The taylor polynomial of many functions has a skip term due to the fact that some derivative of the function is 0, $f^{(i)}(a) = 0$. For example, the sine function can be written as follows:
$$\text{T}_{2n + 2}(\sin(x; 0)) = \sum_\limits{k = 0}^n\frac{(-1)^k}{(2k + 1)!}x^{2k + 1} + E_{2n + 2}(x)$$ where $$E_{2n + 2}(x) = \frac{1}{(2n + 2)!}\int_{0}^x(x - t)^{2n + 2}\sin^{(2n+3)}(t)dt$$
However, we know that the term $2n + 2$ is actually missing in the sum, because it is zero. However, even though the term is missing, we can try to construct a polynomial of 1 degree lower: $2n + 1$ and write the error term differently even though the Taylor polynomial does not change at all:
$$\text{T}_{2n + 1}(\sin(x; 0)) = \sum_\limits{k = 0}^n\frac{(-1)^k}{(2k + 1)!}x^{2k + 1} + E_{2n + 1}(x)$$ where $$E_{2n + 1}(x) = \frac{1}{(2n + 1)!}\int_{0}^x(x - t)^{2n + 1}\sin^{(2n+2)}(t)dt$$
Due to uniqueness of Taylor polynomials, $E_{2n + 1} = E_{2n + 2}$ then (Taylor polynomial is the same, and error is equal for 2 different degrees of the polynomial). Is it the correct conclusion? If it is, why in my textbook and everywhere on the internet people use the term $E_{2n + 2}$ instead of $E_{2n + 1}$, and for other series with skip-terms, like $\arctan$ and $\cos$?