## Precalculus (6th Edition) Blitzer

$\dfrac{2\sqrt 3-1}{2}$
Since, $\cos \dfrac{\pi}{6}=\dfrac{\sqrt 3}{2}$ Now, we have $f(\dfrac{\pi}{6})= 2 \cos \dfrac{\pi}{6}-\cos (2 \times \dfrac{\pi}{6})$ or, $=2 \times \dfrac{\sqrt 3}{2}-\dfrac{1}{2}$ After simplifications, we have or, $=\dfrac{2\sqrt 3-1}{2}$