Lagrange multipliers has nothing to do with this; the relevant term is the Lagrange remainder. This says that there exists a $\xi$ between $0$ and $x$ such that
$$f(x)-T_n(x)=\frac{f^{(n+1)}(\xi) x^{n+1}}{(n+1)!}$$ where $T_n$ is the Maclaurin approximant of $f$ of degree $n$. ("Maclaurin" just means that the point of Taylor expansion is $x=0$.)
In the case of sine specifically, that derivative can be uniformly bounded (regardless of what $\xi$, which we do not know, is) by $1$, so that
$$|f(x)-T_n(x)| \leq \frac{|x|^{n+1}}{(n+1)!}.$$
A relative error estimate like you ask for is problematic because $\sin(x)$ vanishes at certain points, so that your relative error estimate looks like
$$\frac{|f(x)-T_n(x)|}{|\sin(x)|} \leq \frac{|x|^{n+1}}{(n+1)! |\sin(x)|}$$
which is very awkwardly behaved near the zeroes of $\sin$ except for $0$. You can avoid the difficulty by restricting attention to the domain $[-\pi/2,\pi/2]$ and using periodicity to extract values elsewhere.
In the case of $\sin$ and also $\cos$, you can get a slightly better estimate than this one by using the alternating series error estimate, which says that if you have a sequence of positive numbers $a_n$ decreasing to zero, then $\left | \sum_{n=N}^\infty (-1)^n a_n \right | \leq a_N$. For $f=\sin$ you can use this to show that given an odd number $k$:
$$|f(x)-T_k(x)| \leq \frac{|x|^{k+2}}{(k+2)!}$$
provided $|x|<k+2$.
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