# Explaining and using the $N$-term Taylor series for $\sin x$

So I'm given the Taylor Series expansion of the sine function and I've been asked to prove it (Done) and then construct the following by my lecturer:

Explain why the Taylor series containing $N$ terms is: $$\sin x = \sum_{k=0}^{N-1} \frac{(-1)^k}{(2k+1)!}x^{2k+1}+r_{2N-1}(x)$$ with a remainder $r_{2N-1}(x)$ that satisfies: $$|r_{2N-1}(x)| \leqslant \frac{|x|^{2N}}{(2N)!}$$

How many terms of the series do you need to include if you want to compute $\sin x$ with an error of at most $10^{-3}$ for all $x \in [-\pi/2, \pi/2]$?

Compute $\sin(\pi/2)$ from the Taylor Series. How large is the actual error?'

I'm fairly certain I can do the third part, but the first and second have completely thrown me. Can anybody help, or at least point me in the right direction?

Cheers.

• Hello, welcome to Math.SE. Please, try to make the title of your questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. – Lord_Farin Oct 23 '13 at 13:38
• Apologies, I wasn't really sure how to label this though! However I can see your point and in the future I will try to think a little harder. – Mathsishard Oct 23 '13 at 13:44
• Excellent, thanks in advance! You'll have noticed my edit, which provides your question with typeset mathematics. For some basic information on how to do this (besides checking the source of your question) see e.g. here, here, here and here. – Lord_Farin Oct 23 '13 at 13:54

For the second part, you just need to notice that the upper bound for the error term is maximised at the end points of the interval. So as long as you arrange for $r_{2N-1}(\pi/2)$ to be smaller than $10^{-3}$, the same will hold for all the other $x$ in the interval.