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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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    $\begingroup$ 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). $\endgroup$ Jul 7, 2013 at 20:27
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    $\begingroup$ possible duplicate of Evaluate $\cos 18^\circ$ without using the calculator $\endgroup$
    – kba
    Jul 7, 2013 at 20:30
  • $\begingroup$ And now may someone dare and ask about the $\cos \left( 1^{\circ} \right) $? $\endgroup$
    – Ali
    Jul 8, 2013 at 9:31
  • $\begingroup$ Indeed, I should asked wolframalpha first: wolframalpha.com/input/?i=cos%281%2F180+pi%29 $\endgroup$
    – Ali
    Jul 8, 2013 at 9:35
  • $\begingroup$ Useful link: Exact trigonometric constants on Wikipedia. (You can at least check your result there.) $\endgroup$ Jun 8, 2015 at 8:25

6 Answers 6

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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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    $\begingroup$ Shouldn't this line"$$\cos 36=\sqrt{1-\cos^2 36}=\frac{1}{4}(1+\sqrt{5})$$ have a $\sin$ in it? $\endgroup$
    – DJohnM
    Jul 8, 2013 at 4:36
  • $\begingroup$ @User58220 Of course it should; thanks for pointing out. $\endgroup$
    – user5402
    Jul 8, 2013 at 7:22
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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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  • $\begingroup$ This is more a comment than an answer. $\endgroup$
    – user5402
    Jul 7, 2013 at 21:01
  • $\begingroup$ @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. $\endgroup$ Jul 7, 2013 at 21:04
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    $\begingroup$ This is a very nice answer. It even has interesting history. $\endgroup$ Jul 7, 2013 at 21:22
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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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  • $\begingroup$ @PeterTaylor Use it where? $\endgroup$
    – Alraxite
    Jul 7, 2013 at 22:08
  • $\begingroup$ 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. $\endgroup$ Jul 7, 2013 at 22:45
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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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    $\begingroup$ $\frac{1}{60}=\frac{1}{8}\cdot\left(\frac{1}{3}-\frac{1}{5}\right)$ and the method isn't clear. $\endgroup$
    – user5402
    Jul 7, 2013 at 21:00
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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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    $\begingroup$ Is this an answer? $\endgroup$
    – user5402
    Jul 7, 2013 at 20:58
  • $\begingroup$ Expand Sin14Ø and simplify the equation $\endgroup$ Jul 7, 2013 at 21:09
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    $\begingroup$ You mean $\sin(14\phi)$ as a function of $\sin\phi$!! you must be kidding; you have a polynomial of degree 14. $\endgroup$
    – user5402
    Jul 7, 2013 at 21:34

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