Example 5.24. Calculate $\sin 18^{\circ}$, using the first two terms of series (3), and estimate the resulting error.
Solution. Since the Taylor series expansion of $\sin x$ is valid for any $x$, in particular, for $x=\frac{\pi}{10}$ we have

$$
\sin \frac{\pi}{10}=\frac{\pi}{10}-\frac{\frac{\pi}{10}^{3}}{3!}+\frac{\frac{\pi}{10}^{5}}{5!}-\ldots
$$

The obtained series is alternating. By limiting ourselves to the first two terms of this series, i.e., by considering $\sin \frac{\pi}{10}$ to be equal to their sum, we thereby admit an error not exceeding the first discarded term $\frac{\frac{\pi}{10}^{5}}{5!}$ (see 5.3). Since $\frac{\frac{\pi}{10}^{5}}{5!}<0.0001$, we obtain with an accuracy of 0.0001

$$
\sin \frac{\pi}{10} \approx \frac{\pi}{10}-\frac{\frac{\pi}{10}^{3}}{3!}=0.3091
$$