<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] I'm going to use the value of $\cos 18°=\frac{1}{4}\sqrt{10+2\sqrt{5}}$ obtained in this question.
$\sin^2 18°=1-\left(\frac{1}{4}\sqrt{10+2\sqrt{5}}\right)^2=1-\frac{10+2\sqrt{5}}{16}=\frac{6-2\sqrt{5}}{16}$ so $\sin 18°=\frac{1}{4}\sqrt{6-2\sqrt{5}}$
$\sin 36°=2\cos 18°\sin 18°=\frac{1}{4}\sqrt{10-2\sqrt{5}}$
$\cos 36°=\sqrt{1-\sin^2 36°}=\frac{1}{4}(1+\sqrt{5})$
$\cos 6°=\cos(36°-30°)=\cos 36°\cos 30°+\sin 36°\sin 30°=\frac{1}{4}\sqrt{7+\sqrt{5}+\sqrt{30+6\sqrt{5}}}$
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