Answer
The exact value of $\cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)$ is $\frac{-\sqrt{6}+\sqrt{2}}{4}$.
Work Step by Step
Use the difference formula of cosines and evaluate the term as,
$\cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)=\cos \frac{3\pi }{4}\cos \frac{\pi }{6}+\sin \frac{3\pi }{4}\sin \frac{\pi }{6}$
Substitute the values $\cos \frac{3\pi }{4}=-\frac{\sqrt2}{2},\text{ }\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\text{ }\sin \frac{3\pi }{4}=\frac{\sqrt{2}}{2},\text{ and }\sin \frac{\pi }{6}=\frac{1}{2}$.
$\begin{align}
& \cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)=\left( -\frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2} \right)+\left( \frac{\sqrt{2}}{2}\times \frac{1}{2} \right) \\
& =\frac{-\sqrt{6}+{\sqrt{2}}}{4}
\end{align}$
Hence, the exact value of $\cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)$ is equivalent to $\frac{-\sqrt{6}+{\sqrt{2}}}{4}$.
