3. Упростить выражение: 1) sin($$\alpha + \beta$$) + sin($$\alpha - \beta$$); 2) $$\frac{cos(\pi - \alpha) + cos(\frac{3\pi}{2} + \alpha)}{1 + 2cos(-\alpha)sin(-\alpha)}$$.

1) sin($$\alpha + \beta$$) + sin($$\alpha - \beta$$) = (sin$$\alpha$$cos$$\beta$$ + cos$$\alpha$$sin$$\beta$$) + (sin$$\alpha$$cos$$\beta$$ - cos$$\alpha$$sin$$\beta$$) = 2sin$$\alpha$$cos$$\beta$$. 2) $$\frac{cos(\pi - \alpha) + cos(\frac{3\pi}{2} + \alpha)}{1 + 2cos(-\alpha)sin(-\alpha)}$$ = $$\frac{-cos(\alpha) + sin(\alpha)}{1 - 2cos(\alpha)sin(\alpha)}$$ = $$\frac{sin(\alpha) - cos(\alpha)}{1 - sin(2\alpha)}$$ = $$\frac{sin(\alpha) - cos(\alpha)}{sin^2(\alpha) + cos^2(\alpha) - 2sin(\alpha)cos(\alpha)}$$ = $$\frac{sin(\alpha) - cos(\alpha)}{(sin(\alpha) - cos(\alpha))^2}$$ = $$\frac{1}{sin(\alpha) - cos(\alpha)}$$. Ответ: 1) 2sin$$\alpha$$cos$$\beta$$; 2) $$\frac{1}{sin(\alpha) - cos(\alpha)}$$.