Дано: \(\cos \alpha = \frac{1}{4}\)
Используем основное тригонометрическое тождество: \(\sin^2 \alpha + \cos^2 \alpha = 1\)
\(\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - (\frac{1}{4})^2 = 1 - \frac{1}{16} = \frac{15}{16}\)
\(\sin \alpha = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}\)
\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\sqrt{15}}{4}}{\frac{1}{4}} = \sqrt{15}\)
\(\cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}\)
Ответ: \(\sin \alpha = \frac{\sqrt{15}}{4}\); \(\tan \alpha = \sqrt{15}\); \(\cot \alpha = \frac{\sqrt{15}}{15}\).