W = \frac{sin\alpha}{(1+cos\alpha)^2}
Z jedynki trygonometrycznej obliczam cos\alpha.
sin^2\alpha+cos^2\alpha = 1
(\frac{2}{3})^2+cos^2\alpha=1
\frac{4}{9}+cos^2\alpha=1
cos\alpha=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}
cos\alpha=\frac{\sqrt5}{3}
podstawiam
{W=\frac{\frac{2}{3}}{(1-\frac{\sqrt5}{3})^2}=\frac{\frac{2}{3}}{(\frac{3-\sqrt5}{3})^2}=\frac{2}{\not3^1}\cdot \frac{\not9^3}{9-6\sqrt5+5}=\frac{2\cdot 3}{14-6\sqrt5}=\frac{\not2\cdot 3}{\not2(7-3\sqrt5)}=}
{=\frac{3}{7-3\sqrt5}=\frac{3(7+3\sqrt5)}{(7-3\sqrt5)\cdot (7+3\sqrt5)}=\frac{3(7+3\sqrt5)}{49-9\cdot 5}=\frac{3(7+3\sqrt5)}{4}}