Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 1157
Задание 1157
\[\boxed{\mathbf{1157}\mathbf{.}}\]
\[1)\sin\left( \arccos\frac{1}{3} + \arccos\frac{2\sqrt{2}}{3} \right) =\]
\[= \sin\left( \arccos\frac{1}{3} \right) \bullet\]
\[\bullet \cos\left( \arccos\frac{2\sqrt{2}}{3} \right) +\]
\[+ \sin\left( \arccos\frac{2\sqrt{2}}{3} \right) \bullet\]
\[\bullet \cos\left( \arccos\frac{1}{3} \right) =\]
\[= \sqrt{1 - \cos^{2}\left( \arccos\frac{1}{3} \right)} \bullet \frac{2\sqrt{2}}{3} +\]
\[+ \sqrt{1 - \cos^{2}\left( \arccos\frac{2\sqrt{2}}{3} \right)} \bullet \frac{1}{3} =\]
\[= \sqrt{1 - \left( \frac{1}{3} \right)^{2}} \bullet \frac{2\sqrt{2}}{3} +\]
\[+ \sqrt{1 - \left( \frac{2\sqrt{2}}{3} \right)^{2}} \bullet \frac{1}{3} =\]
\[= \sqrt{\frac{9}{9} - \frac{1}{9}} \bullet \frac{2\sqrt{2}}{3} + \sqrt{\frac{9}{9} - \frac{8}{9}} \bullet \frac{1}{3} =\]
\[= \sqrt{\frac{8}{9}} \bullet \frac{2\sqrt{2}}{3} + \sqrt{\frac{1}{9}} \bullet \frac{1}{3} =\]
\[= \frac{2 \bullet \sqrt{2 \bullet 8}}{3 \bullet 3} + \frac{1}{3} \bullet \frac{1}{3} =\]
\[= \frac{2 \bullet \sqrt{16}}{9} + \frac{1}{9} = \frac{2 \bullet 4 + 1}{9} =\]
\[= \frac{8 + 1}{9} = \frac{9}{9} = 1\]
\[2)\cos\left( \arccos\frac{4}{5} - \arccos\frac{3}{5} \right) =\]
\[= \cos\left( \arccos\frac{4}{5} \right) \bullet\]
\[\bullet \cos\left( \arccos\frac{3}{5} \right) + \sin\left( \arccos\frac{4}{5} \right) \bullet\]
\[\bullet \sin\left( \arccos\frac{3}{5} \right) =\]
\[= \frac{4}{5} \bullet \frac{3}{5} + \sqrt{1 - \cos^{2}\left( \arccos\frac{4}{5} \right)} \bullet\]
\[\bullet \sqrt{1 - \cos^{2}\left( \arccos\frac{3}{5} \right)} =\]
\[= \frac{12}{25} + \sqrt{1 - \left( \frac{4}{5} \right)^{2}} \bullet\]
\[\bullet \sqrt{1 - \left( \frac{3}{5} \right)^{2}} = \frac{12}{25} + \sqrt{\frac{25}{25} - \frac{16}{25}} \bullet\]
\[\bullet \sqrt{\frac{25}{25} - \frac{9}{25}} =\]
\[= \frac{12}{25} + \sqrt{\frac{9}{25}} \bullet \sqrt{\frac{16}{25}} = \frac{12}{25} + \frac{3}{5} \bullet\]
\[\bullet \frac{4}{5} = \frac{12}{25} + \frac{12}{25} = \frac{24}{25}\]
