Контрольные задания
Вопрос:Фото задания: 102. Найдите sin a, cos a, tg a и ctg a, если: а) a = \(\frac{2}{3}\)\(\pi\); б) a = \(\frac{3}{4}\)\(\pi\); в) a =…Фото задания

Ответ:

Решение:

  1. При \( \alpha = \frac{2\pi}{3} \):
    • \( \sin \frac{2\pi}{3} = \sin (\pi - \frac{\pi}{3}) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \).
    • \( \cos \frac{2\pi}{3} = \cos (\pi - \frac{\pi}{3}) = -\cos \frac{\pi}{3} = -\frac{1}{2} \).
    • \( \text{tg} \frac{2\pi}{3} = \frac{\sin \frac{2\pi}{3}}{\cos \frac{2\pi}{3}} = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3} \).
    • \( \text{ctg} \frac{2\pi}{3} = \frac{1}{\text{tg} \frac{2\pi}{3}} = \frac{1}{-\sqrt{3}} = -\frac{\sqrt{3}}{3} \).
  2. При \( \alpha = \frac{3\pi}{4} \):
    • \( \sin \frac{3\pi}{4} = \sin (\pi - \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).
    • \( \cos \frac{3\pi}{4} = \cos (\pi - \frac{\pi}{4}) = -\cos \frac{\pi}{4} = -\frac{\sqrt{2}}{2} \).
    • \( \text{tg} \frac{3\pi}{4} = \frac{\sin \frac{3\pi}{4}}{\cos \frac{3\pi}{4}} = \frac{\sqrt{2}/2}{-\sqrt{2}/2} = -1 \).
    • \( \text{ctg} \frac{3\pi}{4} = \frac{1}{\text{tg} \frac{3\pi}{4}} = \frac{1}{-1} = -1 \).
  3. При \( \alpha = \frac{5\pi}{6} \):
    • \( \sin \frac{5\pi}{6} = \sin (\pi - \frac{\pi}{6}) = \sin \frac{\pi}{6} = \frac{1}{2} \).
    • \( \cos \frac{5\pi}{6} = \cos (\pi - \frac{\pi}{6}) = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2} \).
    • \( \text{tg} \frac{5\pi}{6} = \frac{\sin \frac{5\pi}{6}}{\cos \frac{5\pi}{6}} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \).
    • \( \text{ctg} \frac{5\pi}{6} = \frac{1}{\text{tg} \frac{5\pi}{6}} = \frac{1}{-1/\sqrt{3}} = -\sqrt{3} \).

Ответ: а) \( \sin \alpha = \frac{\sqrt{3}}{2}, \cos \alpha = -\frac{1}{2}, \text{tg} \alpha = -\sqrt{3}, \text{ctg} \alpha = -\frac{\sqrt{3}}{3} \); б) \( \sin \alpha = \frac{\sqrt{2}}{2}, \cos \alpha = -\frac{\sqrt{2}}{2}, \text{tg} \alpha = -1, \text{ctg} \alpha = -1 \); в) \( \sin \alpha = \frac{1}{2}, \cos \alpha = -\frac{\sqrt{3}}{2}, \text{tg} \alpha = -\frac{\sqrt{3}}{3}, \text{ctg} \alpha = -\sqrt{3} \).