Страница 155 номер 517, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить: 1) \(\sin 15^\circ\); 2) \(\cos 15^\circ\); 3) \(\operatorname{tg} 22^\circ 30'\); 4) \(\operatorname{ctg} 22^\circ 30'\).
Т.к. \(0^{\circ} < 15^{\circ} < 90^{\circ}\) и \(0^{\circ} < 22^{\circ}30' < 90^{\circ}\), то \(\sin 15^{\circ} > 0\), \(\cos 15^{\circ} > 0\), \(\operatorname{tg} 22^{\circ}30' > 0\), \(\operatorname{ctg} 22^{\circ}30' > 0\).
1) \(\sin 15^{\circ} = \sqrt{\frac{1 - \cos 30^{\circ}}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{1}{2} - \frac{\sqrt{3}}{4}}\);
2) \(\cos 15^{\circ} = \sqrt{\frac{1 + \cos 30^{\circ}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{1}{2} + \frac{\sqrt{3}}{4}}\);
3) \(\operatorname{tg} 22^{\circ}30' = \sqrt{\frac{1 - \cos 45^{\circ}}{1 + \cos 45^{\circ}}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}} = \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}} = \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} = \sqrt{2} - 1\);
4) \(\operatorname{ctg} 22^{\circ}30' = \sqrt{\frac{1 + \cos 45^{\circ}}{1 - \cos 45^{\circ}}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}} = \sqrt{\frac{2 + \sqrt{2}}{2 - \sqrt{2}}} = \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}} = \sqrt{2} + 1\).
