Niech P(x, y) będzie dowolnym punktem różnym od początku układu współrzędnych, leżącym na ramieniu końcowym kąta α ∈ <0°, 180°>. Wówczas: $\sin \alpha =\frac{y}{r},\cos \alpha =\frac{x}{r},\text{gdzie}r=\sqrt{x^2+y^2}$   $\text{tg}\alpha =\frac{y}{x},x\ne 0,\text{czyli}\alpha \ne {90}^{\circ }$   $\text{ctg}\alpha =\frac{x}{y},y\ne 0,\text{czyli}\alpha \ne 0^{\circ },\alpha \ne {180}^{\circ }$

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$\text{a)}A\left(12,16\right)$

$x=12,y=16$  

$r=\sqrt{{12}^2+{16}^2}=\sqrt{144+256}=\sqrt{400}=20$  

Wówczas:

$\sin \alpha =\frac{y}{r}=\frac{16}{20}=\frac{4}{5}$  

$\cos \alpha =\frac{x}{r}=\frac{12}{20}=\frac{3}{5}$  

$\text{tg}\alpha =\frac{y}{x}=\frac{16}{12}=\frac{4}{3}$  

$\text{ctg}\alpha =\frac{x}{y}=\frac{3}{4}$  

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$\text{b)}A\left(-8,6\right)$  

$x=-8,y=6$  

$r=\sqrt{{\left(-8\right)}^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$  

Wówczas:

$\sin \alpha =\frac{y}{r}=\frac{6}{10}=\frac{3}{5}$  

$\cos \alpha =\frac{x}{r}=\frac{-8}{10}=-\frac{4}{5}$  

$\text{tg}\alpha =\frac{y}{x}=\frac{6}{-8}=-\frac{3}{4}$  

$\text{ctg}\alpha =\frac{x}{y}=\frac{-8}{6}=-\frac{4}{3}$  

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$\text{c)}A\left(6,2\right)$  

$x=6,y=2$  

$r=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}$  

Wówczas:

$\sin \alpha =\frac{y}{r}=\frac{2}{2\sqrt{10}}=\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}=\frac{\sqrt{10}}{10}$  

$\cos \alpha =\frac{x}{r}=\frac{6}{2\sqrt{10}}=\frac{3}{\sqrt{10}}=\frac{3}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}=\frac{3\sqrt{10}}{10}$  

$\text{tg}\alpha =\frac{y}{x}=\frac{2}{6}=\frac{1}{3}$  

$\text{ctg}\alpha =\frac{x}{y}=\frac{3}{1}=3$  

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$\text{d)}A\left(-3,9\right)$  

$x=-3,y=9$  

$r=\sqrt{{\left(-3\right)}^2+9^2}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}$  

Wówczas:

$\sin \alpha =\frac{y}{r}=\frac{9}{3\sqrt{10}}=\frac{3}{\sqrt{10}}=\frac{3}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}=\frac{3\sqrt{10}}{10}$  

$\cos \alpha =\frac{x}{r}=\frac{-3}{3\sqrt{10}}=-\frac{1}{\sqrt{10}}=-\frac{1}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}=-\frac{\sqrt{10}}{10}$  

$\text{tg}\alpha =\frac{y}{x}=\frac{9}{-3}=-3$  

$\text{ctg}\alpha =\frac{x}{y}=\frac{-3}{9}=-\frac{1}{3}$  

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$\text{e)}A\left(\sqrt{3},\sqrt{6}\right)$  

$x=\sqrt{3},y=\sqrt{6}$  

$r=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{6}\right)}^2}=\sqrt{3+6}=\sqrt{9}=3$  

Wówczas:

$\sin \alpha =\frac{y}{r}=\frac{\sqrt{6}}{3}$  

$\cos \alpha =\frac{x}{r}=\frac{\sqrt{3}}{3}$  

$\text{tg}\alpha =\frac{y}{x}=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}$  

$\text{ctg}\alpha =\frac{x}{y}=\frac{\sqrt{3}}{\sqrt{6}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$  

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$\text{f)}A\left(-\sqrt{2},\frac{1}{2}\right)$  

$x=-\sqrt{2},y=\frac{1}{2}$  

$r=\sqrt{{\left(-\sqrt{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}=\sqrt{2+\frac{1}{4}}=\sqrt{2\frac{1}{4}}=\sqrt{\frac{9}{4}}=\frac{3}{2}$  

Wówczas:

$\sin \alpha =\frac{y}{r}=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac{1}{2}\cdot \frac{2}{3}=\frac{1}{3}$  

$\cos \alpha =\frac{x}{r}=\frac{-\sqrt{2}}{\frac{3}{2}}=-\sqrt{2}\cdot \frac{2}{3}=-\frac{2\sqrt{2}}{3}$  

$\text{tg}\alpha =\frac{y}{x}=\frac{\frac{1}{2}}{-\sqrt{2}}=-\frac{1}{2\sqrt{2}}=-\frac{1}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=-\frac{\sqrt{2}}{4}$  

$\text{ctg}\alpha =\frac{x}{y}=\frac{-\sqrt{2}}{\frac{1}{2}}=-2\sqrt{2}$