## Precalculus (6th Edition) Blitzer

The exact value of $\cos \left( \frac{2\pi }{3}-\frac{\pi }{6} \right)$ is $0$.
Use the difference formula of cosines and evaluate the term as, $\cos \left( \frac{2\pi }{3}-\frac{\pi }{6} \right)=\cos \frac{2\pi }{3}\cos \frac{\pi }{6}+\sin \frac{2\pi }{3}\sin \frac{\pi }{6}$ Substitute the values $\cos \frac{2\pi }{3}=-\frac{1}{2},\text{ }\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\text{ }\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2},\text{ and }\sin \frac{\pi }{6}=\frac{1}{2}$. \begin{align} & \cos \left( \frac{2\pi }{3}-\frac{\pi }{6} \right)=\left( -\frac{1}{2}\times \frac{\sqrt{3}}{2} \right)+\left( \frac{\sqrt{3}}{2}\times \frac{1}{2} \right) \\ & =0 \end{align} Hence, the exact value of $\cos \left( \frac{2\pi }{3}-\frac{\pi }{6} \right)$ is equivalent to $0$.