Упростить выражение: 1) sin(α + β) + sin(α - β); 2) \frac{cos(π - α) + cos(\frac{3π}{2} + α)}{1 + 2cos(-α)sin(-α)}

1) sin(α + β) + sin(α - β) = (sin α cos β + cos α sin β) + (sin α cos β - cos α sin β) = 2 sin α cos β 2) \frac{cos(π - α) + cos(\frac{3π}{2} + α)}{1 + 2cos(-α)sin(-α)} = \frac{-cos α + sin α}{1 - 2cos α sin α} = \frac{-cos α + sin α}{(sin α - cos α)^2 + 2sinαcosα-2sinαcosα} = \frac{-(cos α - sin α)}{1 - sin 2α}. cos(π - α) = -cos α, cos(\frac{3π}{2} + α) = sin α, cos(-α) = cos α, sin(-α) = -sin α, 2 cos α sin α = sin 2α \\= \frac{sin α - cos α}{1 - 2sin α cos α} = \frac{sin α - cos α}{(sin α - cos α)^2+2sinαcosα-2sinαcosα} = \frac{sin α - cos α}{(sin α - cos α)^2+2sinαcosα-2sinαcosα} = \frac{sin α - cos α}{(sinα-cosα)^2-2sinαcosα+1+2sin α cos α-1}= \frac{-(cos α - sin α)}{1 - sin 2α} = \frac{-(cos α - sin α)}{(sin α - cos α)^2+2sinαcosα-2sinαcosα} = \frac{sin α - cos α}{sin^2 α + cos^2 α - 2sin α cos α}=\frac{sin α - cos α}{(sin α-cos α)^2}= \frac{1}{sin α - cos α} **Ответ:** 1) 2 sin α cos β; 2) \frac{1}{sin α - cos α}