How to compute the error of the Taylor series? $$ \sin(x) \approx x-\frac{x^3}{6},\quad 0\le x \le 1 $$
I still don't know, how does this method work, I would appreciate any help
 A: Consider the general Taylor series:
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots+(-1)^n\frac{x^{2n+1}}{(2n+1)!}+\dots$$
Clearly, for $x\in[0,1]$, we have an alternating series with monotonically decreasing terms, therefore, the partial sum has error bounded by the next term:
$$P_3(x)=x-\frac{x^3}{3!}$$
$$|\sin(x)-P_3(x)|\le\frac{x^5}{5!}$$

To estimate the error for a non-alternating series, I recommend watching this series of Videos from Khan Academy:
Taylor polynomial remainder
A: I thought it might be instructive to present an approach that uses the integral form of the remainder. To that end, we proceed.

We can write $\sin(x)$ as
$$\sin(x)=x-\frac16 x^3+\frac1{24}\int_0^x t^4\cos(x-t)\,dt \tag 1$$


It is a straightforward task to develop $(1)$ by starting with the relationship $\sin(x)=\int_0^x \cos(x-t)\,dt$ and repeatedly integrating by parts.  In fact, this procedure can be generalized to any function $f(x)=f(a)+\int_0^{x-a}f'(x-t)\,dt$ to develop the Taylor's Formula with integral remainder.


Using $(1)$, we have the estimates of the error of $|\sin(x)-x+\frac16x^3|$ for $x\in [0,1]$
$$\begin{align}
\left|\sin(x)-x+\frac16x^3\right|&=\left|\frac1{24}\int_0^x t^4\cos(x-t)\,dt\right|\\\\
&\le \frac1{24}\int_0^x |t^4\,\cos(x-t)|\,dt\\\\
&\le \frac{1}{120}x^5\\\\
&\le \frac1{120}
\end{align}$$
Therefore,
$$\left|\sin(x)-x+\frac16x^3\right|\le \frac1{120}$$
for $x\in[0,1]$.
