Упр.680 ГДЗ Никольский Потапов 9 класс (Алгебра)

Вычислите:
а) $$\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}$$
б) $$\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}$$

Используем формулу

$$2\sin x\cos x=\sin 2x.$$

а)

$$\cos \frac{\pi}{9}\cos \frac{2\pi}{9}\cos \frac{4\pi}{9}$$

Представим первый множитель через синус:

$$\cos \frac{\pi}{9}=\frac{\sin \frac{2\pi}{9}}{2\sin \frac{\pi}{9}}.$$

Тогда

$$\cos \frac{\pi}{9}\cos \frac{2\pi}{9}\cos \frac{4\pi}{9} = \frac{\sin \frac{2\pi}{9}\cos \frac{2\pi}{9}\cos \frac{4\pi}{9}}{2\sin \frac{\pi}{9}} = \frac{\sin \frac{4\pi}{9}\cos \frac{4\pi}{9}}{4\sin \frac{\pi}{9}} = \frac{\sin \frac{8\pi}{9}}{8\sin \frac{\pi}{9}}.$$

Так как $$\sin \frac{8\pi}{9}=\sin \frac{\pi}{9},$$ то

$$\cos \frac{\pi}{9}\cos \frac{2\pi}{9}\cos \frac{4\pi}{9}=\frac{1}{8}.$$

б)

$$\cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}$$

Аналогично:

$$\cos \frac{\pi}{7}=\frac{\sin \frac{2\pi}{7}}{2\sin \frac{\pi}{7}}.$$

Тогда

$$\cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} = \frac{\sin \frac{2\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}}{2\sin \frac{\pi}{7}} = \frac{\sin \frac{4\pi}{7}\cos \frac{4\pi}{7}}{4\sin \frac{\pi}{7}} = \frac{\sin \frac{8\pi}{7}}{8\sin \frac{\pi}{7}}.$$

Но

$$\sin \frac{8\pi}{7}=-\sin \frac{\pi}{7},$$

поэтому

$$\cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}=-\frac{1}{8}.$$

Ответ

а) $$\frac{1}{8}$$; б) $$-\frac{1}{8}$$.