Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 510

\[\boxed{\mathbf{510}\mathbf{.}}\]

\[1)\ \frac{\cos{2a}}{\sin a \bullet \cos a + \sin^{2}a} =\]

\[= ctg\ a - 1\]

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

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

\[= ctg\ a - 1\]

\[\frac{\cos a - \sin a}{\sin a} = ctg\ a - 1\]

\[\frac{\cos a}{\sin a} - 1 = ctg\ a - 1\]

\[ctg\ a - 1 = ctg\ a - 1\]

\[Что\ и\ требовалось\ доказать.\]

\[2)\ \frac{\sin{2a} - 2\cos a}{\sin a - \sin^{2}a} = - 2\ ctg\ a\]

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

\[= - 2\ ctg\ a\]

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

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

\[- 2\ ctg\ a = - 2\ ctg\ a\]

\[Что\ и\ требовалось\ доказать.\]

\[3)\ tg\ a\ \left( 1 + \cos{2a} \right) = \sin{2a}\]

\[tg\ a \bullet 2\cos^{2}a = \sin{2a}\]

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

\[2\sin a \bullet \cos a = \sin{2a}\]

\[\sin{2a} = \sin{2a}\]

\[Что\ и\ требовалось\ доказать.\]

\[4)\ \frac{1 - \cos{2a} + \sin{2a}}{1 + \cos{2a} + \sin{2a}} \bullet ctg\ a =\]

\[= 1\]

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

\[= 1\]

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

\[= 1\]

\[\frac{\sin a}{\cos a} \bullet \frac{\cos a}{\sin a} = 1\]

\[1 = 1\]

\[Что\ и\ требовалось\ доказать.\]

\[5)\ \frac{\left( 1 - 2\cos^{2}a \right)\left( 2\sin^{2}a - 1 \right)}{4\sin^{2}\text{\ a} \bullet \cos^{2}a} =\]

\[= \text{ct}g^{2}\ 2a\]

\[\frac{\left( - \cos{2a} \right)\left( - \cos{2a} \right)}{\sin^{2}{2a}} = ctg^{2}\ 2a\]

\[\frac{\cos^{2}{2a}}{\sin^{2}{2a}} = ctg^{2}\ 2a\]

\[\text{ct}g^{2}\ 2a = ctg^{2}\ 2a\]

\[Что\ и\ требовалось\ доказать.\]

\[6)\ 1 - 2\sin^{2}\left( \frac{\pi}{4} - \frac{a}{2} \right) = \sin a\]

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

\[= \sin a\]

\[\cos\left( 2 \bullet \left( \frac{\pi}{4} - \frac{a}{2} \right) \right) = \sin a\]

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

\[\sin a = \sin a\]

\[Что\ и\ требовалось\ доказать.\]

\[7)\ \frac{\sin a + \sin{2a}}{1 + \cos a + \cos{2a}} = tg\ a\]

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

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

\[\frac{\sin a}{\cos a} = tg\ a\]

\[tg\ a = tg\ a\]

\[Что\ и\ требовалось\ доказать.\]