ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 661

\[= \sqrt{3}\left( \cos\frac{9\pi}{12} + i\sin\frac{9\pi}{12} \right) =\]

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

\[3)\ \frac{\cos{50{^\circ}} + i\sin{50{^\circ}}}{2\left( \cos{20{^\circ}} + i\sin{20{^\circ}} \right)} =\]

\[= \frac{1}{2}\left( \cos(50{^\circ} - 20{^\circ}) + i\sin(50{^\circ} - 20{^\circ}) \right) =\]

\[= \frac{1}{2}\left( \cos{30{^\circ}} + i\sin{30{^\circ}} \right).\]

\[4)\ \frac{2\left( \cos\left( - \frac{5\pi}{12} \right) + i\sin\left( - \frac{5\pi}{12} \right) \right)}{\sqrt{2}\left( \cos\left( - \frac{\pi}{12} \right) + i\sin\left( - \frac{\pi}{12} \right) \right)} =\]

\[= \sqrt{2}\left( \cos\left( - \frac{5\pi}{12} + \frac{\pi}{12} \right) + i\sin\left( - \frac{5\pi}{12} + \frac{\pi}{12} \right) \right) =\]

\[= \sqrt{2}\left( \cos\left( - \frac{4\pi}{12} \right) + i\sin\left( - \frac{4\pi}{12} \right) \right) =\]

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