3. Упростите выражение: 6) \(\frac{2 \sin^2 \alpha}{\sin 2\alpha}\) г) \(\frac{1 + \cos 2\beta}{\cos \beta}\) e) \(\frac{\cos 10^{\circ}}{\cos 5^{\circ} + \sin 5^{\circ}} + \sin 5^{\circ}\)

Решение:

1. 6) Используем формулу синуса двойного угла \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \).
   \(\frac{2 \sin^2 \alpha}{\sin 2\alpha} = \frac{2 \sin^2 \alpha}{2 \sin \alpha \cos \alpha} = \frac{\sin \alpha}{\cos \alpha} = \mathrm{tg} \alpha\)
2. г) Используем формулу косинуса двойного угла \( 1 + \cos 2\beta = 2 \cos^2 \beta \).
   \(\frac{1 + \cos 2\beta}{\cos \beta} = \frac{2 \cos^2 \beta}{\cos \beta} = 2 \cos \beta\)
3. е)
   Приведем к общему знаменателю:
   \(\frac{\cos 10^{\circ}}{\cos 5^{\circ} + \sin 5^{\circ}} + \sin 5^{\circ} = \frac{\cos 10^{\circ} + \sin 5^{\circ}(\cos 5^{\circ} + \sin 5^{\circ})}{\cos 5^{\circ} + \sin 5^{\circ}}\}
   Раскроем скобки в числителе:
   \(\cos 10^{\circ} + \sin 5^{\circ} \cos 5^{\circ} + \sin^2 5^{\circ}\}
   Используем формулы \( \sin x \cos x = \frac{1}{2} \sin 2x \) и \( \sin^2 x = \frac{1 - \cos 2x}{2} \).
   \(\sin 5^{\circ} \cos 5^{\circ} = \frac{1}{2} \sin(2 · 5^{\circ}) = \frac{1}{2} \sin 10^{\circ}\}
   \(\sin^2 5^{\circ} = \frac{1 - \cos(2 · 5^{\circ})}{2} = \frac{1 - \cos 10^{\circ}}{2}\}
   Числитель: \(\cos 10^{\circ} + \frac{1}{2} \sin 10^{\circ} + \frac{1 - \cos 10^{\circ}}{2} = \frac{2 \cos 10^{\circ} + \sin 10^{\circ} + 1 - \cos 10^{\circ}}{2} = \frac{\cos 10^{\circ} + \sin 10^{\circ} + 1}{2}\}
   Знаменатель: \(\cos 5^{\circ} + \sin 5^{\circ}\}
   Умножим знаменатель на \( \sqrt{2} \) и выделим \( \frac{1}{\sqrt{2}} \):
   \(\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 5^{\circ} + \frac{1}{\sqrt{2}} \sin 5^{\circ} \right) = \sqrt{2} (\sin 45^{\circ} \cos 5^{\circ} + \cos 45^{\circ} \sin 5^{\circ}) = \sqrt{2} \sin(45^{\circ} + 5^{\circ}) = \sqrt{2} \sin 50^{\circ}\}
   Воспользуемся тем, что \( \cos x = \sin(90^{\circ}-x) \) и \( \sin x = \cos(90^{\circ}-x) \).
   \(\cos 10^{\circ} = \sin 80^{\circ}\), \(\sin 10^{\circ} = \cos 80^{\circ}\}
   \(\frac{\sin 80^{\circ} + \cos 80^{\circ} + 1}{2}\}
   Этот путь сложен. Попробуем иначе:
   \(\frac{\cos 10^{\circ}}{\cos 5^{\circ} + \sin 5^{\circ}} + \sin 5^{\circ}\}
   Умножим и разделим числитель и знаменатель на \( \cos 5^{\circ} - \sin 5^{\circ} \):
   \(\frac{\cos 10^{\circ}(\cos 5^{\circ} - \sin 5^{\circ})}{(\cos 5^{\circ} + \sin 5^{\circ})(\cos 5^{\circ} - \sin 5^{\circ})} + \sin 5^{\circ} = \frac{\cos 10^{\circ} \cos 5^{\circ} - \cos 10^{\circ} \sin 5^{\circ}}{\cos^2 5^{\circ} - \sin^2 5^{\circ}} + \sin 5^{\circ}\}
   Знаменатель: \(\cos^2 5^{\circ} - \sin^2 5^{\circ} = \cos(2 · 5^{\circ}) = \cos 10^{\circ}.\)
   \(\frac{\cos 10^{\circ} \cos 5^{\circ} - \cos 10^{\circ} \sin 5^{\circ}}{\cos 10^{\circ}} + \sin 5^{\circ} = \cos 5^{\circ} - \sin 5^{\circ} + \sin 5^{\circ} = \cos 5^{\circ}\)
   Перепроверка: \(\frac{\cos 10^{\circ}}{\cos 5^{\circ} + \sin 5^{\circ}} + \sin 5^{\circ} = \frac{\cos 10^{\circ} + \sin 5^{\circ}(\cos 5^{\circ} + \sin 5^{\circ})}{\cos 5^{\circ} + \sin 5^{\circ}} = \frac{\cos 10^{\circ} + \sin 5^{\circ}\cos 5^{\circ} + \sin^2 5^{\circ}}{\cos 5^{\circ} + \sin 5^{\circ}}\}
   \(\frac{\cos 10^{\circ} + \frac{1}{2}\sin 10^{\circ} + \frac{1-\cos 10^{\circ}}{2}}{\cos 5^{\circ} + \sin 5^{\circ}}\} = \frac{\frac{2\cos 10^{\circ} + \sin 10^{\circ} + 1 - \cos 10^{\circ}}{2}}{\cos 5^{\circ} + \sin 5^{\circ}}} = \frac{\cos 10^{\circ} + \sin 10^{\circ} + 1}{2(\cos 5^{\circ} + \sin 5^{\circ})}\}
   Используем \(\cos 10^{\circ} = 2\cos^2 5^{\circ} - 1\) и \(\sin 10^{\circ} = 2\sin 5^{\circ}\cos 5^{\circ}.\)
   \(\frac{2\cos^2 5^{\circ} - 1 + 2\sin 5^{\circ}\cos 5^{\circ} + 1}{2(\cos 5^{\circ} + \sin 5^{\circ})}} = \frac{2\cos^2 5^{\circ} + 2\sin 5^{\circ}\cos 5^{\circ}}{2(\cos 5^{\circ} + \sin 5^{\circ})}} = \frac{2\cos 5^{\circ}(\cos 5^{\circ} + \sin 5^{\circ})}{2(\cos 5^{\circ} + \sin 5^{\circ})}} = \cos 5^{\circ}\)

Ответ: 6) \( \mathrm{tg} \alpha \); г) \( 2 \cos \beta \); е) \( \cos 5^{\circ} \).