Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 460

\[\boxed{\mathbf{460.}}\]

\[1)\cos a,\ если\sin a = \frac{2\sqrt{3}}{5}\]

\[\cos a = \pm \sqrt{1 - \sin^{2}a}\]

\[\cos a = \pm \sqrt{1 - \left( \frac{2\sqrt{3}}{5} \right)^{2}} =\]

\[= \pm \sqrt{\frac{25}{25} - \frac{12}{25}} = \pm \sqrt{\frac{13}{25}} = \pm \frac{\sqrt{13}}{5}\]

\[Ответ:\ \ \pm \frac{\sqrt{13}}{5}.\]

\[2)\sin a,\ если\cos a = - \frac{1}{\sqrt{5}}\]

\[\sin a = \pm \sqrt{1 - \cos^{2}a}\]

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

\[= \pm \sqrt{\frac{5}{5} - \frac{1}{5}} = \pm \sqrt{\frac{4}{5}} = \pm \frac{2}{\sqrt{5}}\]

\[Ответ:\ \ \pm \frac{2}{\sqrt{5}}.\]

\[3)\sin a,\ если\cos a = \frac{2}{3}\]

\[\sin a = \pm \sqrt{1 - \cos^{2}a}\]

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

\[= \pm \sqrt{\frac{9}{9} - \frac{4}{9}} = \pm \sqrt{\frac{5}{9}} = \pm \frac{\sqrt{5}}{3}\]

\[Ответ:\ \ \pm \frac{\sqrt{5}}{3}.\]

\[4)\cos a,\ если\sin a = - \frac{1}{\sqrt{3}}\]

\[\cos a = \pm \sqrt{1 - \sin^{2}a}\]

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

\[= \pm \sqrt{\frac{3}{3} - \frac{1}{3}} = \pm \sqrt{\frac{2}{3}} = \pm \sqrt{\frac{6}{9}} =\]

\[= \pm \frac{\sqrt{6}}{3}\]

\[Ответ:\ \ \pm \frac{\sqrt{6}}{3}.\]