Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 461

\[\boxed{\mathbf{461}\mathbf{.}}\]

\[1)\sin a = \frac{1}{5}\text{\ \ }и\ \ \text{tg}\ a = \frac{1}{\sqrt{24}}\]

\[\cos a = \frac{\sin a}{\text{tg}\ a} = \frac{1}{5}\ :\frac{1}{\sqrt{24}} = \frac{\sqrt{24}}{5}\]

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

\[= \left( \frac{1}{5} \right)^{2} + \left( \frac{\sqrt{24}}{5} \right)^{2} = \frac{1}{25} + \frac{24}{25} =\]

\[= \frac{25}{25} = 1\]

\[Ответ:\ \ могут.\]

\[2)\ ctg\ a = \frac{\sqrt{7}}{3}\text{\ \ }и\ \cos a = \frac{3}{4}\]

\[\sin a = \frac{\cos a}{\text{ctg\ a}} = \frac{3}{4}\ :\frac{\sqrt{7}}{3} =\]

\[= \frac{3}{4} \bullet \frac{3}{\sqrt{7}} = \frac{9}{4\sqrt{7}}\]

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

\[= \left( \frac{9}{4\sqrt{7}} \right)^{2} + \left( \frac{3}{4} \right)^{2} = \frac{81}{112} + \frac{9}{16} =\]

\[= \frac{81 + 63}{112} = \frac{144}{112} \neq 1\]

\[Ответ:\ \ не\ могут.\]