a)

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

 

Pokażemy, że lewa strona jest równa prawej.

$L={\left(\sin \alpha +\cos \alpha \right)}^2={\sin }^2\alpha +2\sin \alpha \cos \alpha +{\cos }^2\alpha ={\sin }^2\alpha +{\cos }^2\alpha +2\sin \alpha \cos \alpha =1+2\sin \alpha \cos \alpha =P$  

 

b)

${\sin }^4\alpha -{\cos }^4\alpha ={\sin }^2\alpha -{\cos }^2\alpha$     

$L={\sin }^4\alpha -{\cos }^4\alpha =\left({\sin }^2\alpha -{\cos }^2\alpha \right)\cdot \left({\sin }^2\alpha +{\cos }^2\alpha \right)=\left({\sin }^2\alpha -{\cos }^2\alpha \right)\cdot 1={\sin }^2\alpha -{\cos }^2\alpha =P$  

 

c)

$\frac{1}{{\cos }^2\alpha }-1={\mathrm{tg}}^2\alpha$  

Założenia:

${\cos }^2\alpha \ne 0$  

$\cos \alpha \ne 0$   

$\alpha \ne {90}^{\circ }$      

 

$L=\frac{1}{{\cos }^2\alpha }-1=\frac{1}{{\cos }^2\alpha }-\frac{{\cos }^2\alpha }{{\cos }^2\alpha }=\frac{\stackrel{\text{korzystamy z jedynki trygonometrycznej}}{\overbrace{1-{\cos }^2\alpha }}}{{\cos }^2\alpha }=\frac{{\sin }^2\alpha }{{\cos }^2\alpha }={\left(\frac{\sin \alpha }{\cos \alpha }\right)}^2={\mathrm{tg}}^2\alpha =P$  

 

 

d)

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

 

$L=2{\sin }^2\alpha -3{\cos }^2\alpha +3=2{\sin }^2\alpha -3{\cos }^2\alpha +3\cdot 1=2{\sin }^2\alpha -3{\cos }^2\alpha +3\cdot \left({\sin }^2\alpha +{\cos }^2\alpha \right)=2{\sin }^2\alpha -3{\cos }^2\alpha +3\cdot {\sin }^2\alpha +3{\cos }^2\alpha =5{\sin }^2\alpha =P$