Answer
\begin{aligned}
\sin 3 \theta =3 \theta-9 \theta^{3}+\ldots \ldots+\frac{(-1)^{n} 3^{n}}{(2 n+1) !} \theta^{2 n+1}
\end{aligned}
Work Step by Step
Since
\begin{aligned}
f(\theta) &=\sin 3 \theta & & f(0)=0 \\
f^{\prime}(\theta) &=3 \cos 3 \theta & & f^{\prime}(0) =3 \\
f^{\prime \prime}(\theta) &=-9 \sin 3 \theta & & f^{\prime \prime}(0) =0 \\
f^{\prime \prime \prime}(\theta) &=-27 \cos 3 \theta & f^{\prime \prime \prime}(0) &=-27 \\
f^{(4)}(\theta) &=81 \sin 3 \theta & f^{(4)}(0) &=0
\end{aligned}Then
\begin{aligned}
\sin 3 \theta &=f(0)+\frac{f^{\prime}(0)}{1 !} \theta+\frac{f^{\prime \prime}(0)}{2 !} \theta^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} \theta^{3}+\ldots \ldots+\frac{f^{(n)}(0)}{n !} \theta^{n} \\
&=3 \theta-9 \theta^{3}+\ldots \ldots+\frac{(-1)^{n} 3^{n}}{(2 n+1) !} \theta^{2 n+1}
\end{aligned}
