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

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

$\frac{1}{9}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{8}{9}$  

$\left|\cos \alpha \right|=\frac{2\sqrt{2}}{3}$  

W drugiej ćwiartce cosinus jest ujemny.

$\cos \alpha =-\frac{2\sqrt{2}}{3}$  

 

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{1}{3}}{-\frac{2\sqrt{2}}{3}}=\frac{1}{3}\cdot \left(-\frac{3}{2\sqrt{2}}\right)=-\frac{1}{2\sqrt{2}}$  

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

${\left(-\frac{\sqrt{15}}{4}\right)}^2+{\cos }^2\alpha =1$  

$\frac{15}{16}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{1}{16}$  

$\left|\cos \alpha \right|=\frac{1}{4}$  

W czwartej ćwiartce cosinus jest dodatni.

$\cos \alpha =\frac{1}{4}$

 

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{-\sqrt{15}}{4}}{\frac{1}{4}}=-\frac{\sqrt{15}}{4}\cdot 4=-\sqrt{15}$  

 

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

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

${\sin }^2\alpha +\frac{3-4\sqrt{3}+4}{9}=1$  

${\sin }^2\alpha =\frac{9-\left(7-4\sqrt{3}\right)}{9}$  

${\sin }^2\alpha =\frac{9-7+4\sqrt{3}}{9}$  

${\sin }^2\alpha =\frac{2+4\sqrt{3}}{9}$  

$\left|\sin \alpha \right|=\frac{\sqrt{2+4\sqrt{3}}}{3}$  

W drugiej ćwiartce cosinus jest ujemny.

$\cos \alpha =\frac{\sqrt{2+4\sqrt{3}}}{3}$  

 

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{2+4\sqrt{3}}}{3}}{\frac{\sqrt{3}-2}{3}}=\frac{\sqrt{2+4\sqrt{3}}}{3}\cdot \frac{3}{\sqrt{3}-2}=\frac{\sqrt{2+4\sqrt{3}}}{\sqrt{3}-2}\cdot \frac{\sqrt{3}+2}{\sqrt{3}+2}=-\left(\sqrt{3}+2\right)\sqrt{2+4\sqrt{3}}$  

 

$d)\mathrm{tg}\alpha =\sqrt{2}$  

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

$\sin \alpha =\sqrt{2}\cos \alpha$  

Z jedynki trygonometrycznej:

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

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

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

$3{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{1}{3}$  

$\left|\cos \alpha \right|=\frac{\sqrt{3}}{3}$  

W trzeciej ćwiartce cosinus jest ujemny:

$\cos \alpha =-\frac{\sqrt{3}}{3}$  

$\sin \alpha =\sqrt{2}\cdot \left(-\frac{\sqrt{3}}{3}\right)=-\frac{\sqrt{6}}{3}$