Answer
The exact value of $\cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)$ is $\frac{1-\sqrt{3}}{2\sqrt{2}}$.
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 $\cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}},\text{ }\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\text{ }\sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}},\text{ and }\sin \frac{\pi }{6}=\frac{1}{2}$.
$\begin{align}
& \cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)=\left( -\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} \right)+\left( \frac{1}{\sqrt{2}}\times \frac{1}{2} \right) \\
& =\frac{1-\sqrt{3}}{2\sqrt{2}}
\end{align}$
Hence, the exact value of $\cos \left( \frac{3\pi }{4}-\frac{\pi }{6} \right)$ is equivalent to $\frac{1-\sqrt{3}}{2\sqrt{2}}$.
