Is there a closed form for $\cos(\frac{\pi}5)$?

Question

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 $\cos(\frac{\pi}5)$?

https://math.la.asu.edu/~surgent/mat170/Exact_Trig_Values.pdf — Gregory Grant, Feb 17 '16 at 02:10
One way to study this: find sides and angles in the regular pentagon inscribed in the unit circle. — GEdgar, Feb 17 '16 at 02:11
In relation to the comment of @GEdgar, Euclid proved that the regular pentagon is constructible. In modern language that construction can be used to derive an expression for $\cos(\pi/5)$ using only arithmetic and square roots. — Lee Mosher, Feb 17 '16 at 02:27
See http://www.cut-the-knot.org/pythagoras/cos36.shtml and http://www.math-only-math.com/exact-value-of-cos-36-degree.html — lab bhattacharjee, Feb 17 '16 at 05:57

score 2 · Accepted Answer · answered Feb 17 '16 at 02:13

2

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

answered Feb 17 '16 at 02:13

S.C.B.

:D How about we say $\cos\frac\pi5=\frac\phi2$? Using the golden ratio is always the best way to go. – Simply Beautiful Art Jan 26 '17 at 01:19
@SimplyBeautifulArt I guess you could say that. The golden ratio arises from the regular pentagon. – S.C.B. Jan 26 '17 at 01:20

score 0 · Answer 2 · answered Feb 17 '16 at 02:11

0

$\cos(\frac{\pi}{5}) = \frac{1 + \sqrt{5}}{4}.$

answered Feb 17 '16 at 02:11

corindo

2 Answers2