Упр.674 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1. Вычислить: 1) $$\left(3\left(\cos\frac{7\pi}{8}+i\sin\frac{7\pi}{8}\right)\right)^4$$; 2) $$\left(\cos 2-+i\sin 20\right)^{12}$$; 3) $$\left(2\left(\cos(-20)+i\sin(-20)\right)\right)^3$$; 4) $$\frac{1}{\left(\cos\frac{\pi}{20}+i\sin\frac{\pi}{20}\right)^5}$$.

1. Применим формулу Муавра:

   $$\left(3\left(\cos \frac{7\pi}{8}+i\sin \frac{7\pi}{8}\right)\right)^4=3^4\left(\cos \frac{28\pi}{8}+i\sin \frac{28\pi}{8}\right).$$

   $$=81\left(\cos \frac{7\pi}{2}+i\sin \frac{7\pi}{2}\right)=81\left(\cos \frac{3\pi}{2}+i\sin \frac{3\pi}{2}\right)=-81i.$$

2. $$\left(\cos 20^\circ+i\sin 20^\circ\right)^{12}=\cos 240^\circ+i\sin 240^\circ.$$

   $$\cos 240^\circ=\cos \frac{4\pi}{3}=-\frac12,\qquad \sin 240^\circ=\sin \frac{4\pi}{3}=-\frac{\sqrt3}{2}.$$

   Следовательно,

   $$\left(\cos 20^\circ+i\sin 20^\circ\right)^{12}=-\frac12-\frac{\sqrt3}{2}i.$$

3. $$\left(2\left(\cos(-20^\circ)+i\sin(-20^\circ)\right)\right)^3=2^3\left(\cos(-60^\circ)+i\sin(-60^\circ)\right).$$

   $$=8\left(\frac12-\frac{\sqrt3}{2}i\right)=4-4\sqrt3\,i.$$

4. $$\frac{1}{\left(\cos \frac{\pi}{20}+i\sin \frac{\pi}{20}\right)^5}=\left(\cos \frac{\pi}{20}+i\sin \frac{\pi}{20}\right)^{-5}.$$

   По формуле Муавра:

   $$=\cos\left(-5\cdot \frac{\pi}{20}\right)+i\sin\left(-5\cdot \frac{\pi}{20}\right)=\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right).$$

   $$=\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i.$$

Ответ

$$1)\ -81i;\quad 2)\ -\frac12-\frac{\sqrt3}{2}i;\quad 3)\ 4-4\sqrt3\,i;\quad 4)\ \frac{\sqrt2}{2}-\frac{\sqrt2}{2}i.$$