Задачи 1) Дано! cos(alpha) = sqrt(3)/2, 0 < alpha < pi/2. Найти! sin(alpha), tg(alpha), ctg(alpha), sin(2*alpha).

1. Находим sin(alpha): sin(alpha) = sqrt(1 - cos^2(alpha)) = sqrt(1 - (sqrt(3)/2)^2) = sqrt(1 - 3/4) = sqrt(1/4) = 1/2.
2. Находим tg(alpha) и ctg(alpha): tg(alpha) = sin(alpha)/cos(alpha) = (1/2)/(sqrt(3)/2) = 1/sqrt(3) = sqrt(3)/3. ctg(alpha) = 1/tg(alpha) = sqrt(3).
3. Находим sin(2*alpha): sin(2*alpha) = 2*sin(alpha)*cos(alpha) = 2*(1/2)*(sqrt(3)/2) = sqrt(3)/2.