ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 95

\[1)\arcsin\frac{1}{\sqrt{3}}\ и\ \arcsin\frac{2}{\sqrt{10}};\]

\[\frac{1}{\sqrt{3}}\ :\frac{2}{\sqrt{10}} = \frac{\sqrt{10}}{2\sqrt{3}} = \frac{\sqrt{10}}{\sqrt{12}} < 1;\]

\[\arcsin\frac{1}{\sqrt{3}} < \arcsin\frac{2}{\sqrt{10}}.\]

\[2)\arcsin\left( - \frac{2}{3} \right)\ и\ \arcsin\left( - \frac{3}{4} \right);\]

\[- \frac{2}{3} + \frac{3}{4} = \frac{- 8 + 9}{12} = \frac{1}{12} > 0;\]

\[\arcsin\left( - \frac{2}{3} \right) > \arcsin\left( - \frac{3}{4} \right).\]

\[3)\arcsin\frac{2}{\sqrt{5}}\ и\ \arcsin\frac{\sqrt{5}}{3};\]

\[\frac{2}{\sqrt{5}}\ :\frac{\sqrt{5}}{3} = \frac{6}{\sqrt{25}} = \frac{6}{5} > 1;\]

\[\arcsin\frac{2}{\sqrt{5}} > \arcsin\frac{\sqrt{5}}{3}.\]

\[4)\arcsin\left( - \frac{\sqrt{2}}{3} \right)\ и\ \arcsin\left( - \frac{3}{4} \right);\]

\[\frac{\sqrt{2}}{3}\ :\frac{3}{4} = \frac{4\sqrt{2}}{9} = \frac{\sqrt{32}}{\sqrt{81}} < 1;\]

\[\arcsin\left( - \frac{\sqrt{2}}{3} \right) > \arcsin\left( - \frac{3}{4} \right).\]