Przykład 1. Niech $\alpha$ będzie kątem ostrym. Uprość podane wyrażenia.

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

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

c) $\mathrm{tg}\alpha \cdot \cos \alpha +\sin \alpha =\frac{\sin \alpha }{\cos \alpha }\cdot \cos \alpha +\sin \alpha =\sin \alpha +\sin \alpha =2\sin \alpha$

d) $\frac{1-{\cos }^2\alpha }{\sin \alpha }\cdot \frac{\cos \alpha }{1-{\sin }^2\alpha }=\frac{{\sin }^2\alpha }{\sin \alpha }\cdot \frac{\cos \alpha }{{\cos }^2\alpha }=\frac{\sin \alpha }{1}\cdot \frac{1}{\cos \alpha }=\frac{\sin \alpha }{\cos \alpha }=\mathrm{tg}\alpha$

 

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

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