Taylor expansion of $\cos{x}$ I found a pdf file on the internet which gives you  known expansions of Taylor's.
There is something I cant understand :
Why is the remainder of $\cos x$ is written like this?
$$\frac{\cos ^{(2n+2)}(c)x^{2n+2}}{(2n+2)!}$$
And not like this:
$$\frac{\cos^{(2n+1)} (c)x^{2n+1}}{(2n+1)!}$$
when the last element was :
$$\frac{(-1)^nx^{2n}} {(2n)!} $$ 

I can't guess what a random, unlinked PDF on the Internet says.  However, this is not the behaviour of the Taylor series of cosine.  For instance, the expansion of $\cos x$ around $x=1$ is $$\cos (1)-(x-1) \sin (1)-\frac{1}{2} (x-1)^2 \cos (1)+\frac{1}{6} (x-1)^3
   \sin (1)+\frac{1}{24} (x-1)^4 \cos (1)-\frac{1}{120} (x-1)^5 \sin
   (1)-\frac{1}{720} (x-1)^6 \cos (1)+\frac{(x-1)^7 \sin
   (1)}{5040}+\frac{(x-1)^8 \cos (1)}{40320}-\frac{(x-1)^9 \sin
   (1)}{362880}-\frac{(x-1)^{10} \cos (1)}{3628800}+\cdots, $$ which you might notice does have the error behaviour you seem to expect: the degree of the error term is equal to the degree of the first omitted term.
The Maclaurin series for cosine, $$1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320}-\frac{x^{10}}{3628800}+\cdots$$ does have the form you cite from the paper and does have the error term you mention.  In the expansion I just showed, the first omitted term has degree $12$, not $11$.  This behaviour is predictable:  cosine is even (i.e., is unchanged on reflection through the line $x=0$).  We would expect similar behaviour for Taylor series expansion around multiples of $\pi$ as well, due to the same reflection symmetry, which we see.  For example, expanding around $x=\pi$, $$-1+\frac{1}{2} (x-\pi )^2-\frac{1}{24} (x-\pi )^4+\frac{1}{720} (x-\pi)^6-\frac{(x-\pi )^8}{40320}+\frac{(x-\pi )^{10}}{3628800}+\cdots.$$
Further, much like sine, cosine has odd symmetry around odd multiples of $\pi/2$, so we would expect the Taylor expansion to contain only odd degree terms,which it does.$$-\left(x-\frac{\pi }{2}\right)+\frac{1}{6} \left(x-\frac{\pi}{2}\right)^3-\frac{1}{120} \left(x-\frac{\pi}{2}\right)^5+\frac{\left(x-\frac{\pi}{2}\right)^7}{5040}-\frac{\left(x-\frac{\pi}{2}\right)^9}{362880}+\cdots$$  Again, the error term has degree matching that of the first omitted term, which here is $11$, not $10$.
