Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 582

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

\[= \sqrt{\frac{8}{9}} \bullet \frac{2\sqrt{2}}{3} + \sqrt{\frac{1}{9}} \bullet \frac{1}{3} =\]

\[= \frac{2 \bullet \sqrt{2 \bullet 8}}{3 \bullet 3} + \frac{1}{3} \bullet \frac{1}{3} =\]

\[= \frac{2 \bullet \sqrt{16}}{9} + \frac{1}{9} = \frac{2 \bullet 4 + 1}{9} =\]

\[= \frac{8 + 1}{9} = \frac{9}{9} = 1\]

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

\[= \frac{12}{25} + \sqrt{1 - \left( \frac{4}{5} \right)^{2}} \bullet \sqrt{1 - \left( \frac{3}{5} \right)^{2}} =\]

\[= \frac{12}{25} + \sqrt{\frac{25}{25} - \frac{16}{25}} \bullet \sqrt{\frac{25}{25} - \frac{9}{25}} =\]

\[= \frac{12}{25} + \sqrt{\frac{9}{25}} \bullet \sqrt{\frac{16}{25}} = \frac{12}{25} + \frac{3}{5} \bullet \frac{4}{5} =\]

\[= \frac{12}{25} + \frac{12}{25} = \frac{24}{25}\]