How to calculate $\cos(6^\circ)$? [closed]

Solution 1:

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 $\sin 18°=\frac{1}{4}\sqrt{6-2\sqrt{5}}$

$\sin 36°=2\cos 18°\sin 18°=\frac{1}{4}\sqrt{10-2\sqrt{5}}$

$\cos 36°=\sqrt{1-\sin^2 36°}=\frac{1}{4}(1+\sqrt{5})$

$\cos 6°=\cos(36°-30°)=\cos 36°\cos 30°+\sin 36°\sin 30°=\frac{1}{4}\sqrt{7+\sqrt{5}+\sqrt{30+6\sqrt{5}}}$

Solution 2:

The ancient astronomer Ptolemy, an Egyptian who wrote a thick book in Greek, faced this problem. See Ptolemy's table of chords. His table remained the most extensive trigonometric table available for well over a thousand years.