a)

$\mathrm{tg}\alpha =3$  

$\frac{\sin \alpha }{\cos \alpha }=3\mid \cdot \cos \alpha$  

$\sin \alpha =3\cos \alpha$  

 

Skorzystajmy z jedynki trygonometrycznej:

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

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

$9{\cos }^2\alpha +{\cos }^2\alpha =1$  

$10{\cos }^2\alpha =1\mid :10$  

${\cos }^2\alpha =\frac{1}{10}$  

$\cos \alpha =\sqrt{\frac{1}{10}}=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}$  

 

$\sin \alpha =3\cos \alpha =3\cdot \frac{\sqrt{10}}{10}=\frac{3\sqrt{10}}{10}$          

 

b)

$\mathrm{tg}\alpha =\frac{4}{5}$  

$\frac{\sin \alpha }{\cos \alpha }=\frac{4}{5}$  

$\sin \alpha =\frac{4}{5}\cos \alpha$  

 

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

${\left(\frac{4}{5}\cos \alpha \right)}^2+{\cos }^2\alpha =1$  

$\frac{16}{25}{\cos }^2\alpha +{\cos }^2\alpha =1$  

$\frac{41}{25}{\cos }^2\alpha =1\mid \cdot 25$  

$41{\cos }^2\alpha =25\mid :41$  

${\cos }^2\alpha =\frac{25}{41}$  

$\cos \alpha =\sqrt{\frac{25}{41}}=\frac{5}{\sqrt{41}}=\frac{5\sqrt{41}}{41}$  

 

$\sin \alpha =\frac{4}{5}\cdot \cos \alpha =\frac{4}{\sout{5}{}_1}\cdot \frac{\sout{5}{}^1\sqrt{41}}{41}=\frac{4\sqrt{41}}{41}$