Podstawowe tożsamości trygonometryczne:

$\text{1)}{\sin }^2\alpha +{\cos }^2\alpha =1,\text{dla}\alpha \in ⟨0^{\circ },{180}^{\circ }⟩$  

$\text{2)}\text{tg}\alpha \cdot \text{ctg}\alpha =1,\text{dla}\alpha \in \left(0^{\circ },{90}^{\circ }\right)\cup \left({90}^{\circ },{180}^{\circ }\right)$  

$\text{3)}\frac{\sin \alpha }{\cos \alpha }=\text{tg}\alpha ,\text{dla}\alpha \in ⟨0^{\circ },{90}^{\circ })\cup ({90}^{\circ },{180}^{\circ }⟩$  

$\text{4)}\frac{\cos \alpha }{\sin \alpha }=\text{ctg}\alpha ,\text{dla}\alpha \in \left(0^{\circ },{180}^{\circ }\right)$  

Tożsamość 1) nazywamy jedynką trygonometryczną.

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$\text{a)}\sin \alpha \cdot \text{ctg}\alpha =\sin \alpha \cdot \frac{\cos \alpha }{\sin \alpha }=\cos \alpha$  

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$\text{b)}\sin \alpha \cdot {\cos }^2\alpha +{\sin }^3\alpha =\sin \alpha \cdot {\cos }^2\alpha +\sin \alpha \cdot {\sin }^2\alpha =\sin \alpha \cdot \underset{=1}{\underbrace{\left({\cos }^2\alpha +{\sin }^2\alpha \right)}}=\sin \alpha$  

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$\text{c)}\frac{\text{tg}\alpha }{\sin \alpha }=\frac{\frac{\sin \alpha }{\cos \alpha }}{\sin \alpha }=\frac{1}{\cos \alpha }$  

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$\text{d)}\frac{1}{\sin \alpha }-\cos \alpha \cdot \text{ctg}\alpha =\frac{1}{\sin \alpha }-\cos \alpha \cdot \frac{\cos \alpha }{\sin \alpha }=$  

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

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$\text{e)}\cos \alpha +\cos \alpha \cdot {\text{tg}}^2\alpha =\cos \alpha +\cos \alpha \cdot {\left(\frac{\sin \alpha }{\cos \alpha }\right)}^2=\cos \alpha +\cos \alpha \cdot \frac{{\sin }^2\alpha }{{\cos }^2\alpha }=$  

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

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$\text{f)}\sin \alpha +\cos \alpha \cdot \text{ctg}\alpha =\sin \alpha +\cos \alpha \cdot \frac{\cos \alpha }{\sin \alpha }=\sin \alpha +\frac{{\cos }^2\alpha }{\sin \alpha }=$  

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

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$\text{g)}\frac{1}{{\sin }^2\alpha }\left(1-{\cos }^2\alpha \right)=\frac{1}{{\sin }^2\alpha }\cdot {\sin }^2\alpha =1$  

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$\text{h)}\left(\cos \alpha +\text{tg}\alpha \cdot \sin \alpha \right)\cdot \text{ctg}\alpha =\left(\cos \alpha +\frac{\sin \alpha }{\cos \alpha }\cdot \sin \alpha \right)\cdot \frac{\cos \alpha }{\sin \alpha }=$  

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

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