We have,

$$\big(\cos(\tfrac{2\pi}{5})^{1/2}+(-\cos(\tfrac{4\pi}{5}))^{1/2}\big)^2 = \tfrac{1}{2}\left(\tfrac{-1+\sqrt{5}}{2}\right)^3\tag{1}$$

$$\big(\cos(\tfrac{2\pi}{7})^{1/3}+\cos(\tfrac{4\pi}{7})^{1/3}+\cos(\tfrac{6\pi}{7})^{1/3}\big)^3 = \frac{5-3\cdot 7^{1/3}}{2}\tag{2}$$

$$\big(\cos(\tfrac{2\pi}{11})^{1/5}+\cos(\tfrac{4\pi}{11})^{1/5}+\dots+\cos(\tfrac{10\pi}{11})^{1/5}\big)^5 = x?\tag{3}$$

Question: What degree is the minimal polynomial of $x$? Since the previous two are deg 2 and 3, I had assumed (3) would be deg 5, but Mathematica does not recognize it as a quintic with small coefficients, nor a 25th deg (even after using 500-decimal digit precision, though I am not sure of the latter result).

  • $\begingroup$ How have you derived the identities? For cubic, we can set $y=(\frac z2)^{\frac13}$ here in my answer math.stackexchange.com/questions/225088/… and then use Vieta's formula $\endgroup$ – lab bhattacharjee Dec 19 '13 at 5:53
  • $\begingroup$ The first is easy, while the second is at the end of mathworld.wolfram.com/TrigonometryAnglesPi7.html. The third is unknown. $\endgroup$ – Tito Piezas III Dec 19 '13 at 16:19
  • $\begingroup$ Are you sure about the first "identity"? In particular, $\cos \frac{4\pi}{5}$ is negative, so taking square root of it results in purely imaginary number; so the sum of two terms is neither real, nor purely imaginary... and thus its square cannot be real. $\endgroup$ – Peter Košinár Dec 19 '13 at 22:40
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    $\begingroup$ @GrigoryM: What Ramanujan formula? Regarding (3), since $y^5=a_i$ has five roots, and we are to do so for five $a_i=\cos(n)$, then I think it has maximum degree $5^5 = 3125$ though it may factor since the $a_i$ are not random. I tested the form $z=(y_1+y_2\zeta^{k}+y_3\zeta^{2k}+y_4\zeta^{3k}+y_5\zeta^{4k})^5$ where the $y_i = \cos(n)^{1/5}$ and $\zeta^5=1$ and tried to form a 125-deg resolvent, but it still doesn't seem to be the minimal deg. $\endgroup$ – Tito Piezas III Dec 20 '13 at 20:03
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    $\begingroup$ I'm not sure there exists a (good) generalization in this direction — but maybe you'll find another generalization of the second formula of interest $\endgroup$ – Grigory M Jan 20 '14 at 8:33

As some people said, the 2nd formula is easy to derive. In Maple, there are commands to get the minimal polynomial of LHS of the 2nd. The following method works for the 2nd formula, but not for the 3rd.

Prove that $\sqrt[3]{\cos\frac{2\pi}{7}} + \sqrt[3]{\cos\frac{4\pi}{7}} + \sqrt[3]{\cos\frac{6\pi}{7}} = \sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$.

$\textit{Proof}$: $\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}$ are the three distinct real roots of $8x^3+4x^2-4x-1=0$. Let $u_1=\sqrt[3]{\cos\frac{2\pi}{7}}, u_2=\sqrt[3]{\cos\frac{4\pi}{7}}, u_3=\sqrt[3]{\cos\frac{6\pi}{7}}$ which are all real. All roots of $8u^9+4u^6-4u^3-1=0$ are $u_1, t u_1, t^2 u_1, u_2, t u_2, t^2 u_2, u_3, t u_3, t^2 u_3$ where $t = e^{2\pi i/3}$.

