530 Найти значение выражения: 1) cos 630° – sin 1470° – ctg 1125°; 2) tg 1800°- sin 495° + cos 945°; 3) 3 cos 3660° + sin (-1560°) + cos (-450°); 4) cos 4455° – cos (-945°) + tg 1035° – ctg (-1500°).

Решение заданий 530: 1) $$cos 630° - sin 1470° - ctg 1125°$$ * $$cos 630° = cos (360° + 270°) = cos 270° = 0$$ * $$sin 1470° = sin (4 \cdot 360° + 30°) = sin 30° = \frac{1}{2}$$ * $$ctg 1125° = ctg (3 \cdot 360° + 45°) = ctg 45° = 1$$ $$0 - \frac{1}{2} - 1 = -\frac{3}{2}$$ 2) $$tg 1800° - sin 495° + cos 945°$$ * $$tg 1800° = tg (5 \cdot 360°) = tg 0° = 0$$ * $$sin 495° = sin (360° + 135°) = sin 135° = \frac{\sqrt{2}}{2}$$ * $$cos 945° = cos (2 \cdot 360° + 225°) = cos 225° = -\frac{\sqrt{2}}{2}$$ $$0 - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$$ 3) $$3 cos 3660° + sin (-1560°) + cos (-450°)$$ * $$cos 3660° = cos (10 \cdot 360° + 60°) = cos 60° = \frac{1}{2}$$ * $$sin (-1560°) = -sin (4 \cdot 360° + 120°) = -sin 120° = -\frac{\sqrt{3}}{2}$$ * $$cos (-450°) = cos (360° - 90°) = cos (-90°) = 0$$ $$3 \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} + 0 = \frac{3 - \sqrt{3}}{2}$$ 4) $$cos 4455° - cos (-945°) + tg 1035° - ctg (-1500°)$$ * $$cos 4455° = cos (12 \cdot 360° + 135°) = cos 135° = -\frac{\sqrt{2}}{2}$$ * $$cos (-945°) = cos (2 \cdot 360° - 225°) = cos (-225°) = -\frac{\sqrt{2}}{2}$$ * $$tg 1035° = tg (2 \cdot 360° + 315°) = tg 315° = -1$$ * $$ctg (-1500°) = -ctg (4 \cdot 360° + 60°) = -ctg 60° = -\frac{\sqrt{3}}{3}$$ $$- \frac{\sqrt{2}}{2} - ( - \frac{\sqrt{2}}{2} ) - 1 - ( - \frac{\sqrt{3}}{3}) = \frac{\sqrt{3}}{3} - 1$$ Ответ: 1) $$\frac{-3}{2}$$ 2) $$- \sqrt{2}$$ 3) $$\frac{3 - \sqrt{3}}{2}$$ 4) $$\frac{\sqrt{3}}{3} - 1$$