Rozwiązanie:
[tex]sin\alpha -cos\alpha =\frac{\sqrt{3} }{4} \\(sin\alpha -cos\alpha )^{2}=\frac{3}{16}\\sin^{2}\alpha -2sin\alpha cos\alpha +cos^{2}\alpha =\frac{3}{16}\\-2sin\alpha cos\alpha =-\frac{13}{16} \\sin\alpha cos\alpha =\frac{13}{32}[/tex]
Zatem:
[tex](sin\alpha +cos\alpha )^{2}+4=5+2sin\alpha cos\alpha =5+2*\frac{13}{32}=5+\frac{13}{16}=5\frac{13}{16}[/tex]
