Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1066

\[\boxed{\mathbf{1066}\mathbf{.}}\]

\[\sin a = \frac{3}{5}\text{\ \ }и\ \ \frac{\pi}{2} < a < \pi:\]

\[угол\ \text{a\ }принадлежит\ II\ \]

\[четверти.\]

\[\cos a = - \sqrt{1 - \sin^{2}a}\]

\[\cos a = - \sqrt{1 - \left( \frac{3}{5} \right)^{2}} = -\]

\[- \sqrt{\frac{25}{25} - \frac{9}{25}} = - \sqrt{\frac{16}{25}} =\]

\[= - \frac{4}{5} = - 0,8.\]

\[1)\sin\frac{a}{2}\]

\[\sin^{2}\frac{a}{2} = \frac{1 - \cos a}{2} = \frac{1 + 0,8}{2} =\]

\[= \frac{1,8}{2} = 0,9 = \frac{9}{10}\]

\[\sin\frac{a}{2} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}\]

\[Ответ:\ \ \frac{3\sqrt{10}}{10}.\]

\[2)\cos\frac{a}{2}\]

\[\cos^{2}\frac{a}{2} = \frac{1 + \cos a}{2} = \frac{1 - 0,8}{2} =\]

\[= \frac{0,2}{2} = 0,1 = \frac{1}{10}\]

\[\cos\frac{a}{2} = \sqrt{\frac{1}{10}} = \sqrt{\frac{10}{100}} = \frac{\sqrt{10}}{10}\]

\[Ответ:\ \ \frac{\sqrt{10}}{10}.\]

\[3)\ tg\frac{a}{2}\]

\[tg^{2}\frac{a}{2} = \frac{1 - \cos a}{1 + \cos a} = \frac{1 + 0,8}{1 - 0,8} =\]

\[= \frac{1,8}{0,2} = \frac{18}{2} = 9\]

\[tg = \sqrt{9} = 3\]

\[Ответ:\ \ 3.\]

\[4)\ ctg\frac{a}{2}\]

\[\text{ct}g^{2}\frac{a}{2} = \frac{1 + \cos a}{1 - \cos a} = \frac{1 - 0,8}{1 + 0,8} =\]

\[= \frac{0,2}{1,8} = \frac{2}{18} = \frac{1}{9}\]

\[ctg = \sqrt{\frac{1}{9}} = \frac{1}{3}\]

\[Ответ:\ \ \frac{1}{3}.\]