Szczegółowe wyjaśnienie:
[tex]\sin\alpha=\dfrac{y}{r}\\\\\cos\alpha=\dfrac{x}{r}\\\\\text{tg}\alpha=\dfrac{y}{x}\\\\\text{ctg}\alpha=\dfrac{x}{y}\\\\r=\sqrt{x^2+y^2}[/tex]
[tex]P(\sqrt6;\ -\sqrt{12})\to x=\sqrt6;\ y=-\sqrt{12}=-\sqrt{4\cdot3}=-2\sqrt3\\\\r=\sqrt{(\sqrt6)^2+(-\sqrt{12})^2}=\sqrt{6+12}=\sqrt{18}=\sqrt{9\cdot2}=3\sqrt2[/tex]
[tex]\sin\alpha=\dfrac{-2\sqrt3}{3\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}=-\dfrac{2\sqrt6}{3\cdot2}=-\dfrac{\sqrt6}{3}\\\\\cos\alpha=\dfrac{\sqrt6}{3\sqrt2}=\dfrac{\sqrt3}{3}\\\\\text{tg}\alpha=\dfrac{\sqrt6}{-2\sqrt3}=-\dfrac{\sqrt2}{2}\\\\\text{ctg}\alpha=\dfrac{-2\sqrt3}{\sqrt6}=-\dfrac{2}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}=-\dfrac{2\sqrt2}{2}=-\sqrt2[/tex]
