Дано: \(\tan \alpha = 2\)
\(\cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{2}\)
Используем формулу: \(1 + \tan^2 \alpha = \frac{1}{\cos^2 \alpha}\)
Тогда, \(\cos^2 \alpha = \frac{1}{1 + \tan^2 \alpha} = \frac{1}{1 + 2^2} = \frac{1}{5}\)
\(\cos \alpha = \sqrt{\frac{1}{5}} = \frac{\sqrt{5}}{5}\)
\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)
Тогда, \(\sin \alpha = \tan \alpha \cdot \cos \alpha = 2 \cdot \frac{\sqrt{5}}{5} = \frac{2 \sqrt{5}}{5}\)
Ответ: \(\sin \alpha = \frac{2 \sqrt{5}}{5}\); \(\cos \alpha = \frac{\sqrt{5}}{5}\); \(\cot \alpha = \frac{1}{2}\).