Cho  góc alpha thỏa sin alpha  = -1/4

Chọn D

\[\sin \alpha = - \frac{1}{4} \Rightarrow co{\mathop{\rm s}\nolimits} \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - {{\left( { - \frac{1}{4}} \right)}^2}} = - \frac{{\sqrt {15} }}{4}\]

\(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} + \sin \alpha .\sin \frac{\pi }{6} = - \frac{{\sqrt {15} }}{4}.\frac{{\sqrt 3 }}{2} + \frac{{ - 1}}{4}.\frac{1}{2} = \frac{{ - 3\sqrt 5 - 1}}{8}\)