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This question already has an answer here:

I was wondering if there was a closed form for $\cos(\frac{\pi}5)$?

We have the following in closed form:

$$\cos(\frac{\pi}2)=0$$

$$\cos(\frac{\pi}3)=\frac12$$

$$\cos(\frac{\pi}4)=\frac{\sqrt{2}}2$$

$$\cos(\frac{\pi}6)=\frac{\sqrt{3}}2$$

$$\cos(\frac{\pi}8)=\frac{\sqrt{2+\sqrt{2}}}2$$

But perhaps a solution to $\cos(\frac{\pi}5)$?

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marked as duplicate by N. F. Taussig, hunter, Watson, Jack's wasted life, Kamil Jarosz Feb 17 '16 at 15:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Draw a isocles triangle $ABC$, with $AB=AC$, $\angle A=\frac{\pi}{5}$.

Draw the angle bisector of $\angle B$, which meets $AC$ on point $P$. Note that $BC=BP=AP$.

Using this, one can calcualte that $AC=\frac{1+\sqrt{5}}{2} \times BC$.

Using this, $\cos \frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$

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  • $\begingroup$ :D How about we say $\cos\frac\pi5=\frac\phi2$? Using the golden ratio is always the best way to go. $\endgroup$ – Simply Beautiful Art Jan 26 '17 at 1:19
  • $\begingroup$ @SimplyBeautifulArt I guess you could say that. The golden ratio arises from the regular pentagon. $\endgroup$ – S.C.B. Jan 26 '17 at 1:20
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$\cos(\frac{\pi}{5}) = \frac{1 + \sqrt{5}}{4}.$

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