Упр.27 ГДЗ Колмогоров 10-11 класс (Алгебра)

а)

$$\sin \frac{\pi}{12}\cos \frac{\pi}{12} =\frac12\cdot 2\sin \frac{\pi}{12}\cos \frac{\pi}{12} =\frac12\sin \frac{\pi}{6} =\frac12\cdot \frac12 =\frac14$$

б)

$$\left(\sin \frac{7\pi}{18}-\sin \frac{\pi}{18}\right):\cos \frac{2\pi}{9}$$
$$\sin \frac{7\pi}{18}-\sin \frac{\pi}{18} =2\sin \frac{\frac{7\pi}{18}+\frac{\pi}{18}}{2}\cos \frac{\frac{7\pi}{18}-\frac{\pi}{18}}{2} =2\sin \frac{\pi}{6}\cos \frac{\pi}{9} =\cos \frac{\pi}{9}$$
$$\frac{\cos \frac{\pi}{9}}{\cos \frac{2\pi}{9}}$$
$$\cos \frac{\pi}{9}=\cos 20^\circ,\qquad \cos \frac{2\pi}{9}=\cos 40^\circ$$
$$\frac{\cos 20^\circ}{\cos 40^\circ}=1$$

в)

$$\left(\sin^2 \frac{\pi}{8}-\cos^2 \frac{\pi}{8}\right)^2 = \left(\frac{1-\cos \frac{\pi}{4}}{2}-\frac{1+\cos \frac{\pi}{4}}{2}\right)^2$$
$$= \left(-\cos \frac{\pi}{4}\right)^2 = \left(-\frac{\sqrt2}{2}\right)^2 = \frac12$$

г)

$$\frac{\cos \frac{11\pi}{12}-\cos \frac{\pi}{12}}{\sin \frac{5\pi}{12}} = \frac{-2\sin \frac{\frac{11\pi}{12}+\frac{\pi}{12}}{2}\sin \frac{\frac{11\pi}{12}-\frac{\pi}{12}}{2}}{\sin \frac{5\pi}{12}}$$
$$= \frac{-2\sin \frac{\pi}{2}\sin \frac{5\pi}{12}}{\sin \frac{5\pi}{12}} =-2$$

Ответ: а) $$\frac14$$; б) $$1$$; в) $$\frac12$$; г) $$-2$$.