Question

arctg(-√3/3), arcsin(-√3/2), arccos(-1/2), arctg(-√3)

Answer 1

$\bf\displaystyle arctg(-\frac{\sqrt{3}}{3}) = -arctg(\frac{\sqrt{3}}{3}) = -\frac{\pi}{6}\\\\arcsin(-\frac{\sqrt{3}}{2})=-arcsin(\frac{\sqrt{3}}{2})=-\frac{\pi}{3}\\\\arccos(-\frac{1}{2})=\pi-arccos(\frac{1}{2})=\pi-\frac{\pi}{3}=\frac{3\pi - \pi}{3}=\frac{2\pi}{3}\\\\arctg(-\sqrt{3})=-arctg(\sqrt{3})=-\frac{\pi}{3}$

Свойства

$\bf arcsin(-a)=-arcsin(a)\\\\arccos(-a)=\pi-arccos(a)\\\\arctg(-a)=-arctg(a)\\\\arcctg(-a)=\pi-arcctg(a)$