Answer
$\frac{\sqrt7}{4}$
Work Step by Step
RECALL:
(1) $\theta = \arccos{x} \longrightarrow \cos{\theta}=x$
(2) $\sin^2{\theta} + \cos^2{\theta} = 1$
Let $\theta = \arccos{(\frac{3}{4})}$
This means that $\cos{\theta} = \frac{3}{4}$.
Since arccosine is defined in Quadrant I and II only, then the angle is in Quadrant I since cosine is positive.
Use the (2) in the recall part above to obtain:
$\sin^2{\theta} + \cos^2{\theta}=1
\\\sin^2{\theta} + \left(\frac{3}{4}\right)^2=1
\\\sin^2{\theta}+\frac{9}{16}=1
\\\sin^2{\theta}=1-\frac{9}{16}
\\\sin^2{\theta}=\frac{7}{16}
\\\sin{\theta} = \pm \sqrt{\frac{7}{16}}
\\\sin{\theta} = \pm \frac{\sqrt{7}}{4}$
Since the angle is in Quadrant I, then sine must be positive.
Thus,
$\sin{\theta} = \frac{\sqrt7}{4}$.
Therefore,
$\sin{(\arccos{(\frac{3}{4}})}=\frac{\sqrt7}{4}$
