Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 714

\[\boxed{\mathbf{714}\mathbf{.}}\]

\[y = \cos x:\]

\[возрастает\ на\ \lbrack\pi;\ 2\pi\rbrack;\]

\[убывает\ на\ \lbrack 0;\ \pi\rbrack.\ \]

\[1)\cos\frac{\pi}{5}\ и\ \sin\frac{\pi}{5};\]

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

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

\[\frac{\pi}{5}\ и\ \frac{3\pi}{10}\ принадлежат\ \lbrack 0;\ \pi\rbrack -\]

\[функция\ убывает;\]

\[\frac{\pi}{5} < \frac{3\pi}{10}\]

\[\cos\frac{\pi}{5} > \cos\frac{3\pi}{10}\]

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

\[2)\sin\frac{\pi}{7}\ и\ \cos\frac{\pi}{7};\]

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

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

\[\frac{5\pi}{14}\ и\ \frac{\pi}{7}\ принадлежат\ \lbrack 0;\ \pi\rbrack -\]

\[функция\ убывает;\]

\[\frac{5\pi}{14} > \frac{\pi}{7}\]

\[\cos\frac{4\pi}{14} < \cos\frac{\pi}{7}\]

\[\sin\frac{\pi}{7} < \cos\frac{\pi}{7}.\]

\[3)\cos\frac{3\pi}{8}\ и\ \sin\frac{5\pi}{8};\]

\[\sin\frac{5\pi}{8} = \cos\left( \frac{\pi}{2} - \frac{5\pi}{8} \right) =\]

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

\[= \cos\left( - \frac{\pi}{8} \right) = \cos\left( \frac{\pi}{8} \right);\]

\[\frac{3\pi}{8}\ и\ \frac{\pi}{8}\ принадлежат\ \lbrack 0;\ \pi\rbrack - \ \]

\[функция\ убывает;\]

\[\frac{3\pi}{8} > \frac{\pi}{8}\]

\[\cos\frac{3\pi}{8} < \cos\frac{\pi}{8}\]

\[\cos\frac{3\pi}{8} < \sin\frac{5\pi}{8}.\]

\[4)\sin\frac{3\pi}{5}\ и\ \cos\frac{\pi}{5};\]

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

\[= \cos\left( \frac{5\pi}{10} - \frac{6\pi}{10} \right) =\]

\[= \cos\left( - \frac{\pi}{10} \right) = \cos\frac{\pi}{10};\]

\[\frac{\pi}{10}\ и\ \frac{\pi}{5}\ принадлежат\ \lbrack 0;\ \pi\rbrack -\]

\[функция\ убывает;\]

\[\frac{\pi}{10} < \frac{\pi}{5},\]

\[\cos\frac{\pi}{10} > \cos\frac{\pi}{5}\]

\[\sin\frac{3\pi}{5} > \cos\frac{\pi}{5}.\]

\[5)\cos\frac{\pi}{6}\ и\ \sin\frac{5\pi}{14};\]

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

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

\[= \cos\frac{2\pi}{14} = \cos\frac{\pi}{7};\]

\[\frac{\pi}{6}\ и\ \frac{\pi}{7}\ принадлежат\ \lbrack 0;\ \pi\rbrack -\]

\[функция\ убывает;\]

\[\frac{\pi}{6} > \frac{\pi}{7}\]

\[\cos\frac{\pi}{6} < \cos\frac{\pi}{7}\]

\[\cos\frac{\pi}{6} < \sin\frac{5\pi}{14}.\]

\[6)\cos\frac{\pi}{8}\ и\ \sin\frac{3\pi}{10};\]

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

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

\[= \cos\frac{2\pi}{10} = \cos\frac{\pi}{5};\]

\[\frac{\pi}{8}\ и\ \frac{\pi}{5}\ принадлежат\ \lbrack 0;\ \pi\rbrack -\]

\[функция\ убывает;\]

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

\[\cos\frac{\pi}{8} > \cos\frac{\pi}{5}\]

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