1454. Expand the function $\sin x$ into a Taylor series at $x_{0}=\frac{\pi}{4}$.
Solution. We find the values of the function and its derivatives at $x=\frac{\pi}{4}$:

$$
\begin{gathered}
f(x)=\sin x, f\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \\
f^{\prime}(x)=\cos x, f^{\prime}\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \\
f^{\prime \prime}(x)=-\sin x, f^{\prime \prime}\left(\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2} \\
f^{\prime \prime \prime}(x)=-\cos x, f^{\prime \prime \prime}\left(\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2} \\
f^{\text {IV }}(x)=\sin x, f^{\text {IV }}\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}
\end{gathered}
$$

Substituting these values into series (1'), we obtain the Taylor series of the given function

$$
\sin x=\frac{\sqrt{2}}{2}\left[1+\frac{x-\frac{\pi}{4}}{1!}-\frac{\left(x-\frac{\pi}{4}\right)^{2}}{2!}-\frac{\left(x-\frac{\pi}{4}\right)^{3}}{3!}+\ldots\right]
$$

Radius of convergence

$$
R=\lim _{n \rightarrow \infty}\left|\frac{a_{n}}{a_{n+1}}\right|=\lim _{n \rightarrow \infty} \frac{(n+1)!}{n!}=\lim _{n \rightarrow \infty}(n+1)=\infty
$$

The sum of the obtained series $S(x)=\sin x$, since condition (3) for the given function is satisfied $\left(\left|f^{(n)}(x)\right| \leqslant 1(n=1,2,3, \ldots)\right)$.