Remainder term for Maclaurin's $\sin x$ expansion We know that for the Maclaurin's series $$\sum_{k=0}^{n}\frac{ f^{k}(0) }{(n+1)!}x^{k}$$ the remainder term is given by the following formula:
$$R_{n} = \frac{\left | f^{(n+1)}(z) \right |x^{n+1} }{(n+1)!}$$
I want to calculate $\sin x$ using Maclaurin's expansion for $$\sin x = \sum^w_{k=0} \frac{(-1)^k}{(2k+1)!} x^{2k+1} $$
and then i want to calculate the remainder term using $w$. Now considering that $n=2w +1$, I think that the remainder term should look like this: 
$$R_{w} = \frac{\left | f^{(2w+1+1)}(z) \right |x^{2w+1+1} }{(2w+1+1)!}$$ where $f(x) = \sin x$.
Would this formula for the remainder term be correct? 
 A: The remainder term after the term of degree $2w+1$ in the power series expansion of $f$ is $$R_w= f^{(2w+2)}(z)x^{2w+2}/(2w+2)!$$ for some $z$ between $0$ and $x$ when $x\ne 0\;$ (or $z=0$ when $x=0.)\;$ Now when $f=\sin$ we have  $f^{(2w+2)}(z)=(-1)^w\sin z.$ We can re-write $R_w$ by noting that $\sin z= z(1-a)=x(1-b)(1-c)=x(1-d)$ where $a,b,c\in (0,1)$, so that $$R_w=(-1)^wx^{2w+3}(1-c)/(2w+2)! \quad \text {with }\;  c\in (0,1).$$ Of course, $c$ depends on $x$ and $w$. When $w$ is sufficiently large, the value of $(1-c)(2w+3)$ is approximately $1$.
A: Hint:
You obtain a greater precision if you take into account the expansion of $\sin x$  at order $2w+1$ is ipso facto an expansion at order $2w+2$.
A: Another hint:
The derivatives cycle through
$\pm\sin(z)$
and
$\pm\cos(z)$.
Also note that,
if 
$a_w
=\frac{x^{2w+1+1} }{(2w+1+1)!}
=\frac{x^{2w+2} }{(2w+2)!}
$,
then
$\frac{a_w}{a_{w-1}}
=\frac{\frac{x^{2w+2} }{(2w+2)!}}{\frac{x^{2w} }{(2w)!}}
=\frac{x^2}{2w(2w+1)}
$.
A: Another way without Taylor remainder term:
The sine and cosine series are alternating. By the Leibniz test and its error estimate, the error is always smaller than the last term (if the last term is smaller than the next-to-last term), i.e., for $x>0$ one has the inclusion
$$
x-\frac{x^3}6\pm …- \frac{x^{4m-1}}{(4m-1)!}\le \sin x\le x-\frac{x^3}6\pm …+ \frac{x^{4n+1}}{(4n+1)!}
$$
for arbitrary $m,n$ and especially for $m=n$ and $m=n+1$ under the condition $x<4n$. 
