Упр.635 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

Задача

Выполнить действия и записать результат в алгебраической форме:
1) $$\frac{i\left(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right)}{\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}}$$;
2) $$\frac{1}{\cos\frac{4\pi}{3}-i\sin\frac{4\pi}{3}}$$;
3) $$\left(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}\right)\frac{(1+\sqrt{3}i)^7}{i}$$;
4) $$\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\left(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}\right)$$;
5) $$\left(t^8+\frac{\sqrt{3}t^5}{4}\right)^5$$;
6) $$\frac{(2i)^7}{\left(-\sqrt{2}+i\sqrt{2}\right)^6}$$.

Подробный ответ

$$i\frac{\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}}{\cos \frac{\pi}{6}+i\sin \frac{\pi}{6}}=i\left(\cos \left(\frac{5\pi}{3}-\frac{\pi}{6}\right)+i\sin \left(\frac{5\pi}{3}-\frac{\pi}{6}\right)\right)$$
$$=i\left(\cos \frac{3\pi}{2}+i\sin \frac{3\pi}{2}\right)=i(0-i)=-i^2=1.$$
$$\frac{1}{\cos \frac{4\pi}{3}-i\sin \frac{4\pi}{3}}=\frac{1}{\cos \left(-\frac{4\pi}{3}\right)+i\sin \left(-\frac{4\pi}{3}\right)}$$
$$=\cos \frac{4\pi}{3}+i\sin \frac{4\pi}{3}=-\frac12-\frac{\sqrt3}{2}i.$$
$$\frac{\left(\cos \frac{\pi}{3}-i\sin \frac{\pi}{3}\right)\left(1+\sqrt3\,i\right)^7}{i}$$
$$=\frac{1}{i}\left(\cos \left(-\frac{\pi}{3}\right)+i\sin \left(-\frac{\pi}{3}\right)\right)\left(2\left(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}\right)\right)^7$$
$$=\frac{1}{i}\cdot 2^7\left(\cos \left(-\frac{\pi}{3}+\frac{7\pi}{3}\right)+i\sin \left(-\frac{\pi}{3}+\frac{7\pi}{3}\right)\right)$$
$$=\frac{128}{i}\left(\cos 2\pi+i\sin 2\pi\right)=\frac{128}{i}=-128i.$$
$$\left(\frac12+\frac{\sqrt3}{2}i\right)\left(\cos \frac{\pi}{3}-i\sin \frac{\pi}{3}\right)$$
$$=\left(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}\right)\left(\cos \left(-\frac{\pi}{3}\right)+i\sin \left(-\frac{\pi}{3}\right)\right)$$
$$=\cos 0+i\sin 0=1.$$
$$\left(\frac{i^8+\sqrt3\,i^5}{4}\right)^5=\left(\frac{1+\sqrt3\,i}{4}\right)^5$$
$$=\left(\frac{2\left(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}\right)}{4}\right)^5 =\left(\frac12\right)^5\left(\cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}\right)$$
$$=\frac{1}{32}\left(\frac12-\frac{\sqrt3}{2}i\right)=\frac{1}{64}-\frac{\sqrt3}{64}i.$$
$$\frac{(2i)^7}{(-\sqrt2+i\sqrt2)^6}=\frac{\left(2\left(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}\right)\right)^7}{\left(2\left(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4}\right)\right)^6}$$
$$=\frac{2^7\left(\cos \frac{7\pi}{2}+i\sin \frac{7\pi}{2}\right)}{2^6\left(\cos \frac{9\pi}{2}+i\sin \frac{9\pi}{2}\right)}$$
$$=2\left(\cos \left(\frac{7\pi}{2}-\frac{9\pi}{2}\right)+i\sin \left(\frac{7\pi}{2}-\frac{9\pi}{2}\right)\right)$$
$$=2(\cos(-\pi)+i\sin(-\pi))=2(-1+0i)=-2.$$