Упр.554 ГДЗ Алимов 10-11 класс (Алгебра)

1)

$$\frac{\sqrt{3}\,(\cos 75^\circ-\cos 15^\circ)}{1-2\sin^2 15^\circ} = \frac{\sqrt{3}\cdot\bigl(-2\sin 45^\circ \sin 30^\circ\bigr)}{\cos 30^\circ}$$

Так как $$\sin 45^\circ=\frac{\sqrt{2}}{2},\quad \sin 30^\circ=\frac12,\quad \cos 30^\circ=\frac{\sqrt{3}}{2},$$
то

$$\frac{\sqrt{3}\cdot(-2)\cdot \frac{\sqrt{2}}{2}\cdot \frac12}{\frac{\sqrt{3}}{2}} = \frac{-\sqrt{6}/2}{\sqrt{3}/2} = -\sqrt{2}.$$

Ответ: $$-\sqrt{2}$$

2)

$$\frac{2\cos^2 \frac{\pi}{8}-1}{1+8\sin^2 \frac{\pi}{8}\cos^2 \frac{\pi}{8}} = \frac{\cos \frac{\pi}{4}}{1+2\sin^2 \frac{\pi}{4}}$$

Так как $$\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2},\quad \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2},$$
то

$$\frac{\cos \frac{\pi}{4}}{1+2\sin^2 \frac{\pi}{4}} = \frac{\frac{\sqrt{2}}{2}}{1+2\cdot \frac12} = \frac{\frac{\sqrt{2}}{2}}{2} = \frac{\sqrt{2}}{4}.$$

Ответ: $$\frac{\sqrt{2}}{4}$$