Упр.34.30 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

Задача

а) $$\frac{8\left(\cos\frac{7\pi}{12}+i\sin\frac{7\pi}{12}\right)}{4\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)}$$;
б) $$(10+10i)\left(\sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)\right)$$;
в) $$\frac{12\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)}{0{,}3\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)}$$;
г) $$\frac{16\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right)}{4-4\sqrt{3}i}$$.

Подробный ответ

а) $$\frac{8\left(\cos \frac{7\pi}{12}+i\sin \frac{7\pi}{12}\right)}{4\left(\cos \left(-\frac{\pi}{4}\right)+i\sin \left(-\frac{\pi}{4}\right)\right)}$$

$$=2\left(\cos \left(\frac{7\pi}{12}+\frac{\pi}{4}\right)+i\sin \left(\frac{7\pi}{12}+\frac{\pi}{4}\right)\right)$$

$$=2\left(\cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6}\right)=2\left(-\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=-\sqrt{3}+i.$$

б) $$\frac{10+10i}{\sqrt{2}\left(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}\right)}$$

Представим числитель в тригонометрической форме:

$$10+10i=10\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right).$$

Тогда

$$\frac{10\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)}{\sqrt{2}\left(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}\right)}$$

$$=10\left(\cos \left(\frac{\pi}{4}-\frac{3\pi}{4}\right)+i\sin \left(\frac{\pi}{4}-\frac{3\pi}{4}\right)\right)$$

$$=10\left(\cos \left(-\frac{\pi}{2}\right)+i\sin \left(-\frac{\pi}{2}\right)\right)=-10i.$$

в) $$\frac{12\left(\cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6}\right)}{0{,}3\left(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}\right)}$$

$$=\frac{12}{0{,}3}\left(\cos \left(\frac{5\pi}{6}-\frac{\pi}{3}\right)+i\sin \left(\frac{5\pi}{6}-\frac{\pi}{3}\right)\right)$$

$$=40\left(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}\right)=40i.$$

г) $$\frac{16\left(\cos \left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)}{4-4\sqrt{3}\,i}$$

Найдём тригонометрическую форму знаменателя:

$$4-4\sqrt{3}\,i=8\left(\cos \left(-\frac{\pi}{3}\right)+i\sin \left(-\frac{\pi}{3}\right)\right).$$

Тогда

$$\frac{16\left(\cos \left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)}{8\left(\cos \left(-\frac{\pi}{3}\right)+i\sin \left(-\frac{\pi}{3}\right)\right)}$$

$$=2\left(\cos \left(-\frac{\pi}{6}+\frac{\pi}{3}\right)+i\sin \left(-\frac{\pi}{6}+\frac{\pi}{3}\right)\right)$$

$$=2\left(\cos \frac{\pi}{6}+i\sin \frac{\pi}{6}\right)=2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=\sqrt{3}+i.$$