ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 458

\[\boxed{\mathbf{458.}}\]

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

\[\text{\ II\ }четверти:\]

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

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

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

\[tg\ a = \frac{\sin a}{\cos a} = \frac{4}{5}\ :\left( - \frac{3}{5} \right) =\]

\[= \frac{4}{5} \bullet \left( - \frac{5}{3} \right) = - \frac{4}{3}\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = 1\ :\left( - \frac{4}{3} \right) = - \frac{3}{4}\]

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

\[в\ III\ четверти:\]

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

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

\[= - \sqrt{\frac{21}{25}} = - \frac{\sqrt{21}}{5}\]

\[tg\ a = \frac{\sin a}{\cos a} = - \frac{2}{5}\ :\left( - \frac{\sqrt{21}}{5} \right) =\]

\[= - \frac{2}{5} \bullet \left( - \frac{5}{\sqrt{21}} \right) = \frac{2}{\sqrt{21}}\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = 1\ :\frac{2}{\sqrt{21}} = \frac{\sqrt{21}}{2}\]