Odpowiedź :
[tex]sin \alpha- cos \alpha=\frac{1}{3}\ \ \ |()^2\\\\sin^2 \alpha-2sin\alpha cos\alpha+ cos^2 \alpha=\frac{1}{9}\\\\1-2sin\alpha cos\alpha=\frac{1}{9}\\\\-2sin\alpha cos\alpha=\frac{1}{9}-1\\\\-2sin\alpha cos\alpha=-\frac{8}{9}\ \ \ |:(-2)\\\\sin\alpha cos\alpha=\frac{4}{9}[/tex]
Odpowiedź:
[tex]sin \alpha -cos \alpha = \frac{1}{3} / ()^2\\ (sin \alpha -cos\alpha)^2 = \frac{1}{9}\\ sin^2\alpha+cos^2\alpha-2sin\alpha*cos\alpha=\frac{1}{9}\\((sin^2\alpha+cos^2\alpha=1 ))\\1 - 2sin\alpha*cos\alpha=\frac{1}{9}\\-2sin\alpha*cos\alpha=-\frac{8}{9}/*(-\frac{1}{2}) \\sin\alpha*cos\alpha=\frac{4}{9}[/tex]
Szczegółowe wyjaśnienie: