1. **State the problem:** We want to find the Maclaurin series expansion of the function $$f(x) = e^{-x \sin m}$$ around $$x=0$$. 2. **Recall the Maclaurin series formula:** The Maclaurin series of a function $$f(x)$$ is given by: $$ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n $$ where $$f^{(n)}(0)$$ is the $$n$$th derivative of $$f$$ evaluated at $$x=0$$. 3. **Use the known expansion for the exponential function:** The exponential function $$e^u$$ can be expanded as: $$ e^u = \sum_{n=0}^\infty \frac{u^n}{n!} $$ Here, $$u = -x \sin m$$. 4. **Substitute $$u$$ into the series:** $$ f(x) = e^{-x \sin m} = \sum_{n=0}^\infty \frac{(-x \sin m)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n (\sin m)^n x^n}{n!} $$ 5. **Write out the first few terms explicitly:** $$ f(x) = 1 - (\sin m) x + \frac{(\sin m)^2 x^2}{2!} - \frac{(\sin m)^3 x^3}{3!} + \frac{(\sin m)^4 x^4}{4!} - \cdots $$ **Final answer:** $$ f(x) = \sum_{n=0}^\infty \frac{(-1)^n (\sin m)^n}{n!} x^n $$