1. **State the problem:** Find the Maclaurin series (Taylor series at $x=0$) for the function $f(x) = \sin 3x$. 2. **Recall the Maclaurin series formula:** $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$$ where $f^{(n)}(0)$ is the $n$th derivative of $f$ evaluated at 0. 3. **Important rule for sine function:** The Maclaurin series for $\sin x$ is $$\sin x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$ 4. **Apply this to $\sin 3x$ by substituting $3x$ for $x$:** $$\sin 3x = \sum_{n=0}^\infty (-1)^n \frac{(3x)^{2n+1}}{(2n+1)!} = \sum_{n=0}^\infty (-1)^n \frac{3^{2n+1} x^{2n+1}}{(2n+1)!}$$ 5. **Write out the first few terms explicitly:** $$\sin 3x = 3x - \frac{3^3 x^3}{3!} + \frac{3^5 x^5}{5!} - \frac{3^7 x^7}{7!} + \cdots = 3x - \frac{27 x^3}{6} + \frac{243 x^5}{120} - \frac{2187 x^7}{5040} + \cdots$$ 6. **Simplify coefficients if desired:** $$\sin 3x = 3x - \frac{27}{6} x^3 + \frac{243}{120} x^5 - \frac{2187}{5040} x^7 + \cdots = 3x - \frac{9}{2} x^3 + \frac{81}{40} x^5 - \frac{729}{1680} x^7 + \cdots$$ **Final answer:** $$\sin 3x = \sum_{n=0}^\infty (-1)^n \frac{3^{2n+1}}{(2n+1)!} x^{2n+1} = 3x - \frac{9}{2} x^3 + \frac{81}{40} x^5 - \frac{729}{1680} x^7 + \cdots$$