\tg\alpha=\frac{1}{5}
\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{1-\cos^2\alpha}{\cos\alpha}
\frac{1-\cos^2\alpha}{\cos\alpha}=\frac{1}{5}
1-\cos^2\alpha=\frac{1}{5}\cos\alpha
-\cos^2\alpha-\frac{1}{5}\cos\alpha+1=0
\Delta=b^2-4ac=(-\frac{1}{5})^2-4*(-1)*1=\frac{1}{25}+4=4\frac{1}{25}=\frac{4}{25}
\sqrt{\Delta}=\sqrt{\frac{4}{25}}=\frac{2}{5}
\cos\alpha=\frac{-b-\sqrt{\Delta}}{2a}=\frac{\frac{1}{5}-\frac{2}{5}}{-2}=\frac{-\frac{1}{5}}{-2}=\frac{1}{5}*\frac{1}{2}=\frac{1}{10}
lub
\cos\alpha=\frac{-b+\sqrt{\Delta}}{2a}=\frac{\frac{1}{5}+\frac{2}{5}}{-2}=\frac{\frac{3}{5}}{-2}=\frac{3}{5}*(-\frac{1}{2})=-\frac{3}{10}
w I ćwiartce wszystkie wartości są dodatnie, a więc
\cos\alpha=\frac{1}{10}
\sin\alpha=\tg\alpha*\cos\alpha=\frac{1}{5}*\frac{1}{10}=\frac{1}{50}