ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 601

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\[1)\cos\left( \arcsin\frac{3}{5} \right) =\]

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

\[= \sqrt{1 - \left( \frac{3}{5} \right)^{2}} = \sqrt{\frac{25}{25} - \frac{9}{25}} =\]

\[= \sqrt{\frac{16}{25}} = \frac{4}{5}\]

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

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

\[= \sqrt{1 - \left( - \frac{4}{5} \right)^{2}} = \sqrt{\frac{25}{25} - \frac{16}{25}} =\]

\[= \sqrt{\frac{9}{25}} = \frac{3}{5}\]

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

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

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

\[= \sqrt{\frac{8}{9}} = \sqrt{\frac{4 \bullet 2}{9}} = \frac{2\sqrt{2}}{3}\]

\[4)\cos\left( \arcsin\frac{1}{4} \right) =\]

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

\[= \sqrt{1 - \left( \frac{1}{4} \right)^{2}} = \sqrt{\frac{16}{16} - \frac{1}{16}} =\]

\[= \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}\]