Монгол Бодлогын Сан

Заавар: $1+\cos60^\circ=2\cos^230^\circ$, $\sin60^\circ=2\cos30^\circ\sin30^\circ$

Бодолт: $$\begin{align*} 1+\cos60^\circ+i\sin60^\circ&=2\cos30^\circ(\cos30^\circ+i\sin30^\circ)\\ &=\sqrt3(\cos30^\circ+i\sin30^\circ) \end{align*}$$ тул $$\begin{align*} (1+\cos60^\circ+i\sin60^\circ)^6&=\{\sqrt3(\cos30^\circ+i\sin30^\circ)\}^6\\ &=(\sqrt{3})^6\cdot(\cos30^\circ+i\sin30^\circ)^6\\ &=27\cdot\{\cos(6\cdot30^\circ)+i\sin(6\cdot30^\circ)\}\\ &=27\cdot(\cos180^\circ+i\sin180^\circ)\\ &=27\cdot(-1)=-27 \end{align*}$$