Rozwiązane

Oblicz wartości funkcji trygonometrycznych kąta α jeżeli na jego ramieniu końcowym leży punkt
A = 1; 4

Szybko, pls

Odpowiedź :

Jola6609viz

[tex]Zadanie\\\\A=(1,4)\\\\r^2=x^{2} +y^2\\\\r^2=1^2+4^2\\\\r^2=1+16\\\\r^2=17\\\\r= \sqrt{17}\\\\\\sin\alpha=\frac{y}{r}=\frac{4}{\sqrt{17} }\cdot\frac{\sqrt{17} }{\sqrt{17} }= \frac{4\sqrt{17} }{17} \\\\cos\alpha=\frac{x}{r}=\frac{1}{\sqrt{17} }=\frac{\sqrt{17} }{17} \\\\ tg\alpha=\frac{y}{x}=\frac{4}{1}=4\\\\ctg\alpha=\frac{x}{y} =\frac{1}{4}[/tex]

Basetlaviz

[tex]A = (1,4) \ \ \rightarrow \ \ x = 1, \ y = 4 \ - \ I. \ cwiartka\\\\r^{2} = x^{2}+y^{2}\\\\r^{2} = 1^{2}+4^{2}\\\\r^{2} = 1+16\\\\r^{2} = 17\\\\r = \sqrt{17}\\\\\\sin\alpha = \frac{y}{r} = \frac{4}{\sqrt{17}}\cdot\frac{\sqrt{17}}{\sqrt{17}} = \frac{4\\\\\sqrt{17}}{17}\\\\cos\alpha = \frac{x}{y} = \frac{1}{\sqrt{17}}\cdot\frac{\sqrt{17}}{\sqrt{17}}=\frac{\sqrt{17}}{17}\\\\tg\alpha = \frac{y}{x} = \frac{4}{1} = 4\\\\ctg\alpha = \frac{x}{y} = \frac{1}{4}[/tex]

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