17.4. Вычислите: а) $\operatorname{arcctg}(-1)+\operatorname{arctg}(-1)$; б) $\arcsin(-\frac{\sqrt2}{2})+\operatorname{arcctg}(-\sqrt3)$; в) $\operatorname{arctg}(-\frac{\sqrt3}{3})-\operatorname{arctg}\frac{\sqrt3}{3}$; г) $\arccos(-\frac12)-\operatorname{arcctg}(-\sqrt3)$.

Question

Вопрос: $Фото задания: 17.4. Вычислите: а) $\operatorname{arcctg}(-1)+\operatorname{arctg}(-1)$; б)…$ Фото задания

Accepted Answer

$\operatorname{arcctg}(-1)=\frac{3\pi}{4}$, а $\operatorname{arctg}(-1)=-\frac{\pi}{4}$.
$\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}$.
$\arcsin(-\frac{\sqrt2}{2})=-\frac{\pi}{4}$, а $\operatorname{arcctg}(-\sqrt3)=\frac{5\pi}{6}$.
$-\frac{\pi}{4}+\frac{5\pi}{6}=-\frac{3\pi}{12}+\frac{10\pi}{12}=\frac{7\pi}{12}$.
$\operatorname{arctg}(-\frac{\sqrt3}{3})=-\frac{\pi}{6}$, $\operatorname{arctg}\frac{\sqrt3}{3}=\frac{\pi}{6}$.
$-\frac{\pi}{6}-\frac{\pi}{6}=-\frac{\pi}{3}$.
$\arccos(-\frac12)=\frac{2\pi}{3}$, а $\operatorname{arcctg}(-\sqrt3)=\frac{5\pi}{6}$.
$\frac{2\pi}{3}-\frac{5\pi}{6}=\frac{4\pi}{6}-\frac{5\pi}{6}=-\frac{\pi}{6}$.

Ответ: а) $\frac{\pi}{2}$; б) $\frac{7\pi}{12}$; в) $-\frac{\pi}{3}$; г) $-\frac{\pi}{6}$.

17.4. Вычислите: а) \(\operatorname{arcctg}(-1)+\operatorname{arctg}(-1)\); б) \(\arcsin(-\frac{\sqrt2}{2})+\operatorname{arcctg}(-\sqrt3)\); в) \(\operatorname{arctg}(-\frac{\sqrt3}{3})-\operatorname{arctg}\frac{\sqrt3}{3}\); г) \(\arccos(-\frac12)-\operatorname{arcctg}(-\sqrt3)\).