Упр.56 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)

Задача

а) cos(?/7)cos(4?/7)cos(5?/7)=1/8;
б) tg(20°)-4sin(20°)sin(50°)=-2sin(20°);
в) 1/sin(10°)-4sin(70°)=2;
г) cos(20°)+2sin^2(55°)-v2sin(65°)=1.

Подробный ответ

а)

$$ \cos \frac{\pi}{7}\cos \frac{4\pi}{7}\cos \frac{5\pi}{7} = \cos \frac{\pi}{7}\cos \frac{4\pi}{7}\cdot \left(-\cos \frac{2\pi}{7}\right) $$

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

$$ = -\frac{\sin \frac{4\pi}{7}\cos \frac{4\pi}{7}}{4\sin \frac{\pi}{7}} = -\frac{\sin \frac{8\pi}{7}}{8\sin \left(\pi+\frac{\pi}{7}\right)} = \frac{\sin \frac{8\pi}{7}}{8\sin \frac{8\pi}{7}} = \frac18. $$

б)

$$ \tg 20^\circ — 4\sin 20^\circ \sin 50^\circ = \frac{\sin 20^\circ}{\cos 20^\circ}-4\sin 20^\circ \sin 50^\circ $$

$$ = \sin 20^\circ\left(\frac1{\cos 20^\circ}-4\sin 50^\circ\right) = \sin 20^\circ\cdot \frac{1-4\sin 50^\circ \cos 20^\circ}{\cos 20^\circ} $$

$$ = \sin 20^\circ\cdot \frac{1-4\cos 40^\circ \cos 20^\circ}{\cos 20^\circ} = \sin 20^\circ\cdot \frac{1-2(\cos 60^\circ+\cos 20^\circ)}{\cos 20^\circ} $$

$$ = \sin 20^\circ\cdot \frac{1-2\cos 60^\circ-2\cos 20^\circ}{\cos 20^\circ} = \sin 20^\circ\cdot \frac{-2\cos 20^\circ}{\cos 20^\circ} = -2\sin 20^\circ. $$

в)

$$ \frac1{\sin 10^\circ}-4\sin 70^\circ = \frac{1-4\sin 10^\circ \sin 70^\circ}{\sin 10^\circ} $$

$$ = \frac{1-2(\cos 60^\circ-\cos 80^\circ)}{\sin 10^\circ} = \frac{1-2\cos 60^\circ+2\cos 80^\circ}{\sin 10^\circ} $$

$$ = \frac{1-1+2\cos 80^\circ}{\sin 10^\circ} = \frac{2\cos 80^\circ}{\sin(90^\circ-80^\circ)} = \frac{2\cos 80^\circ}{\cos 80^\circ} = 2. $$

г)

$$ \cos 20^\circ + 2\sin^2 55^\circ — \sqrt2 \sin 65^\circ = \cos 20^\circ + (1-\cos 110^\circ)-\sqrt2 \sin(45^\circ+20^\circ) $$

$$ = \cos 20^\circ + 1 — \cos 110^\circ — \sqrt2\left(\frac{\sqrt2}{2}\cos 20^\circ+\frac{\sqrt2}{2}\sin 20^\circ\right) $$

$$ = \cos 20^\circ + 1 + \sin 20^\circ — \cos 20^\circ — \sin 20^\circ = 1. $$