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.
 A: 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.
A: 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 $\cos 6^\circ = \cos (36^\circ-30^\circ) = \cos 36^\circ \cos 30^\circ + \sin 36^\circ \sin 30^\circ$.
A: Note that $6^\circ$ is $\frac1{60}$ of the full circle. Since $\frac1{60}=\frac14\cdot(\frac13-\frac15)$, you can obtain the sine and cosine of $6^\circ$ from the well-known values for the regular triangle ($120^\circ$) and the regular pentagon ($72^\circ$), and trigonometic formulas to compute $\cos(\alpha-\beta)$ from the trigonometric functions for $\alpha$ and $\beta$, and the half-angle formulas.
A: Here's one way to get a quick answer; use the taylor series. 6 degrees is 0.104719755 radians. Now the taylor series of cosine is $Cos(\theta)$=$\Sigma (-\theta) ^{2n}/2n!$. Putting the theta value of 0.104719755 gives you the approximation $Cos(6^{\circ})=0.99451688646$. Perfectly accurate up to 5 decimal places. In general, use the 17 degree rule. Sine and Cosine taylor approximations are good to the first order for about 17 degrees off of naught. You will have to excuse some of the odd language, I learned this first as a ship tech.
A: 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 $\sin 18°=\frac{1}{4}\sqrt{6-2\sqrt{5}}$
$\sin 36°=2\cos 18°\sin 18°=\frac{1}{4}\sqrt{10-2\sqrt{5}}$
$\cos 36°=\sqrt{1-\sin^2 36°}=\frac{1}{4}(1+\sqrt{5})$
$\cos 6°=\cos(36°-30°)=\cos 36°\cos 30°+\sin 36°\sin 30°=\frac{1}{4}\sqrt{7+\sqrt{5}+\sqrt{30+6\sqrt{5}}}$
A: Try $\cos \theta = \sin 14\theta$, where $\theta=6^{\circ}$.
