Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 61

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

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

\[\sin\frac{\pi}{9} = \sin\frac{2\pi}{18} < \sin\frac{7\pi}{18};\]

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

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

\[\cos\frac{9\pi}{8} = \sin\left( \frac{5\pi}{2} - \frac{9\pi}{8} \right) = \sin\frac{11\pi}{8};\]

\[\sin\frac{9\pi}{8} > \sin\frac{11\pi}{8};\]

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

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

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

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

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

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

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

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

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