1. **Problem statement:** Find the output $y[n]$ of a system with impulse response $$h[n] = \left(\frac{1}{6}\right)^n u[n]$$ for the input $$x[n] = A \cdot \sin(\omega n)$$ using the Fourier transform. 2. **Recall the system output formula:** The output $y[n]$ is the convolution of $x[n]$ and $h[n]$, or equivalently in the frequency domain, $$Y(e^{j\omega}) = X(e^{j\omega}) H(e^{j\omega})$$ where $X(e^{j\omega})$ and $H(e^{j\omega})$ are the discrete-time Fourier transforms (DTFT) of $x[n]$ and $h[n]$ respectively. 3. **Find $H(e^{j\omega})$:** For $$h[n] = \left(\frac{1}{6}\right)^n u[n],$$ the DTFT is $$H(e^{j\omega}) = \sum_{n=0}^\infty \left(\frac{1}{6}\right)^n e^{-j\omega n} = \frac{1}{1 - \frac{1}{6} e^{-j\omega}} = \frac{1}{1 - \frac{1}{6} e^{-j\omega}}.$$ 4. **Find $X(e^{j\omega})$ for $x[n] = A \sin(\omega n)$:** Using Euler's formula, $$\sin(\omega n) = \frac{e^{j\omega n} - e^{-j\omega n}}{2j},$$ so $$X(e^{j\theta}) = A \cdot \frac{1}{2j} \left( \sum_{n=-\infty}^\infty e^{j\omega n} e^{-j\theta n} - \sum_{n=-\infty}^\infty e^{-j\omega n} e^{-j\theta n} \right).$$ This corresponds to impulses at $\theta = \pm \omega$, so $$X(e^{j\theta}) = \pi A \left[ \delta(\theta - \omega) - \delta(\theta + \omega) \right] / j.$$ 5. **Calculate output in frequency domain:** $$Y(e^{j\theta}) = X(e^{j\theta}) H(e^{j\theta}) = \pi A \frac{H(e^{j\omega}) \delta(\theta - \omega) - H(e^{-j\omega}) \delta(\theta + \omega)}{j}.$$ 6. **Inverse DTFT to get $y[n]$:** Using linearity and the inverse transform, $$y[n] = \frac{A}{2j} \left( H(e^{j\omega}) e^{j\omega n} - H(e^{-j\omega}) e^{-j\omega n} \right).$$ 7. **Simplify $y[n]$:** Since $H(e^{-j\omega}) = \overline{H(e^{j\omega})}$ (complex conjugate), $$y[n] = A \cdot |H(e^{j\omega})| \sin(\omega n + \arg(H(e^{j\omega}))).$$ **Final answer:** $$\boxed{y[n] = A \cdot \left| \frac{1}{1 - \frac{1}{6} e^{-j\omega}} \right| \sin\left( \omega n + \arg\left( \frac{1}{1 - \frac{1}{6} e^{-j\omega}} \right) \right)}.$$