Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1118

\[\boxed{\mathbf{1118}\mathbf{.}}\]

\[1)\cos a\]

\[\sin a = \frac{\sqrt{3}}{3}\text{\ \ }и\ \ \frac{\pi}{2} < a < \pi;\]

\[\text{a\ }принадлежит\ II\ четверти:\]

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

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

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

\[Ответ:\ - \frac{\sqrt{6}}{3}.\]

\[2)\ tg\ a\]

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

\[\text{\ a\ }принадлежит\ III\ четверти:\]

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

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

\[= \sqrt{\frac{9}{5} - \frac{5}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\]

\[Ответ:\ \ \frac{2\sqrt{5}}{5}.\]

\[3)\sin a\]

\[tg\ a = 2\sqrt{2}\text{\ \ }и\ \ 0 < a < \frac{\pi}{2};\]

\[\text{a\ }принадлежит\ I\ четверти:\]

\[\sin a = \frac{\sin a}{\cos a} \bullet \cos a = tg\ a \bullet\]

\[\bullet \cos a = tg\ a \bullet \sqrt{\frac{1}{1 + tg^{2}\text{\ a}}}\]

\[\sin a = 2\sqrt{2} \bullet \sqrt{\frac{1}{1 + \left( 2\sqrt{2} \right)^{2}}} =\]

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

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

\[Ответ:\ \ \frac{2\sqrt{2}}{3}.\]

\[4)\cos a\ \]

\[ctg\ a = \sqrt{2}\text{\ \ }и\ \ \pi < a < \frac{3\pi}{2};\]

\[\text{a\ }принадлежит\ III\ четверти:\]

\[\cos a = \frac{\cos a}{\sin a} \bullet \sin a = ctg\ a \bullet\]

\[\bullet \sin a = - ctg\ a \bullet \sqrt{\frac{1}{1 + ctg^{2}\text{\ a}}}\]

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

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

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

\[Ответ:\ - \frac{\sqrt{6}}{3}.\]