Страница 407 номер 1306, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Упростить выражение (1304—1309). 1) \(\cos^2(\alpha + 2\beta) + \sin^2(\alpha - 2\beta) - 1\); 2) \(\sin^2(\alpha + 2\beta) + \sin^2(\alpha - 2\beta) - 1\).
1) \(\cos^{2}(\alpha + 2\beta) + \sin^{2}(\alpha - 2\beta) - 1 = \cos^{2}(\alpha + 2\beta) - \cos^{2}(\alpha - 2\beta) =\)
\(= \left( \cos(\alpha + 2\beta) - \cos(\alpha - 2\beta) \right)\left( \cos(\alpha + 2\beta) + \cos(\alpha - 2\beta) \right) =\)
\(= - 2\sin\alpha\sin{2\beta} \cdot 2\cos\alpha\cos{2\beta} = - \sin{2\alpha}\sin{4\beta};\)
2) \(\sin^{2}(\alpha + 2\beta) + \sin^{2}(\alpha - 2\beta) - 1 = \sin^{2}(\alpha + 2\beta) - \cos^{2}(\alpha - 2\beta) =\)
\(= \left( \sin(\alpha + 2\beta) - \cos(\alpha - 2\beta) \right)\left( \sin(\alpha + 2\beta) + \cos(\alpha - 2\beta) \right) =\)
\(= \left( \sin\alpha\cos{2\beta} + \cos\alpha\sin{2\beta} - \cos\alpha\cos{2\beta} - \sin\alpha\sin{2\beta} \right) \times\)
\(\times \left( \sin\alpha\cos{2\beta} + \cos\alpha\sin{2\beta} + \cos\alpha\cos{2\beta} + \sin\alpha\sin{2\beta} \right) =\)
\(= \left( \sin\alpha\left( \cos{2\beta} - \sin{2\beta} \right) - \cos\alpha\left( \cos{2\beta} - \sin{2\beta} \right) \right) \times\)
\(\times \left( \sin\alpha\left( \cos{2\beta} + \sin{2\beta} \right) + \cos\alpha\left( \cos{2\beta} + \sin{2\beta} \right) \right) =\)
\(= \left( \sin\alpha - \cos\alpha \right)\left( \sin\alpha + \cos\alpha \right)\left( \cos{2\beta} - \sin{2\beta} \right)\left( \cos{2\beta} + \sin{2\beta} \right) =\)
\(= \left( \sin^{2}\alpha - \cos^{2}\alpha \right)\left( \cos^{2}{2\beta} - \sin^{2}{2\beta} \right) = - \cos{2\alpha} \cdot \cos{4\beta}.\)
