Answer
$\sin (3x)=\Sigma_{n = 0}^{\infty} (-1)^n \dfrac{(x-\pi/2)^{2n}}{(2n!)}$; converges for all $x$ with convergence interval $(-\infty,\infty)$.
Work Step by Step
The Taylor`s Series can be represented as: : $T(x)=\Sigma_{n = 0}^{\infty} \dfrac{f^n (c)}{n!}(x-c)^n$
If we consider $c=0$, then Taylor`s Series, $T(x)$ is also called as Maclaurin Series and the Maclaurin series for $f(x)$ is as given below:
$f(x)=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0) x^2}{2!}+\dfrac{f^{\prime \prime \prime}(0) x^3}{3!}+\dfrac{f^{\prime \prime \prime \prime}(0) x^4}{4!}+..............$
From Table $2$, we see that $f(x)=\cos x=\Sigma_{n = 0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n!)}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}.....$; converges for $x$
Now, Taylor`s Series becomes: $f(x)=\sin x=\cos (x-\pi/2)=\Sigma_{n = 0}^{\infty} (-1)^n \dfrac{(x-\pi/2)^{2n}}{(2n!)}$; converges for $x$
Thus, we have the Taylor`s Series: $\sin (3x)=\Sigma_{n = 0}^{\infty} (-1)^n \dfrac{(x-\pi/2)^{2n}}{(2n!)}$; converges for all $x$ with convergence interval $(-\infty,\infty)$.
