Страница 151 номер 510, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Доказать тождество: 1) \(\frac{\cos 2\alpha}{\sin \alpha \cos \alpha + \sin^2 \alpha} = \operatorname{ctg} \alpha - 1\); 2) \(\frac{\sin 2\alpha - 2 \cos \alpha}{\sin \alpha - \sin^2 \alpha} = -2 \operatorname{ctg} \alpha\); 3) \(\operatorname{tg} \alpha (1 + \cos 2\alpha) = \sin 2\alpha\); 4) \(\frac{1 - \cos 2\alpha + \sin 2\alpha}{1 + \cos 2\alpha + \sin 2\alpha} \cdot \operatorname{ctg} \alpha = 1\); 5) \(\frac{(1 - 2 \cos^2 \alpha)(2 \sin^2 \alpha - 1)}{4 \sin^2 \alpha \cos^2 \alpha} = \operatorname{ctg}^2 2\alpha\); 6) \(1 - 2 \sin^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \sin \alpha\); 7) \(\frac{\sin \alpha + \sin 2\alpha}{1 + \cos \alpha + \cos 2\alpha} = \operatorname{tg} \alpha\).
1) \(\frac{\cos 2\alpha}{\sin \alpha \cos \alpha + \sin^{2} \alpha} = \operatorname{ctg} \alpha - 1\)
\(\dfrac{\cos 2\alpha}{\sin \alpha \cos \alpha + \sin^{2} \alpha} = \dfrac{(\cos \alpha - \sin \alpha)(\cos \alpha + \sin \alpha)}{\sin \alpha(\cos \alpha + \sin \alpha)} =\)
\(= \dfrac{\cos \alpha}{\sin \alpha} - \dfrac{\sin \alpha}{\sin \alpha} = \operatorname{ctg} \alpha - 1;\)
2) \(\frac{\sin 2\alpha - 2\cos \alpha}{\sin \alpha - \sin^{2} \alpha} = -2\operatorname{ctg} \alpha\)
\(\dfrac{\sin 2\alpha - 2\cos \alpha}{\sin \alpha - \sin^{2} \alpha} = \dfrac{2\cos \alpha(\sin \alpha - 1)}{\sin \alpha(1 - \sin \alpha)} = -\dfrac{2\cos \alpha}{\sin \alpha} = -2\operatorname{ctg} \alpha;\)
3) \(\operatorname{tg} \alpha \cdot (1 + \cos 2\alpha) = \sin 2\alpha\)
\(\operatorname{tg} \alpha \cdot (1 + \cos 2\alpha) =\)
\(= \operatorname{tg} \alpha \cdot (1 - \sin^{2} \alpha + \cos^{2} \alpha) = \dfrac{\sin \alpha}{\cos \alpha} \cdot 2\cos^{2} \alpha =\)
\(= 2\sin \alpha \cos \alpha = \sin 2\alpha;\)
4) \(\frac{1 - \cos 2\alpha + \sin 2\alpha}{1 + \cos 2\alpha + \sin 2\alpha} \cdot \operatorname{ctg} \alpha = 1\)
\(\dfrac{1 - \cos 2\alpha + \sin 2\alpha}{1 + \cos 2\alpha + \sin 2\alpha} \cdot \operatorname{ctg} \alpha =\)
\(= \dfrac{1 - \cos^{2} \alpha + \sin^{2} \alpha + 2\sin \alpha \cos \alpha}{1 - \sin^{2} \alpha + \cos^{2} \alpha + 2\sin \alpha \cos \alpha} \cdot \operatorname{ctg} \alpha =\)
\(= \dfrac{2\sin \alpha(\sin \alpha + \cos \alpha)}{2\cos \alpha(\cos \alpha + \sin \alpha)} \cdot \dfrac{\cos \alpha}{\sin \alpha} = 1;\)
5) \(\frac{(1 - 2\cos^{2} \alpha)(2\sin^{2} \alpha - 1)}{4\sin^{2} \alpha \cos^{2} \alpha} = \operatorname{ctg}^{2} 2\alpha\)
\(\dfrac{(1 - 2\cos^{2} \alpha)(2\sin^{2} \alpha - 1)}{4\sin^{2} \alpha \cos^{2} \alpha} = \dfrac{(\cos^{2} \alpha + \sin^{2} \alpha - 2\cos^{2} \alpha)(2\sin^{2} \alpha - \sin^{2} \alpha - \cos^{2} \alpha)}{\sin^{2} 2\alpha} =\)
\(= \dfrac{(-\cos 2\alpha)(-\cos 2\alpha)}{\sin^{2} 2\alpha} = \dfrac{\cos^{2} 2\alpha}{\sin^{2} 2\alpha} = \operatorname{ctg}^{2} 2\alpha;\)
6) \(1 - 2\sin^{2}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \sin \alpha\)
\(1 - 2\sin^{2}\left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right) =\)
\(= \sin^{2}\left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right) + \cos^{2}\left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right) - 2\sin^{2}\left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right) =\)
\(= \cos\left(\dfrac{\pi}{2} - \alpha\right) = \cos\dfrac{\pi}{2}\cos \alpha + \sin\dfrac{\pi}{2}\sin \alpha = \sin \alpha;\)
7) \(\frac{\sin \alpha + \sin 2\alpha}{1 + \cos \alpha + \cos 2\alpha} = \operatorname{tg} \alpha\)
\(\dfrac{\sin \alpha + \sin 2\alpha}{1 + \cos \alpha + \cos 2\alpha} = \dfrac{\sin \alpha(1 + 2\cos \alpha)}{\cos \alpha + \cos^{2} \alpha + \sin^{2} \alpha + \cos^{2} \alpha - \sin^{2} \alpha} =\)
\(= \dfrac{\sin \alpha(1 + 2\cos \alpha)}{\cos \alpha(1 + 2\cos \alpha)} = \operatorname{tg} \alpha.\)
