Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1127

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\[1)\ \frac{2\sin{2a} - \sin{4a}}{2\sin{2a} + \sin{4a}} = tg^{2}\text{\ a}\]

\[\frac{2\sin{2a} - 2 \bullet \sin{2a} \bullet \cos{2a}}{2\sin{2a} + 2 \bullet \sin{2a} \bullet \cos{2a}} =\]

\[= tg^{2}\text{\ a}\]

\[\frac{2\sin{2a} \bullet \left( 1 - \cos{2a} \right)}{2\sin{2a} \bullet \left( 1 + \cos{2a} \right)} = tg^{2}\text{\ a}\]

\[\frac{1 - \cos{2a}}{1 + \cos{2a}} = tg^{2}\text{\ a}\]

\[tg^{2}\ \left( \frac{2a}{2} \right) = tg^{2}\text{\ a}\]

\[tg^{2}\ a = tg^{2}\text{\ a}\]

\[Что\ и\ требовалось\ доказать.\]

\[2)\ \frac{2\cos{2a} - \sin{4a}}{2\cos{2a} + \sin{4a}} =\]

\[= tg^{2}\left( \frac{\pi}{4} - a \right)\]

\[\frac{2\cos{2a} - 2 \bullet \sin{2a} \bullet \cos{2a}}{2\cos{2a} + 2 \bullet \sin{2a} \bullet \cos{2a}} =\]

\[= tg^{2}\left( \frac{\pi}{4} - a \right)\]

\[\frac{2\cos{2a} \bullet \left( 1 - \sin{2a} \right)}{2\cos{2a} \bullet \left( 1 + \sin{2a} \right)} =\]

\[= tg^{2}\left( \frac{\pi}{4} - a \right)\]

\[\frac{1 - \sin{2a}}{1 + \sin{2a}} = tg^{2}\left( \frac{\pi}{4} - a \right)\]

\[\frac{1 - \cos\left( \frac{\pi}{2} - 2a \right)}{1 + \cos\left( \frac{\pi}{2} - 2a \right)} = tg^{2}\left( \frac{\pi}{4} - a \right)\]

\[tg^{2}\left( \frac{1}{2} \bullet \left( \frac{\pi}{2} - 2a \right) \right) =\]

\[= tg^{2}\left( \frac{\pi}{4} - a \right)\]

\[tg^{2}\left( \frac{\pi}{4} - a \right) = tg^{2}\left( \frac{\pi}{4} - a \right)\]

\[Что\ и\ требовалось\ доказать.\]