Answer
$\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(3x)^{2n+1}}{(2n+1)!}$
Work Step by Step
Here, we have: $f(x)=\sin 3x \implies f(0)=0\\ f'(x)=3 \cos 3x \implies f'(0)=3\\ f''(x)=-9 \sin 3x \implies f''(0)=0$
and so on.
The Taylor series of a function $f(x)$ at $x=c$ can be written as:
$T(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+.......\dfrac{f^n(a)}{n!}(x-a)^n$
Hence, the required Taylor series is:
$T(x)=0+\dfrac{3}{1!}(x-0)+\dfrac{0}{2!} (x-0)^2+....=3x-9x^3=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(3x)^{2n+1}}{(2n+1)!}$
