Answer
$sin~(arccos~\frac{1}{4}) = \frac{\sqrt{15}}{4}$
Work Step by Step
$\theta = arccos(\frac{1}{4})$
$cos~\theta = \frac{1}{4} = \frac{adjacent}{hypotenuse}$
Note that $\theta$ is in quadrant I. We can find the value of the opposite side:
$opposite = \sqrt{4^2-1^2} = \sqrt{15}$
We can find the value of $sin~\theta$:
$sin~\theta = \frac{opposite}{hypotenuse}$
$sin~\theta = \frac{\sqrt{15}}{4}$
Therefore, $sin~(arccos~\frac{1}{4}) = \frac{\sqrt{15}}{4}$
