Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 503

\[\boxed{\mathbf{503}\mathbf{.}}\]

\[1)\sin a = \frac{3}{5}\text{\ \ }и\ \ \frac{\pi}{2} < a < \pi\]

\[\text{\ a\ }принадлежит\ второй\ \]

\[четверти:\]

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

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

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

\[\sin{2a} = 2\sin a \bullet \cos a\]

\[\sin{2a} = 2 \bullet \frac{3}{5} \bullet \left( - \frac{4}{5} \right) = - \frac{24}{25}\]

\[Ответ:\ - \frac{24}{25}.\]

\[2)\cos a = - \frac{4}{5}\text{\ \ }и\ \ \pi < a < \frac{3\pi}{2}\]

\[\text{a\ }принадлежит\ третьей\ \]

\[четверти:\]

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

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

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

\[\sin{2a} = 2\sin a \bullet \cos a\]

\[\sin{2a} = 2 \bullet \left( - \frac{3}{5} \right) \bullet \left( - \frac{4}{5} \right) = \frac{24}{25}\]

\[Ответ:\ \ \frac{24}{25}\text{.\ }\]