remainder to determine how many terms of the Taylor series of cos expanded
at 0 are needed to estimate $cos(\frac{21\pi}{5})$ . Accuracy to 3 decimal places
So I have a lot of rough work on paper. I know math.stackexchange dosn't really answer assignment questions but I am looking for a head start on even where to start. 
If you guys want, i can type up my notes here. Bascically the f'(c) term from the lagrange formula can be 1. I know my inequality has something to do with < 0.001. I need taylor's expansion (obviously to find the n I need). 
I just can't put it together. This is my first real-analysis course.
 A: The Wikipedia article Lagrange remainder has a good discussion.  The Taylor series is a polynomial that approximates a function near a point, calculated by taking derivatives at the point.  The error term is just that-the difference between the function and the polynomial.  There are various forms, given there, for the error of the Taylor series approximation.  All of them are right, but generally not very useful except as bounds, like you have in this problem.  The difficulty is that you don't know the value of $\xi$, just that it is between $a$ and $x$.
I think one of the points of the problem is to see that when $(x-a)$ gets large you need a lot of terms for accuracy.  If you calculate the number of terms at $21\pi/5$ and then again at $\pi/5$ it will be evident.
A: You are right that the derivative term in the error is bounded by 1, so you just need the $\frac{(x-a)^{(n+1)}}{(n+1)!}$ less than .001.  As $21\pi/5$ is rather large, it will take a lot of terms.
A: To cut down the number of terms it seems reasonable to use
$$\cos \left( \frac {21\pi}{5} \right) = \cos \left( \frac{\pi}{5} \right),$$
so we look for the smallest $n$ such that
$$ \frac{1}{(n+1)!} \left( \frac{\pi}{5} \right)^{n+1} < \frac{1}{1000}.$$
