Страница 407 номер 1308, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Упростить выражение (1304—1309). 1) \(\frac{4\sin^2 \alpha - \sin^2 2\alpha}{4 - 4\sin^2 \alpha - \sin^2 2\alpha}\); 2) \(\frac{\operatorname{tg}^2 2\alpha \operatorname{tg}^2 \alpha - 1}{\operatorname{tg}^2 \alpha - \operatorname{tg}^2 2\alpha}\).
1) \(\frac{4\sin^{2}\alpha - \sin^{2}{2\alpha}}{4 - 4\sin^{2}\alpha - \sin^{2}{2\alpha}} = \frac{4\sin^{2}\alpha - 4\sin^{2}\alpha\cos^{2}\alpha}{4\cos^{2}\alpha - 4\sin^{2}\alpha\cos^{2}\alpha} =\)
\(= \dfrac{4\sin^{2}\alpha\left( 1 - \cos^{2}\alpha \right)}{4\cos^{2}\alpha\left( 1 - \sin^{2}\alpha \right)} = \dfrac{\sin^{4}\alpha}{\cos^{4}\alpha} = \operatorname{tg}^{4}\alpha;\)
2) \(\frac{\operatorname{tg}^{2}{2\alpha}\operatorname{tg}^{2}\alpha - 1}{\operatorname{tg}^{2}\alpha - \operatorname{tg}^{2}{2\alpha}} = \left( \left( \frac{2\operatorname{tg} \alpha}{1 - \operatorname{tg}^{2}\alpha} \right)^{2}\operatorname{tg}^{2}\alpha - 1 \right)\ :\left( \operatorname{tg}^{2}\alpha - \left( \frac{2\operatorname{tg} \alpha}{1 - \operatorname{tg}^{2}\alpha} \right)^{2} \right) =\)
\(= \dfrac{4\operatorname{tg}^{4}\alpha - 1 + 2\operatorname{tg}^{2}\alpha - \operatorname{tg}^{4}\alpha}{\left( 1 - \operatorname{tg}^{2}\alpha \right)^{2}} \cdot \dfrac{\left( 1 - \operatorname{tg}^{2}\alpha \right)^{2}}{\operatorname{tg}^{2}\alpha - 2\operatorname{tg}^{4}\alpha + \operatorname{tg}^{6}\alpha - 4\operatorname{tg}^{2}\alpha} =\)
\[= \frac{3\operatorname{tg}^{4}\alpha + 2\operatorname{tg}^{2}\alpha - 1}{\operatorname{tg}^{6}\alpha - 2\operatorname{tg}^{4}\alpha - 3\operatorname{tg}^{2}\alpha} = \left| \begin{array}{l} \operatorname{tg}^{2}\alpha = z \\ z_{1{,}2} = \frac{-2 \pm \sqrt{4 + 12}}{6},\ z_{1} = \frac{1}{3},\ z_{2} = - 1 \\ z_{3{,}4} = \frac{2 \pm \sqrt{4 + 12}}{2},z_{3} = 3,\ z_{4} = - 1\ \end{array} \right| =\]
\(= \dfrac{3\left( \operatorname{tg}^{2}\alpha + 1 \right)\left( \operatorname{tg}^{2}\alpha - \frac{1}{3} \right)}{\operatorname{tg}^{2}\alpha\left( \operatorname{tg}^{2}\alpha + 1 \right)\left( \operatorname{tg}^{2}\alpha - 3 \right)} = \dfrac{3\operatorname{tg}^{2}\alpha - 1}{\operatorname{tg}^{2}\alpha\left( \operatorname{tg}^{2}\alpha - 3 \right)} = \dfrac{\left( 3\sin^{2}\alpha - \cos^{2}\alpha \right)\cos^{2}\alpha}{\sin^{2}\alpha\left( \sin^{2}\alpha - 3\cos^{2}\alpha \right)} =\)
\(= \operatorname{ctg} \alpha \cdot \dfrac{\cos\alpha\left( \cos^{2}\alpha - 3\sin^{2}\alpha \right)}{\sin\alpha\left( 3\cos^{2}\alpha - \sin^{2}\alpha \right)} =\)
\(= \operatorname{ctg} \alpha \cdot \dfrac{\cos{2\alpha}\cos\alpha - \sin{2\alpha}\sin\alpha}{\sin{2\alpha}\cos\alpha + \cos{2\alpha}\sin\alpha} =\)
\(= \operatorname{ctg} \alpha \cdot \dfrac{\cos{3\alpha}}{\sin{3\alpha}} = \operatorname{ctg} \alpha \cdot \operatorname{ctg} {3\alpha}.\)
