Is there a formula for the cosine of 1/5 of an angle? I'm trying to find a formula for $\cos \frac{x}{5} $ as follow.
By the elementary trigonometric identity, $\cos 5x = 16\cos^5 x - 20\cos^3 x + 5\cos x$
By putting $x=  \frac{x}{5}$ , and using the change of variable $ y = \cos \frac{x}{5} $, we get the quintic equation $ 16y^5 - 20y^3 + 5y - \cos x = 0$.
If we can solve this equation by radicals, then we can get a formula for $\cos \frac{x}{5} $
However, I can not see how we can resolve this equation. Is this a solvable quintic equation? If so, how can we solve it by radicals?
 A: Depending on how you look at this situation, it is solvable by radicals. Let’s take your cosine, and as @mweiss does, let’s call it $C$. I should point out that the case $C=1$ is very special, since $16x^5-20x^3+5x-1=(x-1)(4x^2+2x-1)^2$, probably giving the Wolfram-generated roots that I (wrongly) disparaged in a comment.
In the general case, set $S$ to be the corresponding sine, $S=\sqrt{1-C^2}$, then $\zeta=C+iS$ is a point on the unit circle, and you’re talking about $\zeta^{1/5}$ and your answer is the real part of this. If you accept the fifth root of a real number as something understandable, I don’t think it can be justified to disparage the fifth root of a complex number.
To direct an abstract eye onto the problem, let’s take a general number $\kappa$, which will play the role of $C$, and look at the field $F_0=\Bbb Q(\kappa)$. Then we make a quadratic extension, adjoining $\sigma=\sqrt{1-\kappa^2}$, so that we have $F_1=F_0(\sigma)$, and then adjoin a fifth root of unity, $\omega_5$, where
$$
\omega_5=\frac{-1+\sqrt5}4+\frac i2\sqrt{\frac{5+\sqrt5}2}\,,
$$
and $F_2=F_1(\omega_5)$ is at worst quartic over $F_1$, with Galois group cyclic of order $4$, $2$, 0r $1$, depending on what $\kappa$ was.
Then adjoin $(\kappa+i\sigma)^{1/5}$ to get the final field $F_3$, and again this extension is Galois, with cyclic group of order $5$ or $1$. So, starting with $F_0=\Bbb Q(\kappa)$, we get a chain of radical extensions,
$$
F_0\subset F_1\subset F_2\subset F_3\,,
$$
each inclusion being a cyclic Galois extension. If you call $\xi=(\kappa+i\sigma)^{1/5}$, then its real part is $\frac12(\xi+1/\xi)$, the desired cosine.
A: It is possible to make a one fifth formula, but not with a generic form. See this wikipedia article

A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation $x^3 - \frac{3x+d}{4}=0$, where x is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle. However, the discriminant of this equation is positive, so this equation has three real roots (of which only one is the solution for the cosine of the one-third angle). None of these solutions is reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots. 

A: This is not a definitive answer, but it is strong circumstantial evidence that this quintic is not solvable:
If you ask Wolfram Alpha about the function $f(t)=16t^5-20t^3+5t-C$, it will give you explicit solutions for the two irrational roots of $t$ in each of the special cases where $C=\pm1$.  In terms of your problem, then, we have explicit values for $\cos(\theta/5)$ when $\cos(\theta)=\pm1$.
If you ask it for the roots of $f(t)=16t^5-20t^3+5t-0.5$, it will give you explicit formulas for all five real roots.  In other words, it can compute explicit values for $\cos(\theta/5)$ when $\cos(\theta)=0.5$.
But comparing the explicit formulas for the $C=\pm1$ case to the explicit formulas for the $c=0.5$ case, there doesn't seem to be some general form of which those special cases are instances.
Moreover if you ask for the real roots of $f(t)=16t^5-20t^3+5t-0.6$ you just get decimal approximations.  All of this strongly suggests that there is not much hope of finding an explicit formula for the roots of $f(t)=16t^5-20t^3+5t-C$ expressed in terms of $C$.  (Certainly if there is such a formula, Alpha doesn't know it.) 
A: p = 3, trisection
p >= 3
Python explanation
