Страница 407 номер 1303, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Доказать тождество: 1) \(\frac{1 - \cos(2\pi - 2\alpha)}{1 - \cos^2(\alpha + \pi)} = 2\); 2) \(\frac{\sin^2(\alpha + 90^\circ)}{1 + \sin(-\alpha)} = 1 + \cos(\alpha - 90^\circ)\).
1) \(\frac{1 - \cos(2\pi - 2\alpha)}{1 - \cos^{2}(\alpha + \pi)} = 2\)
\(\dfrac{1 - \cos(2\pi - 2\alpha)}{1 - \cos^{2}(\alpha + \pi)} = \dfrac{1 - \cos{2\alpha}}{1 - \cos^{2}\alpha} = \dfrac{2\sin^{2}\alpha}{\sin^{2}\alpha} = 2;\)
2) \(\frac{\sin^{2}(\alpha + 90^{\circ})}{1 + \sin(-\alpha)} = 1 + \cos(\alpha - 90^{\circ})\)
\(\dfrac{\sin^{2}(\alpha + 90^{\circ})}{1 + \sin(-\alpha)} = \dfrac{\cos^{2}\alpha}{1 - \sin\alpha} = \dfrac{1 - \sin^{2}\alpha}{1 - \sin\alpha} = 1 + \sin\alpha,\)
\(1 + \cos(\alpha - 90^{\circ}) = 1 + \cos(90^{\circ} - \alpha) = 1 + \sin\alpha.\)
