[tex]sin\alpha+cos\alpha=1\frac{1}{3}\ \ \ |()^2\\(sin\alpha+cos\alpha)^2=(1\frac{1}{3})^2\\sin^2\alpha+2 sin\alpha cos\alpha+sin^2\alpha=(\frac{4}{3})^2\\1+2 sin\alpha cos\alpha=\frac{16}{9}\\2sin\alpha cos\alpha=\frac{16}{9}-\frac{9}{9}\\2sin\alpha cos\alpha=\frac{7}{9}[/tex]
[tex](sin\alpha-cos\alpha)^2=sin^2\alpha-2 sin\alpha cos\alpha+cos^2\alpha=1-2 sin\alpha cos\alpha=1-\frac{7}{9}=\frac{2}{9}[/tex]
