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 $10$.
This is just the Taylor–Lagrange formula. Recall that:
If $f$ is of class $C^N$ on $[a,b]$ and $N+1$ times differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that $$f(b)=\sum_{k=0}^N\frac{f^{(k)}(a)}{k!}(b-a)^k\;+\;\frac{f^{(N+1)}(c)}{(N+1)!}(b-a)^{N+1}.$$
Now, let $x\in\mathbb{R}^*$, and apply the Taylor–Lagrange formula to $\cos$ on $[0,x]$ (with $a=0$ and $b=x$) with $N=2n+1$. You'll obtain $c\in(0,x)$ such that $$\cos(x)=\sum_{k=0}^{2n+1}\frac{\cos^{(k)}(0)}{k!}x^k\;+\;\frac{\cos^{(2n+2)}(c)}{(2n+2)!}x^{2n+2},$$ i.e., using the fact that the odd derivatives of $\cos$ at $0$ are nil and that $\cos^{(2p)}(0)=(-1)^p$, $$\cos(x)=\sum_{p=0}^{n}\frac{(-1)^p}{(2p)!}x^{2p}\;+\;\frac{\cos^{(2n+2)}(c)}{(2n+2)!}x^{2n+2}.$$
You can also apply it with $N=2n$ and you obtain $c\in(0,x)$ (likely distinct from the previous one, of course) such that $$\cos(x)=\sum_{k=0}^{2n}\frac{\cos^{(k)}(0)}{k!}x^k\;+\;\frac{\cos^{(2n+1)}(c)}{(2n+1)!}x^{2n+1},$$ i.e., $$\cos(x)=\sum_{p=0}^{n}\frac{(-1)^p}{(2p)!}x^{2p}\;+\;\frac{\cos^{(2n+1)}(c)}{(2n+1)!}x^{2n+1}.$$
The general formula for the Taylor expansion of $\cos x$ is $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$$ So the powers of $x$ and the factorial at the denominator are always even.
