Answer
The required value is $-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right)$
Work Step by Step
By using the sum formula $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$ and trigonometric values to compute the value of the given expression, we get:
$\begin{align}
& \sin \left( \frac{3\pi }{4}+\frac{5\pi }{6} \right)=\sin \frac{3\pi }{4}\cos \frac{5\pi }{6}+\cos \frac{3\pi }{4}\sin \frac{5\pi }{6} \\
& =\frac{\sqrt{2}}{2}.\frac{-\sqrt{3}}{2}+\frac{-\sqrt{2}}{2}.\frac{1}{2} \\
& =-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\
& =-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right)
\end{align}$
Hence, the required value is $-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right)$
