Упр.1181 ГДЗ Никольский Потапов 9 класс (Алгебра)

а) sin(10°)cos(20°)cos(40°)=1/8; б) cos(20°)cos(40°)cos(80°)=1/8;
в) sin(?/5)cos(2?/5)=(1/4)tg(?/5); г) sin(18°)cos(36°)=1/4.

1. а) Используем формулу $$\sin 2x=2\sin x\cos x$$:

   $$ \sin 10^\circ \cos 20^\circ \cos 40^\circ = \frac{\sin 20^\circ \cos 20^\circ \cos 40^\circ}{2\cos 10^\circ} = \frac{\sin 40^\circ \cos 40^\circ}{4\cos 10^\circ \cos 20^\circ} $$

   $$ = \frac{\sin 80^\circ}{8\sin 80^\circ} = \frac18. $$

2. б) Аналогично:

   $$ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{\sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ}{\sin 20^\circ} = \frac{\sin 40^\circ \cos 40^\circ \cos 80^\circ}{2\sin 20^\circ} $$

   $$ = \frac{\sin 80^\circ \cos 80^\circ}{4\sin 20^\circ \cos 20^\circ} = \frac{\sin 160^\circ}{8\sin 160^\circ} = \frac18. $$

3. в) Преобразуем левую часть:

   $$ \sin \frac{\pi}{5}\cos \frac{2\pi}{5} = \frac{\sin \frac{2\pi}{5}\cos \frac{2\pi}{5}}{2\cos \frac{\pi}{5}} = \frac{\sin \frac{4\pi}{5}}{4\cos \frac{\pi}{5}}. $$

   Так как $$\sin \frac{4\pi}{5}=\sin \frac{\pi}{5}$$ и $$\tg \frac{\pi}{5}=\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}},$$ то

   $$ \sin \frac{\pi}{5}\cos \frac{2\pi}{5} = \frac14 \tg \frac{\pi}{5}. $$

4. г) Используем формулу $$\sin 2x=2\sin x\cos x$$:

   $$ \sin 18^\circ \cos 36^\circ = \frac{\sin 36^\circ \cos 36^\circ}{2\cos 18^\circ} = \frac{\sin 72^\circ}{4\cos 18^\circ}. $$

   Так как $$\sin 72^\circ=\cos 18^\circ,$$ получаем

   $$ \sin 18^\circ \cos 36^\circ=\frac14. $$

Ответ

а) $$\frac18$$; б) $$\frac18$$; в) $$\frac14 \tg \frac{\pi}{5}$$; г) $$\frac14$$.