Упр.56 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)
- Докажите справедливость равенства:
а) $$\cos\frac{\pi}{7}\cos\frac{4\pi}{7}\cos\frac{5\pi}{7}=\frac{1}{8}$$;
б) $$\tg 20^\circ-4\sin 20^\circ\sin 50^\circ=-2\sin 20^\circ$$;
в) $$\frac{1}{\sin 10^\circ}-4\sin 70^\circ=2$$;
г) $$\cos 20^\circ+2\sin^2 55^\circ-\sqrt{2}\sin 65^\circ=1$$.
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) $$\tg 20^\circ-4\sin20^\circ\sin50^\circ=\frac{\sin20^\circ}{\cos20^\circ}-4\sin20^\circ\sin50^\circ$$
$$=\sin20^\circ\left(\frac1{\cos20^\circ}-4\sin50^\circ\right)=\sin20^\circ\cdot\frac{1-4\sin50^\circ\cos20^\circ}{\cos20^\circ}$$
$$=\sin20^\circ\cdot\frac{1-4\cos40^\circ\cos20^\circ}{\cos20^\circ}=\sin20^\circ\cdot\frac{1-2(\cos60^\circ+\cos20^\circ)}{\cos20^\circ}$$
$$=\sin20^\circ\cdot\frac{1-2\cdot\frac12-2\cos20^\circ}{\cos20^\circ}=\sin20^\circ\cdot(-2)=-2\sin20^\circ.$$
в) $$\frac1{\sin10^\circ}-4\sin70^\circ=\frac{1-4\sin10^\circ\sin70^\circ}{\sin10^\circ}$$
$$=\frac{1-2(\cos60^\circ-\cos80^\circ)}{\sin10^\circ}=\frac{1-1+2\cos80^\circ}{\sin10^\circ}$$
$$=\frac{2\cos80^\circ}{\sin(90^\circ-80^\circ)}=\frac{2\cos80^\circ}{\cos80^\circ}=2.$$
г) $$\cos20^\circ+2\sin^2 55^\circ-\sqrt2\sin65^\circ$$
$$=\cos20^\circ+\left(1-\cos110^\circ\right)-\sqrt2\sin(45^\circ+20^\circ)$$
$$=\cos20^\circ+1-\cos110^\circ-\sqrt2\left(\frac{\sqrt2}{2}\cos20^\circ+\frac{\sqrt2}{2}\sin20^\circ\right)$$
$$=\cos20^\circ+1+\sin20^\circ-\cos20^\circ-\sin20^\circ=1.$$
Ответ
а) $$\frac18$$; б) $$-2\sin20^\circ$$; в) $$2$$; г) $$1$$.
