Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1029

\[\boxed{\mathbf{1029.}}\]

\[1)\sin\left( a + \frac{\pi}{6} \right)\]

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

\[\ в\ III\ четверти:\]

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

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

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

\[Получаем:\]

\[\sin\left( a + \frac{\pi}{6} \right) = \sin a \bullet \cos\frac{\pi}{6} +\]

\[+ \cos a \bullet \sin\frac{\pi}{6}\]

\[\sin\left( a + \frac{\pi}{6} \right) = - \frac{4}{5} \bullet \frac{\sqrt{3}}{2} - \frac{3}{5} \bullet \frac{1}{2} =\]

\[= - \frac{4\sqrt{3}}{10} - \frac{3}{10} = - \frac{4\sqrt{3} + 3}{10}\]

\[Ответ:\ \ - \frac{4\sqrt{3} + 3}{10}.\]

\[2)\sin\left( \frac{\pi}{4} - a \right)\]

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

\[во\text{\ II\ }четверти:\]

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

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

\[= - \sqrt{\frac{7}{9}} = - \frac{\sqrt{7}}{3}\]

\[Получаем:\]

\[\sin\left( \frac{\pi}{4} - a \right) = \sin\frac{\pi}{4} \bullet \cos a -\]

\[- \cos\frac{\pi}{4} \bullet \sin a\]

\[\sin\left( \frac{\pi}{4} - a \right) = \frac{\sqrt{2}}{2}\left( - \frac{\sqrt{7}}{3} \right) -\]

\[- \frac{\sqrt{2}}{2} \bullet \frac{\sqrt{2}}{3} = - \frac{\sqrt{14}}{6} - \frac{2}{6} =\]

\[= - \frac{\sqrt{14} + 2}{6}\]

\[Ответ:\ \ - \frac{\sqrt{14} + 2}{6}.\]