$a)\cos 2\alpha =0,8$  

Z jedynki trygonometrycznej:

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

${\sin }^{2}2\alpha =0,36$

$\left|\sin 2\alpha \right|=0,6$

$\text{Skoro}\alpha \in \left(0,\frac{\pi }{2}\right)\text{to}2\alpha <\pi \text{a więc}\sin 2\alpha >0$  

$\sin 2\alpha =0,6$

$\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha =0,8$

${\cos }^2\alpha =0,8+{\sin }^2\alpha$

Z jedynki trygonometrycznej:

${\sin }^2\alpha +0,8+{\sin }^2\alpha =1$    

$2{\sin }^2\alpha =0,2$

${\sin }^2\alpha =\frac{1}{10}$  

$\left|\sin \alpha \right|=\frac{\sqrt{10}}{10}$

$\text{Dla}\alpha \in \left(0,\frac{\pi }{2}\right)\sin \alpha ,\cos \alpha >0$  

$\sin \alpha =\frac{\sqrt{10}}{10}$

${\cos }^2\alpha =1-\frac{1}{10}=\frac{9}{10}$

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

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{\sqrt{10}}{10}}{\frac{3}{\sqrt{10}}}=\frac{\sqrt{10}}{10}\cdot \frac{\sqrt{10}}{3}=\frac{10}{10\cdot 3}=\frac{10}{30}=\frac{1}{3}$  

$b)\cos 2\alpha =-\frac{3}{5}$

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

${\sin }^{2}2\alpha =\frac{16}{25}$

$\left|\sin 2\alpha \right|=\frac{4}{5}$

$\text{Dla}\alpha \in \left(\frac{\pi }{2},\pi \right)\text{to}\pi <2\alpha <2\pi \text{a więc}\sin 2\alpha <0$

$\sin 2\alpha =-\frac{4}{5}$  

$\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha =-\frac{3}{5}$

${\cos }^2\alpha =-\frac{3}{5}+{\sin }^2\alpha$

Z jedynki trygonometrycznej:

${\sin }^2\alpha +-\frac{3}{5}+{\sin }^2\alpha =1$    

$2{\sin }^2\alpha =\frac{8}{5}$

${\sin }^2\alpha =\frac{4}{5}$  

$\left|\sin \alpha \right|=\frac{2}{\sqrt{5}}$

$\text{Dla}\alpha \in \left(\frac{\pi }{2},\pi \right)\sin \alpha >0$  

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

Z jedynki trygonometrycznej:

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

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

$\left|\cos \alpha \right|=\frac{\sqrt{5}}{5}$

$\text{Dla}\alpha \in \left(\frac{\pi }{2},\pi \right)\cos \alpha <0$

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

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{2}{\sqrt{5}}}{-\frac{\sqrt{5}}{5}}=-\frac{2}{\sqrt{5}}\cdot \frac{5}{\sqrt{5}}=-2\cdot \frac{5}{5}=-2$