2α będzie kątem z II ćwiartki
[tex]sin^2 2\alpha+cos^22\alpha=1\\\\(\frac{11}{25})^2+cos^2\alpha=1\\\\\frac{121}{625}+cos^22\alpha=\frac{625}{625}\\\\cos^22\alpha=\frac{625}{625}-\frac{121}{625}\\\\cos^22\alpha=\frac{504}{625}\\\\cos2\alpha=-\sqrt{\frac{504}{625}}\\\\cos\2\alpha=-\frac{\sqrt{504}}{25}\\\\cos\2\alpha=-\frac{6\sqrt{14}}{25}[/tex]
