Wyznaczam sinα
[tex]cos\alpha=2sin\alpha\ \ \ |:2\\\\sin\alpha=\frac{1}{2}cos\alpha[/tex]
Obliczam cosα
[tex]sin^2\alpha+cos^2\alpha=1\\\\(\frac{1}{2}cos\alpha)^2+cos^2\alpha=1\\\\\frac{1}{4}cos^2\alpha+cos^2\alpha=1\\\\\frac{5}{4}cos^2\alpha=1\ \ \ |:\frac{5}{4}\\\\cos^2\alpha=\frac{4}{5}\\\\cos\alpha=\sqrt{\frac{4}{5}}\\\\cos\alpha=\frac{2}{\sqrt5}\\\\cos\alpha=\frac{2\sqrt5}{5}[/tex]
