Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 627

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

\[\cos\frac{\pi}{9} > 0; - \sin\frac{\pi}{9} < 0;\]

\[\varphi = - \frac{\pi}{9}.\]

\[z = \cos\left( - \frac{\pi}{9} \right) + i\sin\left( - \frac{\pi}{9} \right).\]

\[2)\ 12\left( - \cos\frac{\pi}{8} + i\sin\frac{\pi}{8} \right)\]

\[- \cos\frac{\pi}{8} < 0;\text{\ \ \ }\sin\frac{\pi}{8} > 0;\]

\[\varphi = \pi - \frac{\pi}{8} = \frac{7\pi}{8}.\]

\[z = 12\left( \cos\frac{7\pi}{8} + i\sin\frac{7\pi}{8} \right).\]

\[3)\ \sqrt{2}\left( - \cos\frac{\pi}{7} - i\sin\frac{\pi}{7} \right)\]

\[- \cos\frac{\pi}{7} < 0;\ \ \ - \sin\frac{\pi}{7} < 0;\]

\[\varphi = \pi + \frac{\pi}{7} = \frac{8\pi}{7}.\]

\[z = \sqrt{2}\left( \cos\frac{8\pi}{7} + i\sin\frac{8\pi}{7} \right).\]

\[4)\ 3\left( \sin\frac{\pi}{5} + i\cos\frac{\pi}{5} \right)\]

\[\sin\frac{\pi}{5} > 0;\text{\ \ \ }\cos\frac{\pi}{5} > 0;\]

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

\[z = 3\left( \cos\frac{3\pi}{10} + i\sin\frac{3\pi}{10} \right).\]