Answer
$$\frac{{\sqrt {15} }}{{\text{4}}}$$
Work Step by Step
$$\eqalign{
& \sin \left( {\arccos \frac{1}{4}} \right) \cr
& {\text{Let }}\theta = \arccos \frac{1}{4}{\text{, thus}} \cr
& \cos \theta = \frac{1}{4} \cr
& {\text{Recall that }}\cos \theta = \frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}} \cr
& {\text{adjacent side}} = 1 \cr
& {\text{hypotenuse}} = 4 \cr
& {\text{opposite side }} = \sqrt {{4^2} - {1^2}} = \sqrt {15} \cr
& {\text{Then}} \cr
& \sin \theta = \tan \left( {\arccos \frac{1}{4}} \right) = \frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}} \cr
& \sin \left( {\arccos \frac{1}{4}} \right) = \frac{{\sqrt {15} }}{{\text{4}}} \cr} $$
