- Znanija.Site

Ответы 1

$1)\qquad arcctg(ctgx)=x\; ,\; \; esli\; \; -\frac{\pi}{2}\ \textless \ x\ \textless \ \frac{\pi}{2}\\\\arcctg(tg\frac{27\pi}{7})=arcctg(tg(3\pi +\frac{6\pi}{7}))=arcctg(tg\frac{6\pi}{7})=\\\\=[\; \frac{6\pi}{7}otin (-\frac{\pi}{2},\frac{\pi}{2})\; ]=arcctg(tg(\frac{\pi}{2}+\frac{5\pi}{14}))=arcctg(-ctg\frac{5\pi}{14})=\\\\=[\; \frac{5\pi}{14}\in (-\frac{\pi}{2},\frac{\pi}{2})\; ,\; arcctg(- \alpha )=\pi -arcctg \alpha \; ]=\\\\=\pi -arcctg(ctg\frac{5\pi}{14})=\pi -\frac{5\pi}{14}=\frac{9\pi}{14}$ $2)\qquad arctg(tgx)=x\; ,\; -\frac{\pi}{2}\ \textless \ x\ \textless \ \frac{\pi}{2}\\\\arctg(ctg(-\frac{21\pi}{5}))=arctg(-ctg(4\pi +\frac{\pi}{5}))=\\\\=[\; arctg(- \alpha )=-arctg \alpha \; ]= -arctg(ctg\frac{\pi}{5})=\\\\= -arctg(ctg(\frac{\pi}{2}-\frac{3\pi}{10}))= -arctg(tg\frac{3\pi}{10})=\\\\=[\; \frac{3\pi}{10}\in (-\frac{\pi}{2},\frac{\pi}{2})\; ]=-\frac{3\pi}{10}$
- Автор:
  bambi32
- 5 лет назад
- 0
Добавить свой ответ