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I tried to find the exact value of $\cos(\frac{2\pi}{17})$ with WolframAlpha but only obtained a decimal approximation. Is there any way to find this exact value with WolframAlpha?

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  • $\begingroup$ Are you requiring a rational number? Or rather do you have reason to believe it is a rational number? $\endgroup$ – Tony Hellmuth Jun 5 '18 at 4:15
  • $\begingroup$ @TonyHellmuth, I know it is not a rational number; however, WolframAlpha can find the exact value of some irrational numbers. Am not sure why it refuses to give me the exact value this time. $\endgroup$ – Zuriel Jun 5 '18 at 4:19
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    $\begingroup$ I'm not sure why WolframAlpha can't solve that one, but Mathematica can using ToRadicals and you get $\dfrac{1}{4 \sqrt{\frac{2}{\sqrt{17}-\sqrt{2 \left(17-\sqrt{17}\right)}+\sqrt{2 \left(6 \sqrt{17}+\sqrt{2 \left(17-\sqrt{17}\right)}-\sqrt{34 \left(17-\sqrt{17}\right)}+8 \sqrt{2 \left(\sqrt{17}+17\right)}+34\right)}+15}}}$ $\endgroup$ – Joseph Eck Jun 5 '18 at 4:32
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    $\begingroup$ Just click on "More" under alternate forms. $\endgroup$ – Moo Jun 5 '18 at 4:51
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    $\begingroup$ @Moo, it worked! Thank you so much!! $\endgroup$ – Zuriel Jun 5 '18 at 4:53
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Wolfram give the results - see link below.

http://mathworld.wolfram.com/TrigonometryAnglesPi17.html

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  • $\begingroup$ Thanks! I was wondering if WolframAlpha can compute the exact value $\cos(\frac{2\pi}{17})$ directly; if not, why not? $\endgroup$ – Zuriel Jun 5 '18 at 4:37
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First, the value as Gauss wrote it, from Galois Theory by David A. Cox. Cox also shows how Gauss would have derived this, in exrecises in the previous chapter.

enter image description here

Instead of your number, it is easier to double it, and get a root of $$ x^8 + x^7 - 7 x^6 - 6 x^5 + 15 x^4 + 10 x^3 - 10 x^2 - 4 x + 1 $$

see page 18 of Reuschle (1875)

enter image description here enter image description here

Note that $2 \cos \frac{2 \pi}{17} \approx 1.8649444588$

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  • $\begingroup$ Thank you! It is not exactly what I was asking but it is helpful to know how to compute the exact value by hand. $\endgroup$ – Zuriel Jun 6 '18 at 21:31

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