Do ramienia końcowego kąta alfa należy punkt P= ( minus pierwiastek z 5, pierwiastek z 6). Oblicz wartości funkcji trygonometrycznych kąta alfa

Odpowiedź :

[tex]P = (-\sqrt{5}, \sqrt{6}) \ \ \rightarrow \ \ x = -\sqrt{5}, \ \ y = \sqrt{6}\\\\\\r^{2} = x^{2}+y^{2}\\\\r^{2} = (-\sqrt{5})^{2}+(\sqrt{6})^{2}\\\\r^{2} = 5+6\\\\r^{2} = 11\\\\r = \sqrt{11}[/tex]

[tex]sin\alpha = \frac{y}{r}=\frac{\sqrt{6}}{\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}} = \frac{\sqrt{6\cdot11}}{11}=\frac{\sqrt{66}}{11}\\\\cos\alpha = \frac{x}{r} = \frac{-\sqrt{5}}{\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}}=\frac{-\sqrt{5\cdot11}}{11} = -\frac{\sqrt{55}}{11}\\\\tg\alpha = \frac{y}{x} = \frac{\sqrt{6}}{-\sqrt{5}}\cdot\frac{-\sqrt{5}}{-\sqrt{5}} = \frac{-\sqrt{6\cdot5}}{5} = -\frac{\sqrt{30}}{5}[/tex]

[tex]ctg\alpha = \frac{x}{y} = \frac{-\sqrt{5}}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}} = \frac{-\sqrt{5\cdot6}}{6} = -\frac{\sqrt{30}}{6}[/tex]