Przypomnijmy: Wartości funkcji trygonometrycznych kąta ostrego są dodatnie.

$\text{a)}$  

$\sin \alpha =0,1$  

$\sin \alpha =\frac{1}{10}$  

Korzystając z jedynki trygonometrycznej mamy:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

${\left(\frac{1}{10}\right)}^2+{\cos }^2\alpha =1$  

$\frac{1}{100}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{99}{100}$  

$\cos \alpha =\frac{\sqrt{99}}{10}$  

$\cos \alpha =\frac{3\sqrt{11}}{10}$  

Zatem:

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{1}{10}}{\frac{3\sqrt{11}}{10}}=\frac{1}{{\cancel{10}}^1}\cdot \frac{{\cancel{10}}^1}{3\sqrt{11}}=\frac{1}{3\sqrt{11}}=\frac{1\cdot \sqrt{11}}{3\sqrt{11}\cdot \sqrt{11}}=\frac{\sqrt{11}}{3\cdot 11}=\frac{\sqrt{11}}{33}$  

$\mathrm{ctg}\alpha =\frac{1}{\mathrm{tg}\alpha }=\frac{33}{\sqrt{11}}=\frac{33\sqrt{11}}{11}=3\sqrt{11}$  

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$\text{b)}$  

$\cos \alpha =0,9$  

$\cos \alpha =\frac{9}{10}$  

Korzystając z jedynki trygonometrycznej mamy:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

${\sin }^2\alpha +{\left(\frac{9}{10}\right)}^2=1$  

${\sin }^2\alpha +\frac{81}{100}=1$  

${\sin }^2\alpha =\frac{19}{100}$  

$\sin \alpha =\frac{\sqrt{19}}{10}$  

Zatem:

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{19}}{10}}{\frac{9}{10}}=\frac{\sqrt{19}}{9}$   

$\mathrm{ctg}\alpha =\frac{1}{\mathrm{tg}\alpha }=\frac{9}{\sqrt{19}}=\frac{9\sqrt{19}}{19}$  

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$\text{c)}$  

$\cos \alpha =\frac{1}{6}$    

Korzystając z jedynki trygonometrycznej mamy:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

${\sin }^2\alpha +{\left(\frac{1}{6}\right)}^2=1$  

${\sin }^2\alpha +\frac{1}{36}=1$  

${\sin }^2\alpha =\frac{35}{36}$  

$\sin \alpha =\frac{\sqrt{35}}{6}$  

Zatem: 

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{35}}{6}}{\frac{1}{6}}=\sqrt{35}$     

$\mathrm{ctg}\alpha =\frac{1}{\mathrm{tg}\alpha }=\frac{1}{\sqrt{35}}=\frac{\sqrt{35}}{35}$  

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$\text{d)}$  

$\sin \alpha =\frac{\sqrt{3}}{3}$   

Korzystając z jedynki trygonometrycznej mamy:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

${\left(\frac{\sqrt{3}}{3}\right)}^2+{\cos }^2\alpha =1$  

$\frac{3}{9}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{6}{9}$  

$\cos \alpha =\frac{\sqrt{6}}{3}$

Zatem:

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{6}}{3}}=\frac{\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{18}}{6}=\frac{3\sqrt{2}}{6}=\frac{\sqrt{2}}{2}$     

$\mathrm{ctg}\alpha =\frac{1}{\mathrm{tg}\alpha }=\frac{2}{\sqrt{2}}=\sqrt{2}$  

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$\text{e)}$  

$\mathrm{tg}\alpha =\frac{8}{15}$  

$\frac{\sin \alpha }{\cos \alpha }=\frac{8}{15}$  

$8\cos \alpha =15\sin \alpha$  

$\cos \alpha =\frac{15}{8}\sin \alpha$    

=sinalpha/cosalpha

${\sin }^2\alpha +{\cos }^2\alpha =1$  

${\sin }^2\alpha +\frac{225}{64}{\sin }^2\alpha =1$  

$\frac{289}{64}{\sin }^2\alpha =1$  

$\sin \alpha =\frac{8}{17}$  

Zatem:

$\cos \alpha =\frac{15}{8}\sin \alpha =\frac{15}{8}\cdot \frac{8}{17}=\frac{15}{17}$   

$\mathrm{ctg}\alpha =\frac{15}{8}$    

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$\text{f)}$  

$\mathrm{tg}\alpha =\sqrt{5}$  

$\frac{\sin \alpha }{\cos \alpha }=\sqrt{5}$  

$\sin \alpha =\sqrt{5}\cos \alpha$  

Korzystając z jedynki trygonometrycznej mamy:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

$5{\cos }^2\alpha +{\cos }^2\alpha =1$   

${\cos }^2\alpha =\frac{1}{6}$  

$\cos \alpha =\frac{1}{\sqrt{6}}$

$\cos \alpha =\frac{\sqrt{6}}{6}$  

 Zatem:

$\sin \alpha =\sqrt{5}\cdot \frac{\sqrt{6}}{6}=\frac{\sqrt{30}}{6}$  

$\mathrm{ctg}\alpha =\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$  

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$\text{g)}$  

$\mathrm{ctg}\alpha =\frac{1}{2}$  

$\mathrm{tg}\alpha =2$  

$\frac{\sin \alpha }{\cos \alpha }=2$  

$\sin \alpha =2\cos \alpha$  

Korzystając z jedynki trygonometrycznej:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

$4{\cos }^2\alpha +{\cos }^2\alpha =1$   

${\cos }^2\alpha =\frac{1}{5}$  

$\cos \alpha =\frac{1}{\sqrt{5}}$  

$\cos \alpha =\frac{\sqrt{5}}{5}$  

Zatem:

$\sin \alpha =2\cdot \frac{\sqrt{5}}{5}=\frac{2\sqrt{5}}{5}$  

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$\text{h)}$  

$\mathrm{ctg}\alpha =\frac{11}{5}$  

$\mathrm{tg}\alpha =\frac{5}{11}$  

$\frac{\sin \alpha }{\cos \alpha }=\frac{5}{11}$  

$\sin \alpha =\frac{5}{11}\cos \alpha$  

Korzystając z jedynki trygonometrycznej:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

$\frac{25}{121}{\cos }^2\alpha +{\cos }^2\alpha =1$  

$\frac{146}{121}{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{121}{146}$  

$\cos \alpha =\frac{11}{\sqrt{146}}$  

$\cos \alpha =\frac{11\sqrt{146}}{146}$  

Zatem:

$\sin \alpha =\frac{5}{11}\cdot \frac{11}{\sqrt{146}}=\frac{5}{\sqrt{146}}=\frac{5\sqrt{146}}{146}$