Страница 165 номер 554, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить (553—554). 1) \(\frac{\sqrt{3} (\cos 75^\circ - \cos 15^\circ)}{1 - 2 \sin^2 15^\circ}\); 2) \(\frac{2 \cos^2 \frac{\pi}{8} - 1}{1 + 8 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}}\).
1) \(\frac{\sqrt{3}(\cos 75^{\circ} - \cos 15^{\circ})}{1 - 2 \sin^{2} 15^{\circ}} = \frac{-\sqrt{3} \cdot 2 \sin 45^{\circ} \sin 30^{\circ}}{\cos^{2} 15^{\circ} - \sin^{2} 15^{\circ}} =\)
\(= -\frac{2\sqrt{3} \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2}}{\cos 30^{\circ}} = -\frac{\sqrt{3} \cdot \sqrt{2}}{2} \cdot \frac{2}{\sqrt{3}} = -\sqrt{2};\)
2) \(\frac{2 \cos^{2} \frac{\pi}{8} - 1}{1 + 8 \sin^{2} \frac{\pi}{8} \cos^{2} \frac{\pi}{8}} = \frac{\cos^{2} \frac{\pi}{8} - \sin^{2} \frac{\pi}{8}}{1 + 2 \sin^{2} \frac{\pi}{4}} = \frac{\cos \frac{\pi}{4}}{1 + 1} = \frac{\sqrt{2}}{4}.\)
