ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1177

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\[1)\sin\left( \arcsin\frac{1}{3} + \arccos\frac{2\sqrt{2}}{3} \right) =\]

\[= \sin\left( \arcsin\frac{1}{3} \right) \bullet\]

\[\bullet \cos\left( \arccos\frac{2\sqrt{2}}{3} \right) +\]

\[+ \sin\left( \arccos\frac{2\sqrt{2}}{3} \right) \bullet\]

\[\bullet \cos\left( \arcsin\frac{1}{3} \right) =\]

\[= \frac{1}{3} \bullet \frac{2\sqrt{2}}{3} +\]

\[+ \sqrt{1 - \cos^{2}\left( \arccos\frac{2\sqrt{2}}{3} \right)} \bullet\]

\[\bullet \sqrt{1 - \sin^{2}\left( \arcsin\frac{1}{3} \right)} =\]

\[= \frac{2\sqrt{2}}{9} + \sqrt{1 - \left( \frac{2\sqrt{2}}{3} \right)^{2}} \bullet\]

\[\bullet \sqrt{1 - \left( \frac{1}{3} \right)^{2}} = \frac{2\sqrt{2}}{9} +\]

\[+ \sqrt{\frac{9}{9} - \frac{8}{9}} \bullet \sqrt{\frac{9}{9} - \frac{1}{9}} =\]

\[= \frac{2\sqrt{2}}{9} + \sqrt{\frac{1}{9}} \bullet \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{9} + \frac{\sqrt{8}}{9} =\]

\[= \frac{2\sqrt{2}}{9} + \frac{2\sqrt{2}}{9} = \frac{4\sqrt{2}}{9}\]

\[Ответ:\ \ \frac{4\sqrt{2}}{9}.\]

\[2)\cos\left( \arcsin\frac{3}{5} + \arccos\frac{4}{5} \right) =\]

\[= \cos\left( \arcsin\frac{3}{5} \right) \bullet\]

\[\bullet \cos\left( \arccos\frac{4}{5} \right) - \sin\left( \arcsin\frac{3}{5} \right) \bullet\]

\[\bullet \sin\left( \arccos\frac{4}{5} \right) =\]

\[= \sqrt{1 - \sin^{2}\left( \arcsin\frac{3}{5} \right)} \bullet \frac{4}{5} -\]

\[- \frac{3}{5} \bullet \sqrt{1 - \cos^{2}\left( \arccos\frac{4}{5} \right)} =\]

\[= \sqrt{1 - \left( \frac{3}{5} \right)^{2}} \bullet \frac{4}{5} - \frac{3}{5} \bullet\]

\[\bullet \sqrt{1 - \left( \frac{4}{5} \right)^{2}} = \sqrt{\frac{25}{25} - \frac{9}{25}} \bullet \frac{4}{5} -\]

\[- \frac{3}{5} \bullet \sqrt{\frac{25}{25} - \frac{16}{25}} =\]

\[= \sqrt{\frac{16}{25}} \bullet \frac{4}{5} - \frac{3}{5} \bullet \sqrt{\frac{9}{25}} = \frac{4}{5} \bullet \frac{4}{5} -\]

\[- \frac{3}{5} \bullet \frac{3}{5} = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}\]

\[Ответ:\ \ \frac{7}{25}.\]