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

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

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

• math.la.asu.edu/~surgent/mat170/Exact_Trig_Values.pdf – Gregory Grant Feb 17 '16 at 2:10
• One way to study this: find sides and angles in the regular pentagon inscribed in the unit circle. – GEdgar Feb 17 '16 at 2:11
• $\frac{1+ \sqrt{5}}{4}$ – Ivan S. Guerra Feb 17 '16 at 2:12
• 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 2:27
• – lab bhattacharjee Feb 17 '16 at 5:57

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, $\cos \frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$
• :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 1:19
$\cos(\frac{\pi}{5}) = \frac{1 + \sqrt{5}}{4}.$