$a)\sin \alpha =\frac{2}{3}$  

Z jedynki trygonometrycznej:

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

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

$\frac{4}{9}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{5}{9}$  

$\cos \alpha =\frac{\sqrt{5}}{3}$  

Iloczyn wynosi:

$\sin \alpha \cdot \cos \alpha =\frac{2}{3}\cdot \frac{\sqrt{5}}{3}=\frac{2\sqrt{5}}{9}$  

 

$b)\cos \alpha =\frac{1}{5}$  

Z jedynki trygonometrycznej:

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

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

${\sin }^2\alpha +\frac{1}{25}=1$  

${\sin }^2\alpha =\frac{24}{25}$  

Różnica kwadratów wynosi:

${\sin }^2\alpha -{\cos }^2\alpha =\frac{24}{25}-\frac{1}{25}=\frac{23}{25}$