<s>[INST] How to calculate $\cos(6^\circ)$? Do you know any method to calculate $\cos(6^\circ)$ ? 
I tried lots of trigonometric equations, but not found any suitable one for this problem.
 [/INST] If you grant the use of $\cos 18^\circ$ = $\dfrac{\sqrt{10+2\sqrt5}}{4}$, you may proceed as follows: 
First, calculate $\sin 18^\circ$.
Then, calculate $\cos 36^\circ$ and $\sin 36^\circ$ using $\cos 2\theta = 2\cos^2 \theta - 1$ and $\sin 2\theta = 2\sin \theta \cos \theta$ for $\theta = 18^\circ$.
Finally, use $\cos 6^\circ = \cos (36^\circ-30^\circ) = \cos 36^\circ \cos 30^\circ + \sin 36^\circ \sin 30^\circ$.
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