Szczegółowe wyjaśnienie:
[tex]P(x,\ y)\\r=\sqrt{x^2+y^2}\\\\\sin\alpha=\dfrac{y}{r}\\\\\cos\alpha=\dfrac{x}{r}\\\\\text{tg}\alpha=\dfrac{y}{x}\\\\\text{ctg}\alpha=\dfrac{x}{y}[/tex]
Mamy
[tex]P(-2,-5)[/tex]
Podstawiamy:
[tex]r=\sqrt{(-2)^2+(-5)^2}=\sqrt{4+25}=\sqrt{29}\\\\\sin\alpha=\dfrac{-5}{\sqrt{29}}\cdot\dfrac{\sqrt{29}}{\sqrt{29}}=-\dfrac{5\sqrt{29}}{29}\\\\\cos\alpha=\dfrac{-2}{\sqrt{29}}\cdot\dfrac{\sqrt{29}}{\sqrt{29}}=-\dfrac{2\sqrt{29}}{29}\\\\\text{tg}\alpha=\dfrac{-5}{-2}=\dfrac{5}{2}\\\\\text{ctg}\alpha=\dfrac{-2}{-5}=\dfrac{2}{5}[/tex]
