# 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.

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Please see here for help writing math with MathJax, and see here for help formatting posts with Markdown. Some MathJax advice: Named math operators should appear upright with a following space, and the common ones have their own MathJax code for this purpose (e.g. \sin, \log - see entry 11 in the MathJax guide). –  Zev Chonoles Jul 7 '13 at 20:27
possible duplicate of Evaluate $\cos 18^\circ$ without using the calculator –  kba Jul 7 '13 at 20:30
And now may someone dare and ask about the $\cos \left( 1^{\circ} \right)$? –  Ali Jul 8 '13 at 9:31
Indeed, I should asked wolframalpha first: wolframalpha.com/input/?i=cos%281%2F180+pi%29 –  Ali Jul 8 '13 at 9:35
Useful link: Exact trigonometric constants on Wikipedia. (You can at least check your result there.) –  Martin Sleziak Jun 8 at 8:25

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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Shouldn't this line"$$\cos 36=\sqrt{1-\cos^2 36}=\frac{1}{4}(1+\sqrt{5})$$ have a $\sin$ in it? –  DJohnM Jul 8 '13 at 4:36
@User58220 Of course it should; thanks for pointing out. –  metacompactness Jul 8 '13 at 7:22

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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@PeterTaylor Use it where? –  Alraxite Jul 7 '13 at 22:08
I withdraw that comment. When I can't tell the difference between a half and a third, I should probably take it as a sign to step away from the mathematical website. –  Peter Taylor Jul 7 '13 at 22:45

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.

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This is more a comment than an answer. –  metacompactness Jul 7 '13 at 21:01
@metacompactness : It is an answer in that the linked article does say something about how to compute such things. Of course it leaves a fair amount of work to do a particular case. –  Michael Hardy Jul 7 '13 at 21:04
This is a very nice answer. It even has interesting history. –  Baby Dragon Jul 7 '13 at 21:22

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.

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$\frac{1}{60}=\frac{1}{8}\cdot\left(\frac{1}{3}-\frac{1}{5}\right)$ and the method isn't clear. –  metacompactness Jul 7 '13 at 21:00

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.

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Try $\cos \theta = \sin 14\theta$, where $\theta=6^{\circ}$.

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Is this an answer? –  metacompactness Jul 7 '13 at 20:58
Expand Sin14Ø and simplify the equation –  DOCTOR NGILAZI BANDA JOSHUA Jul 7 '13 at 21:09
You mean $\sin(14\phi)$ as a function of $\sin\phi$!! you must be kidding; you have a polynomial of degree 14. –  metacompactness Jul 7 '13 at 21:34