Страница 408 номер 1319, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Доказать тождество (1312—1320). 1) \(\frac{1 - 2\sin^2 \alpha}{1 + \sin 2\alpha} = \frac{1 - \operatorname{tg} \alpha}{1 + \operatorname{tg} \alpha}\); 2) \(\frac{1}{4\sin^2 \alpha \cos^2 \alpha} = 1 + \frac{(1 - \operatorname{tg}^2 \alpha)^2}{4\operatorname{tg}^2 \alpha}\); 3) \(\operatorname{tg}\left(\frac{\pi}{4} + \alpha\right) = \frac{1 + \sin 2\alpha}{\cos 2\alpha}\); 4) \(\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha} = \operatorname{ctg}^2\left(\frac{\pi}{4} + \alpha\right)\).
1) \(\frac{1 - 2\sin^{2}\alpha}{1 + \sin{2\alpha}} = \frac{1 - \operatorname{tg} \alpha}{1 + \operatorname{tg} \alpha}\)
\(\dfrac{1 - 2\sin^{2}\alpha}{1 + \sin{2\alpha}} = \dfrac{\cos^{2}\alpha - \sin^{2}\alpha}{\left( \cos\alpha + \sin\alpha \right)^{2}} =\)
\(= \dfrac{\cos\alpha - \sin\alpha}{\cos\alpha + \sin\alpha} = \dfrac{1 - \frac{\sin\alpha}{\cos\alpha}}{1 + \dfrac{\sin\alpha}{\cos\alpha}} = \dfrac{1 - \operatorname{tg} \alpha}{1 + \operatorname{tg} \alpha};\)
2) \(\frac{1}{4\sin^{2}\alpha\cos^{2}\alpha} = 1 + \frac{\left( 1 - \operatorname{tg}^{2}\alpha \right)^{2}}{4\operatorname{tg}^{2}\alpha}\)
\(\dfrac{1}{4\sin^{2}\alpha\cos^{2}\alpha} = \dfrac{\sin^{2}{2\alpha} + \cos^{2}{2\alpha}}{\sin^{2}{2\alpha}} = 1 + \operatorname{ctg}^{2}{2\alpha};\)
\(1 + \dfrac{\left( 1 - \operatorname{tg}^{2}\alpha \right)^{2}}{4\operatorname{tg}^{2}\alpha} = 1 + \left( \dfrac{1}{\operatorname{tg} {2\alpha}} \right)^{2} = 1 + \operatorname{ctg}^{2}{2\alpha};\)
3) \(\operatorname{tg} \left( \frac{\pi}{4} + \alpha \right) = \frac{1 + \sin{2\alpha}}{\cos{2\alpha}}\)
\(\operatorname{tg} \left( \dfrac{\pi}{4} + \alpha \right) = \dfrac{\sin\left( \frac{\pi}{4} + \alpha \right)}{\cos\left( \dfrac{\pi}{4} + \alpha \right)} =\)
\(= \dfrac{\sin\frac{\pi}{4}\cos\alpha + \cos\frac{\pi}{4}\sin\alpha}{\cos\dfrac{\pi}{4}\cos\alpha - \sin\dfrac{\pi}{4}\sin\alpha} = \dfrac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha};\)
\(\dfrac{1 + \sin{2\alpha}}{\cos{2\alpha}} = \dfrac{\left( \sin\alpha + \cos\alpha \right)^{2}}{\cos^{2}\alpha - \sin^{2}\alpha} = \dfrac{\sin\alpha + \cos\alpha}{\cos\alpha - \sin\alpha};\)
4) \(\frac{1 - \sin{2\alpha}}{1 + \sin{2\alpha}} = \operatorname{ctg}^{2}\left( \frac{\pi}{4} + \alpha \right)\)
\(\dfrac{1 - \sin{2\alpha}}{1 + \sin{2\alpha}} = \dfrac{\left( \cos\alpha - \sin\alpha \right)^{2}}{\left( \cos\alpha + \sin\alpha \right)^{2}};\)
\(\operatorname{ctg}^{2}\left( \dfrac{\pi}{4} + \alpha \right) = \dfrac{\cos^{2}\left( \frac{\pi}{4} + \alpha \right)}{\sin^{2}\left( \dfrac{\pi}{4} + \alpha \right)} =\)
\(= \dfrac{\left( \cos\frac{\pi}{4}\cos\alpha - \sin\frac{\pi}{4}\sin\alpha \right)^{2}}{\left( \sin\dfrac{\pi}{4}\cos\alpha + \cos\dfrac{\pi}{4}\sin\alpha \right)^{2}} = \dfrac{\left( \cos\alpha - \sin\alpha \right)^{2}}{\left( \cos\alpha + \sin\alpha \right)^{2}}.\)
