Задача 56903 ...
Условие
Решение
По формуле Муавра:
=cos(3*(2π/9))+i*sin(3*(2π/9))=cos(2π/3)+i*sin(2π/3)=[b](1/2)+i*sqrt(3)/2)[/b]
2)
[m]\frac{6-6i}{-\sqrt{3}+i}=\frac{-6+6i}{\sqrt{3}-i}=\frac{(-6+6i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)}=[/m]
[m]=\frac{-6\sqrt{3}+6\sqrt{3}i-6i+(6i)^2}{(\sqrt{3})^2-(i)^2}=\frac{(-6-6\sqrt{3})+(-6+6\sqrt{3})i}{4}=-\frac{3+3\sqrt{3}}{2}+\frac{-3+3\sqrt{3}}{2}i[/m]
[m]r=\sqrt{(-\frac{3+3\sqrt{3}}{2})^2+(\frac{-3+3\sqrt{3}}{2})^2}=\sqrt{18}=3\sqrt{2}[/m]
[m]sin θ =-\frac{3+3\sqrt{3}}{6\sqrt{2}}[/m]
[m]cos θ =\frac{-3+3\sqrt{3}}{6\sqrt{2}}[/m]
Пойдем другим путем ( cм. скрин)
[m]\frac{6-6i}{-\sqrt{3}+i}=\frac{-6+6i}{\sqrt{3}-i}=\frac{(-6+6i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)}=\frac{(-6+6i)(\sqrt{3}+i)}{(\sqrt{3})^2-(i)^2}=\frac{(-6+6i)(\sqrt{3}+i)}{4}[/m]
[m]-6+6i=6\sqrt{2}\cdot (cos ( \frac{3π}{4})+isin( \frac{3π}{4}))[/m]
[m]\sqrt{3}+i=2\cdot (cos ( \frac{π}{6})+isin( \frac{π}{6}))[/m]
[m](-6+6i)(\sqrt{3}+i)=12\sqrt{2}\cdot (cos( \frac{3π}{4}+ \frac{π}{6})+isin( \frac{3π}{4}+\frac{π}{6})=12\sqrt{2}\cdot (cos\frac{11π}{12}+isin\frac{11π}{12})[/m]
[m]\frac{6-6i}{-\sqrt{3}+i}=\frac{(-6+6i)(\sqrt{3}+i)}{4}=3\sqrt{2}\cdot (cos\frac{11π}{12}+isin\frac{11π}{12})[/m]
[m]r=3\sqrt{2}[/m]
[m] θ =\frac{11π}{12}[/m]
