Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 515

\[\boxed{\mathbf{515}\mathbf{.}}\]

\[\cos a = 0,6\ \ и\ \ 0 < a < \frac{\pi}{2}:\ \]

\[угол\ \text{a\ }принадлежит\ \]

\[I\ четверти.\]

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

\[\sin^{2}\frac{a}{2} = \frac{1 - \cos a}{2} = \frac{1 - 0,6}{2} =\]

\[= \frac{0,4}{2} = \frac{2}{10} = \frac{1}{5}\]

\[\sin\frac{a}{2} = \sqrt{\frac{1}{5}} = \sqrt{\frac{5}{25}} = \frac{\sqrt{5}}{5}\]

\[Ответ:\ \ \frac{\sqrt{5}}{5}.\]

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

\[\cos^{2}\frac{a}{2} = \frac{1 + \cos a}{2} = \frac{1 + 0,6}{2} =\]

\[= \frac{1,6}{2} = 0,8 = \frac{4}{5}\]

\[\cos\frac{a}{2} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\]

\[Ответ:\ \ \frac{2\sqrt{5}}{5}.\]

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

\[tg^{2}\frac{a}{2} = \frac{1 - \cos a}{1 + \cos a} = \frac{1 - 0,6}{1 + 0,6} =\]

\[= \frac{0,4}{1,6} = \frac{4}{16} = \frac{1}{4}\]

\[tg = \sqrt{\frac{1}{4}} = \frac{1}{2}\]

\[Ответ:\ \ \frac{1}{2}.\]

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

\[\text{ct}g^{2}\frac{a}{2} = \frac{1 + \cos a}{1 - \cos a} = \frac{1 + 0,6}{1 - 0,6} =\]

\[= \frac{1,6}{0,4} = \frac{16}{4} = 4\]

\[ctg = \sqrt{4} = 2\]

\[Ответ:\ \ 2.\]