Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ .

I am sure this is derived from using roots of unity and Euler's complex number function, but I am very uncomfortable in these areas so some help would be great. It is evident that $(a + b + c)^2 - 2(ab + ac + bc) = a^2 + b^2 + c^2$ . So, using a polynomial of degree 3 and the coefficients on the $x^2$ and $x$ terms will get where we need to be.


First of all, by noticing $\cot^2(\pi - x) = \cot^2(x)$, we can write this identity as $$\sum_{k=1,3,5} \cot^2(\frac{2\pi k}{14}) = 5$$

By writing $\cot^2(x) = \frac{1}{\sin^2(x)} - 1$, and using symmetries of $\cos$ and $\sin$ ($\cos(x)=\cos(-x)$, $\sin(\frac{\pi}{2}-x)=\cos x$), we can write this sum as follows:

$$\sum_{k=1,3,5} \frac{1}{\cos^2(\frac{\pi k}{14})} = 8$$

If we let $a_i = \cos(\frac{\pi (2i-1)}{14}), i=1,2,3$, we can write this expression as $$(*) \frac{(\sum_{i<j} a_i a_j)^2 - 2\prod a_i \sum a_i}{(\prod a_i)^2}$$

The 7'th Chebyshev Polynomial (of the first kind) vanishes exactly on $\cos(\frac{2k-1}{14}\pi)$, $1\le k \le 7$. Those roots are actually $\pm (a_1, a_2, a_3)$ and $0$, each is a simple root.

We can compute the polynomial recursively and find that it equals $$T_7(x) = 64x^7-112x^5+56x^3-7x=x(64x^6-112x^4+56x^2-7)$$ We'll work with $P_7(x)=\frac{T_7(x)}{64x}$, a monic polynomial with roots $\pm(a_1,a_2,a_3)$.

This shows, by using Vieta and the symmetry of roots (it requires some manipulation on symmetric polynomials):

  1. $(\prod a_i)^2=\frac{7}{64}$ (by considering coefficient of $x^0$)
  2. $(\sum_{i<j} a_i a_j)^2 - 2\prod a_i \sum a_i = \frac{56}{64}$ (by considering coefficient of $x^2$ - this one required some computation)

So the sum $(*)$ equals $\frac{56}{64} / \frac{7}{64} = 8$, which implies your identity. $\blacksquare$

EDIT: I'll describe some of the philosophy behind the answer.

The first half - I knew I wanted to you Chebyshev polynomials in some way (because its roots are related to the expression), so I did basic manipulations that helped me use the coefficients of the Chebyshev polynomial. I didn't know apriori that there are any 'good' manipulations, but I hoped and it indeed worked out.

The second half - What I really wanted is a polynomial $Q(x)$ whose roots are $a_1,a_2,a_3$. Unfortunately, I had managed only to construct the polynomial $P_7(x)$ which equals $-Q(x)Q(-x)$. Fortunately, the coefficients of $P_7$ encode enough information about the coefficients of $Q$. Explicitly, by comparing coefficients: $$P_7[X^k] = \sum_{i+j=k} (-1)^{1+j} Q[X^i]Q[X^j]$$ I used this for $k=0,2$ and it was enough. $k=0$ gave $P_7(0)=-Q(0)^2$, i.e. we have the product of the $a_i$! (up to sign, but we don't even need it.)

$k=2$ gave $P_7[X^2] = Q[X^1]^2-2Q[X^2]Q[X^0]$, which luckily was exactly the missing ingredient in calculating the rational expression $(*)$, so that's it.

EDIT 2: I feel that I need to expand on the "theory" of Chebyshev polynomial, because using it might scare people away.

The $n$'th Chebyshev polynomial of the first kind is the unique polynomial satisfying $T_n(\cos (\theta)) = \cos(n\theta)$, for any $\theta$. Evidently, $\cos(\frac{\pi}{2n}(2k+1))$ is a root for any $k$ - just plug $\theta = \frac{\pi}{2n}(2k+1))$. As $\deg T_n = n$ (see the next paragraph), there can be no other roots.