Let $(u-u_1)(u-u_2)(u-u_3)=u^3+a_2u^2+a_1u - 1/2$. We have \begin{align} (u-tu_1)(u-tu_2)(u-tu_3) &=u^3+ta_2u^2+t^2a_1u - 1/2, \\ (u-t^2u_1)(u-t^2u_2)(u-t^2u_3) &=u^3+t^2a_2u^2+ta_1u - 1/2. \end{align} It follows from \begin{align} &8u^9+4u^6-4u^3-1 \\ =\ & 8(u^3+a_2u^2+a_1u - 1/2)(u^3+ta_2u^2+t^2a_1u - 1/2)(u^3+t^2a_2u^2+ta_1u - 1/2) \end{align} that \begin{align} a_2^3-3a_1a_2-2 &=0, \\ 4a_1^3+6a_1a_2+5 &= 0. \end{align} From these two equations, we get $a_1 = \frac{a_2^3-2}{3a_2}$ and $4a_2^9+30a_2^6+75a_2^3-32=0$. Let $f(v)=4v^9+30v^6+75v^3-32$. We have $f'(v)=9v^2(2v^3+5)^2\ge 0$. Thus, $f(v)=0$ has a unique real root. It is easy to check that $v=-\sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$ is a root of $f(v)=0$. Thus, $a_2 = -\sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$. Thus, $u_1+u_2+u_3 = -a_2 = \sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$. This completes the proof.

In general, suppose $x_1,x_2,x_3$ are the three distinct real roots of $x^3+bx^2+cx+d^3=0$, where $b,c,d$ are rational numbers. We have \begin{equation} \sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3} = \sqrt[3] {-\frac{3}{2}\Big(\sqrt[3]{Q+4\sqrt{\Delta}}+\sqrt[3]{Q-4\sqrt{\Delta}}\Big) -b-6d }, \end{equation} where $Q = 24bd^2+36d^3+4bc+24cd, \Delta = -4b^3d^3-27d^6+18bcd^3+b^2c^2-4c^3$. Here, $\Delta$ is the discriminant of $x^3+bx^2+cx+d^3=0$.

For the 3rd, similarly, $x_k = \cos\frac{2k\pi}{11}, k=1,\cdots, 5$ are the five distinct real roots of $32x^5+16x^4-32x^3-12x^2+6x+1=0$. Let $u_1 = \sqrt[5]{\cos\frac{2\pi}{11}}, u_2= \sqrt[5]{\cos\frac{4\pi}{11}}, u_3 = \sqrt[5]{\cos\frac{6\pi}{11}}, u_4 = \sqrt[5]{\cos\frac{8\pi}{11}}, u_5= \sqrt[5]{\cos\frac{10\pi}{11}}$. All roots of $32u^{25}+16u^{20}-32u^{15}-12u^{10}+6u^5+1=0$ are $t^j u_k, k=1,2,3,4, 5; j=0,1,2,3,4$ where $t= e^{2\pi i/5}$.

Let $\prod_{k=1}^5 (u-u_k) = u^5 + a_4u^4+a_3u^3+a_2u^2+a_1u+\frac{1}{2}$. Similarly, we have four equations of \begin{align} 0 & = a_4^5-5a_3a_4^3+5a_2a_4^2+5a_3^2a_4-5a_1a_4-5a_2a_3+2, \\ 0 & = 8a_1^5-20a_1^3a_2+10a_1^2a_3+10a_1a_2^2-5a_1a_4-5a_2a_3+1,\\ 0 & = -10a_1a_2a_4^3+10a_1a_3^2a_4^2+10a_2^2a_3a_4^2-10a_2a_3^3a_4+2a_3^5+10a_1^2a_4^2-10a_1a_2a_3a_4\\ &\quad-10a_1a_3^3-10a_2^3a_4+10a_2^2a_3^2-5a_3a_4^3+10a_1^2a_3+10a_1a_2^2+10a_2a_4^2+10a_3^2a_4\\ &\quad -15a_1a_4-15a_2a_3+7,\\ 0 & = -40a_1^3a_3a_4+40a_1^2a_2^2a_4+40a_1^2a_2a_3^2-40a_1a_2^3a_3+8a_2^5-40a_1^3a_2+20a_1^2a_4^2\\ &\quad-20a_1a_2a_3a_4-20a_1a_3^3-20a_2^3a_4+20a_2^2a_3^2+40a_1^2a_3+40a_1a_2^2+10a_2a_4^2\\ &\quad +10a_3^2a_4-30a_1a_4-30a_2a_3+13. \end{align}

However, it seems no hope for determination of $a_4$.

  • 2
    $\begingroup$ I don't mean to be rude, by I find this utter unreadable. Could you perhaps make use of some displayed math environments (using $$ (math here) $$) or the align environment (\begin{align} \end{align}) to make it easier to follow? You can also find other formatting tips in the MathJax tutorial. $\endgroup$ – Xander Henderson Aug 17 '18 at 3:19

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