Страница 146 номер 482, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить, не пользуясь таблицами: 1) \(\cos 57^\circ 30' \cos 27^\circ 30' + \sin 57^\circ 30' \sin 27^\circ 30'\); 2) \(\cos 19^\circ 30' \cos 25^\circ 30' - \sin 19^\circ 30' \sin 25^\circ 30'\); 3) \(\cos\frac{7\pi}{9} \cos\frac{11\pi}{9} - \sin\frac{7\pi}{9} \sin\frac{11\pi}{9}\); 4) \(\cos\frac{8\pi}{7} \cos\frac{\pi}{7} + \sin\frac{8\pi}{7} \sin\frac{\pi}{7}\).
1) \(\cos 57^{\circ}30' \cos 27^{\circ}30' + \sin 57^{\circ}30' \sin 27^{\circ}30' =\)
\(= \cos(57^{\circ}30' - 27^{\circ}30') = \cos 30^{\circ} = \dfrac{\sqrt{3}}{2};\)
2) \(\cos 19^{\circ}30' \cos 25^{\circ}30' - \sin 19^{\circ}30' \sin 25^{\circ}30' =\)
\(= \cos(19^{\circ}30' + 25^{\circ}30') = \cos 45^{\circ} = \dfrac{\sqrt{2}}{2};\)
3) \(\cos\frac{7\pi}{9} \cos\frac{11\pi}{9} - \sin\frac{7\pi}{9} \sin\frac{11\pi}{9} = \cos\left(\frac{7\pi}{9} + \frac{11\pi}{9}\right) = \cos 2\pi = 1\);
4) \(\cos\frac{8\pi}{7} \cos\frac{\pi}{7} + \sin\frac{8\pi}{7} \sin\frac{\pi}{7} = \cos\left(\frac{8\pi}{7} - \frac{\pi}{7}\right) = \cos\pi = -1\).
