ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 726

\[\boxed{\mathbf{726}\mathbf{.}}\]

\[y = \sin x:\]

\[возрастает\ на\left\lbrack - \frac{\pi}{2};\ \frac{\pi}{2} \right\rbrack;\]

\[убывает\ на\left\lbrack \frac{\pi}{2};\ \frac{3\pi}{2} \right\rbrack.\ \]

\[1)\sin\frac{\pi}{9}\ и\ \cos\frac{\pi}{9};\]

\[\cos\frac{\pi}{9} = \sin\left( \frac{\pi}{2} - \frac{\pi}{9} \right) =\]

\[= \sin\left( \frac{9\pi}{18} - \frac{2\pi}{18} \right) = \sin\frac{7\pi}{18};\]

\[\frac{\pi}{9}\ и\ \frac{7\pi}{18}\ принадлежат\left\lbrack - \frac{\pi}{2};\ \frac{\pi}{2} \right\rbrack -\]

\[\ функция\ возрастает;\]

\[\frac{\pi}{9} < \frac{7\pi}{18}\]

\[\sin\frac{\pi}{9} < \sin\frac{7\pi}{18}\]

\[\sin\frac{\pi}{9} < \cos\frac{\pi}{9}.\]

\[2)\sin\frac{9\pi}{8}\ и\ \cos\frac{9\pi}{8};\]

\[\cos\frac{9\pi}{8} = \sin\left( \frac{\pi}{2} - \frac{9\pi}{8} \right) =\]

\[= \sin\left( \frac{4\pi}{8} - \frac{9\pi}{8} \right) = \sin\left( - \frac{5\pi}{8} \right) =\]

\[= \sin\left( \pi + \frac{5\pi}{8} \right) = \sin\frac{13\pi}{8};\]

\[\frac{9\pi}{8}\ и\ \frac{13\pi}{8}\ принадлежат\left\lbrack \frac{\pi}{2};\ \frac{3\pi}{2} \right\rbrack -\]

\[\ функция\ убывает;\]

\[\frac{9\pi}{8} < \frac{13\pi}{8}\]

\[\sin\frac{9\pi}{8} > \sin\frac{13\pi}{18}\]

\[\sin\frac{9\pi}{8} > \cos\frac{9\pi}{8}.\]

\[3)\sin\frac{\pi}{5}\ и\ \cos\frac{5\pi}{14};\]

\[\cos\frac{5\pi}{14} = \sin\left( \frac{\pi}{2} - \frac{5\pi}{14} \right) =\]

\[= \sin\left( \frac{7\pi}{14} - \frac{5\pi}{14} \right) = \sin\frac{2\pi}{14} = \sin\frac{\pi}{7};\]

\[\frac{\pi}{5}\ и\ \frac{\pi}{7}\ принадлежат\left\lbrack - \frac{\pi}{2};\ \frac{\pi}{2} \right\rbrack -\]

\[функция\ возрастает;\]

\[\frac{\pi}{5} > \frac{\pi}{7}\]

\[\sin\frac{\pi}{5} > \sin\frac{\pi}{7}\]

\[\sin\frac{\pi}{5} > \cos\frac{5\pi}{14}.\]

\[4)\sin\frac{\pi}{8}\ и\ \cos\frac{3\pi}{10};\]

\[\cos\frac{3\pi}{10} = \sin\left( \frac{\pi}{2} - \frac{3\pi}{10} \right) =\]

\[= \sin\left( \frac{5\pi}{10} - \frac{3\pi}{10} \right) = \sin\frac{2\pi}{10} = \sin\frac{\pi}{5};\]

\[\frac{\pi}{8}\ и\ \frac{\pi}{5}\ принадлежат\left\lbrack - \frac{\pi}{2};\ \frac{\pi}{2} \right\rbrack - \ \]

\[функция\ возрастает;\]

\[\frac{\pi}{8} < \frac{\pi}{5}\]

\[\sin\frac{\pi}{8} < \sin\frac{\pi}{5}\]

\[\sin\frac{\pi}{8} < \cos\frac{3\pi}{10}.\]