Страница 141 номер 470, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Доказать тождество: 1) \((1 - \cos 2\alpha)(1 + \cos 2\alpha) = \sin^2 2\alpha\); 2) \(\frac{\sin \alpha - 1}{\cos^2 \alpha} = -\frac{1}{1 + \sin \alpha}\); 3) \(\cos^4 \alpha - \sin^4 \alpha = \cos^2 \alpha - \sin^2 \alpha\); 4) \((\sin^2 \alpha - \cos^2 \alpha)^2 + 2\cos^2 \alpha \sin^2 \alpha = \sin^4 \alpha + \cos^4 \alpha\); 5) \(\frac{\sin \alpha}{1 + \cos \alpha} + \frac{1 + \cos \alpha}{\sin \alpha} = \frac{2}{\sin \alpha}\); 6) \(\frac{\sin \alpha}{1 - \cos \alpha} = \frac{1 + \cos \alpha}{\sin \alpha}\); 7) \(\frac{1}{1 + \operatorname{tg}^2 \alpha} + \frac{1}{1 + \operatorname{ctg}^2 \alpha} = 1\); 8) \(\operatorname{tg}^2 \alpha - \sin^2 \alpha = \operatorname{tg}^2 \alpha \sin^2 \alpha\).
1) \((1 - \cos 2\alpha)(1 + \cos 2\alpha) = \sin^{2} 2\alpha\)
\(1 - \cos^{2} 2\alpha = \sin^{2} 2\alpha;\)
2) \(\frac{\sin \alpha - 1}{\cos^{2} \alpha} = -\frac{1}{1 + \sin \alpha}\)
\(\dfrac{\sin \alpha - 1}{1 - \sin^{2} \alpha} = -\dfrac{1}{1 + \sin \alpha},\)
\(\dfrac{\sin \alpha - 1}{(1 - \sin \alpha)(1 + \sin \alpha)} = -\dfrac{1}{1 + \sin \alpha},\)
\(-\dfrac{1}{1 + \sin \alpha} = -\dfrac{1}{1 + \sin \alpha};\)
3) \(\cos^{4} \alpha - \sin^{4} \alpha = \cos^{2} \alpha - \sin^{2} \alpha\)
\((\cos^{2} \alpha - \sin^{2} \alpha)(\cos^{2} \alpha + \sin^{2} \alpha) = \cos^{2} \alpha - \sin^{2} \alpha,\)
\(\cos^{2} \alpha - \sin^{2} \alpha = \cos^{2} \alpha - \sin^{2} \alpha;\)
4) \((\sin^{2} \alpha - \cos^{2} \alpha)^{2} + 2\cos^{2} \alpha \sin^{2} \alpha = \sin^{4} \alpha + \cos^{4} \alpha\)
\(\sin^{4} \alpha - 2\cos^{2} \alpha \sin^{2} \alpha + \cos^{4} \alpha + 2\cos^{2} \alpha \sin^{2} \alpha = \sin^{4} \alpha + \cos^{4} \alpha,\)
\(\sin^{4} \alpha + \cos^{4} \alpha = \sin^{4} \alpha + \cos^{4} \alpha;\)
5) \(\frac{\sin \alpha}{1 + \cos \alpha} + \frac{1 + \cos \alpha}{\sin \alpha} = \frac{2}{\sin \alpha}\)
\(\dfrac{\sin^{2} \alpha + (1 + \cos \alpha)^{2}}{(1 + \cos \alpha)\sin \alpha} = \dfrac{2}{\sin \alpha},\)
\(\dfrac{\sin^{2} \alpha + 1 + 2\cos \alpha + \cos^{2} \alpha}{(1 + \cos \alpha)\sin \alpha} = \dfrac{2}{\sin \alpha},\)
\(\dfrac{2(1 + \cos \alpha)}{(1 + \cos \alpha)\sin \alpha} = \dfrac{2}{\sin \alpha},\)
\(\dfrac{2}{\sin \alpha} = \dfrac{2}{\sin \alpha};\)
6) \(\frac{\sin \alpha}{1 - \cos \alpha} = \frac{1 + \cos \alpha}{\sin \alpha}\)
\(\dfrac{\sin^{2} \alpha}{(1 - \cos \alpha)\sin \alpha} = \dfrac{1 + \cos \alpha}{\sin \alpha},\)
\(\dfrac{1 - \cos^{2} \alpha}{(1 - \cos \alpha)\sin \alpha} = \dfrac{1 + \cos \alpha}{\sin \alpha};\)
\(\dfrac{1 + \cos \alpha}{\sin \alpha} = \dfrac{1 + \cos \alpha}{\sin \alpha};\)
7) \(\frac{1}{1 + \operatorname{tg}^{2} \alpha} + \frac{1}{1 + \operatorname{ctg}^{2} \alpha} = 1\)
\(\dfrac{1}{\dfrac{\cos^{2} \alpha}{\cos^{2} \alpha} + \dfrac{\sin^{2} \alpha}{\cos^{2} \alpha}} + \dfrac{1}{\dfrac{\sin^{2} \alpha}{\sin^{2} \alpha} + \dfrac{\cos^{2} \alpha}{\sin^{2} \alpha}} = 1,\)
\(\dfrac{\cos^{2} \alpha}{\cos^{2} \alpha + \sin^{2} \alpha} + \dfrac{\sin^{2} \alpha}{\sin^{2} \alpha + \cos^{2} \alpha} = 1,\)
\(\cos^{2} \alpha + \sin^{2} \alpha = 1;\)
8) \(\operatorname{tg}^{2} \alpha - \sin^{2} \alpha = \operatorname{tg}^{2} \alpha \sin^{2} \alpha\)
\(\sin^{2} \alpha \left(\dfrac{1}{\cos^{2} \alpha} - 1\right) = \operatorname{tg}^{2} \alpha \sin^{2} \alpha,\)
\(\sin^{2} \alpha \left(\dfrac{1 - \cos^{2} \alpha}{\cos^{2} \alpha}\right) = \operatorname{tg}^{2} \alpha \sin^{2} \alpha,\)
\(\sin^{2} \alpha \operatorname{tg}^{2} \alpha = \operatorname{tg}^{2} \alpha \sin^{2} \alpha.\)
