Odpowiedź :
[tex](sin\alpha-cos\alpha)^2=sin^2\alpha-2sin\alpha cos\alpha+cos^2\alpha=1-2sin\alpha cos\alpha=\\\\1-2\cdot\frac{1}{6}=1-\frac{1}{3}=\frac{2}{3}[/tex]
[tex](sin\alpha-cos\alpha)^2=sin^2\alpha-2sin\alpha cos\alpha+cos^2\alpha=1-2sin\alpha cos\alpha=\\\\1-2\cdot\frac{1}{6}=1-\frac{1}{3}=\frac{2}{3}[/tex]