Answer
$\dfrac{\sqrt 2-\sqrt 6}{4}$
Work Step by Step
Here, we have $\cos (\dfrac{17 \pi}{12})=\cos (\dfrac{12 \pi}{12}+\dfrac{5 \pi}{12})$
This gives: $\cos (\pi+\dfrac{5\pi}{12})=-\cos (\dfrac{5\pi}{12})$
or, $-\cos (\dfrac{5\pi}{12})=-\cos (\dfrac{3\pi}{12}+\dfrac{2\pi}{12})=-\cos (\dfrac{\pi}{4}+\dfrac{\pi}{6})$
or, $-(\cos \dfrac{\pi}{4} \cos \dfrac{\pi}{6}-\sin \dfrac{\pi}{4} \sin \dfrac{\pi}{6})=-(\dfrac{\sqrt 2}{2}\cdot \dfrac{\sqrt 3 }{2}- \dfrac{\sqrt 2}{2} \cdot \dfrac{1}{2})$
Thus, $\cos(\dfrac{17 \pi}{12})=\dfrac{\sqrt 2-\sqrt 6}{4}$