Why is $T_n$ necessarily a polynomial? Well, for $n=0$ we have $T_0 = 1$, and for $n=1$ we have $T_1(x)=x$. For $n=2$ we already need some trigonometry: $\cos(2\theta)=2\cos^2(\theta)-1$, so $T_2(x)=2x^2-1$. We can define $T_n$ recursively by trigonometric insights: $$\cos(\alpha)+\cos(\beta)=2\cos(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})$$ $$\implies \cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos(n\theta)\cos(\theta)$$ $$\implies T_{n+1}(x) + T_{n-1}(x) = 2T_{n}(x)x$$

This is how I calculated $T_7$. In practice I just used the recurrence relation $T_{n+1}(x) = 2T_{n}(x)x-T_{n-1}$ and the table here. There are some shortcuts, since the leading coefficient of $T_n$ is $2^{n-1}$ and the last coefficient is $0$ when $n$ is odd.

| cite | improve this answer | |
  • $\begingroup$ Interestingly enough, because of the symmetry of the roots, setting $x=\frac{1}{y^2}$ and using Vieta's formula on the $y^2$ term yields the same result. I assume that the other answer given also used Chebyshev polynomials of the first kind? I appreciate the explanation on the recursive nature of $cos(n\theta)$ $\endgroup$ – Alex Dec 27 '12 at 2:01
  • $\begingroup$ @Alex - I think that answer used something else, namely the addition identity for $\tan$. Anyway, your comment made me realize a simplification and generalization of the solution - If you want to compute $\sum_{k=1, k< \frac{n+1}{2}}^{n} \frac{1}{\cos^2(\frac{\pi( 2k-1)}{2n})}$, you can just take the $n$'th Chebyshev Polynomial (divide it by $x$ if $n$ is odd), "reverse" it (this makes its roots be the reciprocals of the original roots), make it monic, call it $P$. Then, your sum is just $\frac{1}{2}(a_1^2+2a_2)$, where $a_i$ is the coefficient of $X^{\deg P - i}$ in $P$. [cont. below] $\endgroup$ – Ofir Dec 27 '12 at 9:40
  • $\begingroup$ [cont.] But $a_1=0$, so this becomes simply $a_2$, as you noticed. To calculate this $a_2$, you need a formula for the coefficients of $x^2,x^0$ (when $n$ is even) of the Chebyshev Polynomial, and $x^3,x^1$ (when $n$ is odd). Explicit formulas can be given by using the recurrence relation of the polynomials (differentiating it and plugging $x=0$ might help). $\endgroup$ – Ofir Dec 27 '12 at 9:41
  • $\begingroup$ [cont.] Fix: It's $\frac{1}{2}(a_1^2-2a_2)$. But $a_1=0$ (from symmetry), so this becomes simply $-a_2$, as you noticed. To calculate this $a_2$, you need a formula for the coefficients of $x_2,x_0$ (when n is even) of the Chebyshev Polynomial, and $x_3,x_1$ (when n is odd). Explicit formulas can be given by using the recurrence relation of the polynomials (differentiating it and plugging $x=0$ might help). So, in general, your sum if a quotient of 2 coefficients of the Chebyshev Polynomial. The coefficient of $x^0$ is $1,0,-1,0,\cdots$. [cont.] $\endgroup$ – Ofir Dec 27 '12 at 10:15
  • $\begingroup$ The coefficient of $x^1$ is $(-1)^{\frac{n-1}{2}}n$ when $n$ is odd, and $0$ when $n$ is even. The coefficients of $x^2$ obey the recurrence $b_{n+1}+b_{n-1}=2a_{n}$ where $a_n$ is the coefficient of $x^1$. Similarly, the coefficients of $x^3$ obey the recurrence $c_{n+1}+c_{n-1}=2b_{n}$ where $b_n$ is the coefficient of $x^2$. $\endgroup$ – Ofir Dec 27 '12 at 10:17

Using this,

the roots of $\displaystyle z^3+z^2-3z-1=0\qquad (1)$ are $\displaystyle 2\cos\frac{2\pi}7, 2\cos\frac{4\pi}7, 2\cos\frac{6\pi}7$

If $\displaystyle\cot^2\frac{r\pi}7=u, \cos\frac{2r\pi}7=\frac{1-\tan^2\frac{r\pi}7}{1+\tan^2\frac{r\pi}7}=\frac{\cot^2\frac{r\pi}7-1}{\cot^2\frac{r\pi}7+1}=\frac{u-1}{u+1}$

$\displaystyle\implies\frac{2(u-1)}{u+1}=2\cos\frac{2r\pi}7($ where $r=1,2,3)$ will satisfy $(1)$

$\displaystyle\implies \left(\frac{2(u-1)}{u+1}\right)^3+\left(\frac{2(u-1)}{u+1}\right)^2-3\left(\frac{2(u-1)}{u+1}\right)-1=0$

On simplification, $\displaystyle 7 u^3-35 u^2+21 u-1=0$ whose roots are $\displaystyle\cot^2\frac{r\pi}7($ where $r=1,2,3)$

Now, use Vieta's formulas, to find $\displaystyle \sum \cot^2\frac{r\pi}7=\frac{35}7$

| cite | improve this answer | |

As $$\tan(2n+1)s=\frac{t^{2n+1}-\binom{2n+1}2t^{2n-1}+\cdots}{\binom{2n+1}1t^{2n}-\binom{2n+1}3t^{2n-2}+\cdots}$$ where $t=\tan s$

