Here is a typical Calculus problem involving Taylor polynomials. The solution and my comment follow.
Use the linear Taylor polynomial associated with the sine function to estimate the value of $\sin(50^{\circ})$ with an error of less than $0.02$.
Solution
Lagrange's remainder term for the linear Taylor polynomial associated with the sine function at $60^{\circ}$ is \begin{equation*} R_{1}(x) = - \frac{\sin{\theta}}{2!} \, \left(x - \frac{\pi}{3}\right)^{2} . \end{equation*} for some real number $\theta$ between $\pi/3$ and $x$. $50^{\circ}$ is equivalent to $5\pi/18$ radians, and \begin{equation*} \frac{\pi}{3} - \frac{5\pi}{18} = \frac{\pi}{18} < \frac{1}{5} . \end{equation*} Since the sine function is bounded by $1$, the evaluation of the linear Taylor polynomial $T_{1}(x)$ at $5\pi/18$ is an estimate for $\sin(50^{\circ})$ with an error of less than $0.02$. \begin{equation*} T_{1}\left(\frac{5\pi}{18}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} \left(\frac{5\pi}{18} - \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - \frac{\pi}{36} . \end{equation*}
Comment
Isn't this answer disingenuous? If the author of the Calculus textbook is pretending that we do not have a calculator to evaluate $\sin(50^{\circ})$, we also should have a calculator to give estimates for $\sqrt{3}$ ... or for $\pi$.
Approximating square roots with Newton's method is very efficient (and is probably close to what your calculator does) so there's no pretense there. Pi is just a constant, so there's no pretense there (although it's an approximation of pi that's stored in the calculator, of course).
Are Taylor series really how your calculator computes sines? Probably not, but for a very long time, they were the best thing we had, and DID get used to compute a lot of things, even when the computations were done by hand rather than with a calculator.