Maclaurin Series Approximation of $\sin{x}$ Use first ten terms of the Maclaurin series for $\sin{x}$ to find an approximation to the values of both $\sin{\left(\frac{6\pi}{7}\right)}$ and $\sin{\left(\frac{20\pi}{7}\right)}$?
One can say that $\sin{\left(\frac{6\pi}{7}\right)}=\sin{\left(\frac{20\pi}{7}\right)}$. 
What is the error in each approximation according to Taylors Inequality? 
Edit: How would yo suggest using this Maclauren Series to approximate a value for sin(theta) where theta is potentially very large?
What I've done thus far:
I know the $\sin{x}$ series is sum(n=0)$\frac{(-1)^n(x^{2n+1})}{(2n+1)!}$ and I have found the first ten terms, but I am unsure how to proceed.
 A: Well your series should be 
$$\sum_{n=0}^{\infty}\frac{(-1)^n(x^{2n+1})}{(2n+1)!}$$
$$ = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\dots$$
simply plug in $x=\frac{6\pi}{7}$ and $x=\frac{20\pi}{7}$. Then compare the difference (if the series were perfectly accurate they would be the same, because $\frac{20\pi}{7}=\frac{6\pi}{7}+2\pi$)
The error is told by Taylor's inequality 
$$|R_n(x)|\leq\frac{M|x-a|^{n+1}}{(n+1)!}$$
where $n$ in this case is the number of terms in your series approximation, and $M$ is the maximum value of what you're looking at (so in the case of sine, the maximum value is 1).
edit: $M$ is actually the maximum $|f^{(n+1)}(x)|\leq M$ where $f$ is the function you're approximating. But in this case that is still $1$, because that will be the maximum of any nth-derivative of $\sin(x)$. 
edit 2: yeah it's simply a matter of taking more terms. But it doesn't take long before you need a ridiculous amount of terms to approximate it. For example, if we are looking at something as small as $4\pi$, doing 10 terms gives us an error bound of $30913.828$ by the formula above. It takes up to 35 terms just to get to an error $|R_{35}(x)|\leq 0.0287$.
