În triunghiul ABC, dreptunghic în A, avem:
[tex]\it AB^2 + AC^2= BC^2\ \ \ \ \ \ \ (*)\ \ \ [ Teorema\ \ lui\ \ Pitagora ][/tex]
[tex]\it \left.\begin{aligned}sinB=\dfrac{AC}{BC} \Rightarrow sin^2B=\dfrac{AC^2}{BC^2}\\ \\ cosB=\dfrac{AB}{BC} \Rightarrow cos^2B=\dfrac{AB^2}{BC^2}\end{aligned}\right\} \Rightarrow sin^2B+cos^2B=\dfrac{AC^2}{BC^2}+\dfrac{AB^2}{BC^2}=\\ \\ \\ =\dfrac{AC^2+AB^2}{BC^2}\ \stackrel{(*)}{=}\ \dfrac{BC^2}{BC^2}=1[/tex]
[tex]\it Pentru\ \ m(\widehat B)=x\ \Rightarrow\ sin^2x+cos^2x=1[/tex]