So, $$\tan 7s=\frac{t^7-21t^5+35t^3-7t}{7t^6-35t^4+21t-1}$$

If we put $7s=\pi,t^7-21t^5+35t^3-7t=0--->(1)$ whose roots are $\tan\frac{r\pi}7$ where $r=0,1,2,3,4,5,6$

So, the roots of $t^6-21t^4+35t^2-7=0--->(2)$ are $\tan\frac{r\pi}7$ where $r=1,2,3,4,5,6$

If we put $z=\frac1t$ (as $t\ne0,$) $\frac1{z^6}-\frac{21}{z^4}+\frac{35}{z^2}-7=0\implies z^6-5z^4+3z^2-\frac17=0$ whose roots are $\cot\frac{r\pi}7$ where $r=1,2,3,4,5,6$

So, $$z^6-5z^4+3z^2-\frac17=\prod_{1\le r\le 6}(z-\cot\frac{r\pi}7)$$

But as $\cot\frac{(7-r)\pi}7=\cot(\pi-\frac{r\pi}7)=-\cot\frac{r\pi}7$,

so $\prod_{1\le r\le 6}(z-\cot\frac{r\pi}7)$ $=\prod_{1\le r\le 3}(z-\cot\frac{r\pi}7)\prod_{4\le r\le 6}(z-\cot\frac{r\pi}7)$ $=\prod_{1\le r\le 3}(z-\cot\frac{r\pi}7)\prod_{3\ge u\ge 1}(z+\cot\frac{u\pi}7)$ (putting $7-r=u$)

$=\prod_{1\le r\le 3}(z^2-\cot^2\frac{r\pi}7)$

So,$z^6-5z^4+3z^2-\frac17$ $=z^6-z^4\sum_{1\le r\le 3}\cot^2\frac{r\pi}7$ $+z^2(\cot^2\frac{\pi}7\cot^2\frac{2\pi}7+\cot^2\frac{\pi}7\cot^2\frac{2\pi}7+\cot^2\frac{2\pi}7\cot^2\frac{3\pi}7)-\prod_{1\le r\le 3}\cot^2\frac{r\pi}7$

Comparing the coefficients of $z^4,$ we get the required identity.

Alternatively, If we put $\cot^2\frac{n\pi}7=y$ where $n=1,2,3$ or $n=7-1,7-2,7-3$ we get $y=\frac1{t^2}$ (as $\cot\frac{(7-r)\pi}7=-\cot\frac{r\pi}7$)

Replacing $t^2=\frac1y$ in $(2)$ we get $\frac1{y^3}-\frac{21}{y^2}+\frac{35}y-7=0$

or $7y^3-35y^2+21y-1=0$

So, $\sum_{1\le n\le 3}\cot^2\frac{n\pi}7=\frac{35}7=5$ using Vieta's Formulae.

| cite | improve this answer | |

The following approach is intimately related to Ofir's; you can think of this as the linear algebraic variation of his route.

Consider the perturbed Toeplitz tridiagonal matrix


which has the characteristic polynomial


(where $U_n(x)$ is the Chebyshev polynomial of the second kind) and the eigenvalues

$$\mu_k=4\cos^2\left(\frac{k\pi}{2n+1}\right),\qquad k=1\dots n$$

We find then that the matrix $\mathbf H=4\mathbf I-\mathbf T$ has the eigenvalues

$$\eta_k=4\sin^2\left(\frac{k\pi}{2n+1}\right),\qquad k=1\dots n$$

Now, take the case $n=3$:

$$\mathbf H=\begin{pmatrix}2&1&\\1&2&1\\&1&3\end{pmatrix}$$

We have

$$\mathbf U=\mathbf H^{-1}=\frac17\begin{pmatrix}5&-3&1\\-3&6&-2\\1&-2&3\end{pmatrix}$$

The eigenvalues of $\mathbf U$ are

$$\xi_k=\frac14\csc^2\left(\frac{k\pi}{7}\right),\qquad k=1\dots 3$$

which means the eigenvalues of $\mathbf W=4\mathbf U-\mathbf I$ are

$$\lambda_k=\cot^2\left(\frac{k\pi}{7}\right),\qquad k=1\dots 3$$

Now, with

$$\mathbf W=\frac17\begin{pmatrix}13&-12&4\\-12&17&-8\\4&-8&5\end{pmatrix}$$

you can see that the trace of $\mathbf W$ (which is also the sum of the eigenvalues of $\mathbf W$) is $5$.

| cite | improve this answer | |
  • $\begingroup$ It's way too difficult $\endgroup$ – Alex Jul 9 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.