Proving $\frac{1}{\cos^2\frac{\pi}{7}}+ \frac {1}{\cos^2\frac {2\pi}{7}}+\frac {1}{\cos^2\frac {3\pi}{7}} = 24$

Question

Someone gave me the following problem, and using a calculator I managed to find the answer to be $24$.

Calculate $$\frac {1}{\cos^2\frac{\pi}{7}}+ \frac{1}{\cos^2\frac{2\pi}{7}}+\frac {1}{\cos^2\frac{3\pi}{7}}\,.$$

The only question left is, Why? I've tried using Euler's Identity, using a heptagon with Law of Cosine w/ Ptolemy's, etc. but the fact that the cosine values are all squared and in the denominator keeps getting me stuck.

If $\zeta=e^{\frac{2\pi i}{7}}$, then the required expression is $$4\left(\frac{\zeta^2}{(\zeta+1)^2}+\frac{\zeta^4}{(\zeta^2+1)^2}+\frac{\zeta^6}{(\zeta^3+1)^2}\right).$$ How do we simplify this result further?

Answer 1

The roots of $x^6+x^5+\ldots+x+1$ over $\mathbb{C}$ are $x=\exp\left(\frac{2\pi\text{i}}{7}\right)$ for $k=1,2,\ldots,6$. Let $y:=x+\frac{1}{x}$. Then, $$\frac{x^6+x^5+\ldots+x+1}{x^3}=\left(y^3-3y\right)+\left(y^2-2\right)+y+1=y^3+y^2-2y-1\,.$$ Hence, the roots of $y^3+y^2-2y-1$ are $y=y_k:=2\,\cos\left(\frac{2k\pi}{7}\right)$ for $k=1,2,3$. Observe that $$S:=\sum_{k=1}^3\,\frac{1}{\cos^{2}\left(\frac{k\pi}{7}\right)}=\sum_{k=1}^3\,\frac{2}{1+\cos\left(\frac{2k\pi}{7}\right)}=4\,\sum_{k=1}^3\,\frac{1}{2+y_k}\,.$$ Since $y_k^3+y_k^2-2y_k-1=0$, we have $$y_k^2-y_k=\frac{1}{2+y_k}$$ for all $k=1,2,3$. Consequently, $$S=4\,\sum_{k=1}^3\,\left(y_k^2-y_k\right)\,.$$ The rest should be easy.

---

In general, let $n$ be a nonnegative integer and we are evaluating the sums $\displaystyle \sum_{k=0}^{2n}\,\frac{1}{\cos^{2}\left(\frac{k\pi}{2n+1}\right)}$ and $\displaystyle \sum_{k=1}^{n}\,\frac{1}{\cos^{2}\left(\frac{k\pi}{2n+1}\right)}$. The roots of $x^{2n+1}-1$ over $\mathbb{C}$ are $x=x_k:=\exp\left(\frac{2k\pi}{2n+1}\right)$, for $k=0,1,2,\ldots,2n$. Observe that $$\frac{1}{1+x_k}=\frac{1}{2}\,\left(\frac{1+x_k^{2n+1}}{1+x_k}\right)=\frac{1}{2}\,\sum_{j=0}^{2n}\,(-1)^j\,x_k^j=\frac{2n+1}{2}-\frac{1}{2}\,\sum_{j=1}^{2n}\,\left(1-\left(-x_k\right)^j\right)\,.$$ That is, $$\frac{1}{\left(1+x_k\right)^2}=\frac{2n+1}{2}\left(\frac{1}{1+x_k}\right)-\frac{1}{2}\,\sum_{j=1}^{2n}\,\sum_{i=0}^{j-1}\,(-1)^i\,x_k^i\,,$$ or equivalently, $$\frac{1}{\left(1+x_k\right)^2}=\frac{2n+1}{4}\,\sum_{j=0}^{2n}\,(-1)^j\,x_k^j-\frac{1}{2}\,\sum_{j=1}^{2n}\,\sum_{i=0}^{j-1}\,(-1)^i\,x_k^i\,.$$ Consequently, $$\frac{x_k}{\left(1+x_k\right)^2}=\frac{2n+1}{4}\,\sum_{j=0}^{2n}\,(-1)^j\,x_k^{j+1}-\frac{1}{2}\,\sum_{j=1}^{2n}\,\sum_{i=0}^{j-1}\,(-1)^i\,x_k^{i+1}=\frac{2n+1}{4}+f\left(x_k\right)$$ for some polynomial $f(x)$ of degree at most $2n$ without the constant term. Then, $$\sum_{k=0}^{2n}\,\frac{x_k}{\left(1+x_k\right)^2}=\frac{(2n+1)^2}{4}+\sum_{k=1}^{2n}\,f\left(x_k\right)\,.$$ It is evident that $\displaystyle\sum_{k=0}^{2n}\,f\left(x_k\right)=0$. Furthermore, $$\frac{x_k}{\left(1+x_k\right)^2}=\frac{1}{2}\,\left(\frac{1}{1+\cos\left(\frac{2k\pi}{2n+1}\right)}\right)=\frac{1}{4}\,\left(\frac{1}{\cos^2\left(\frac{k\pi}{2n+1}\right)}\right)\,.$$ Ergo, $$\frac{1}{4}\,\sum_{k=0}^{2n}\,\frac{1}{\cos^2\left(\frac{k\pi}{2n+1}\right)}=\sum_{k=0}^{2n}\,\frac{x_k}{\left(1+x_k\right)^2}=\frac{(2n+1)^2}{4}\,.$$ This shows that $$\sum_{k=0}^{2n}\,\frac{1}{\cos^2\left(\frac{k\pi}{2n+1}\right)}=(2n+1)^2\,.$$ Furthermore, we have $$\sum_{k=1}^n\,\frac{1}{\cos^2\left(\frac{k\pi}{2n+1}\right)}=\frac{(2n+1)^2-1}{2}=2n(n+1)\,.$$

Answer 2

In this lovely answer, @joriki establishes the identity $$ \sum _{l=1}^{n}\tan^2 \frac {l\pi } {2n+1}=n(2n+1)\;. $$ With $n=3$ this gives $$ \tan^2\frac\pi7+\tan^2\frac{2\pi}7 + \tan^2\frac{3\pi}7=21. $$ The desired result follows from the identity $\sec^2\theta=1+\tan^2\theta$.

Answer 3

First notice that since $\cos(x)=\cos(\pi-x)$, we have

$$1+2\left(\frac{1}{\cos(\frac{\pi}{7})^2}+\frac{1}{\cos(\frac{2\pi}{7})^2}+\frac{1}{\cos(\frac{3\pi}{7})^2}\right)=\sum_{k=0}^6 \frac{1}{\cos(\frac{k\pi}{7})^2}$$

Now, $x\mapsto 2x$ is a bijection of the integers mod $7$, so we may make the summands $\cos(\frac{2\pi k}{7})^{-2}$.

Using $\cos(\frac{2\pi k}{n})=(\zeta^k+\zeta^{-k})/2$ where $\zeta=e^{2\pi i/n}$ combined with the geometric sum formula

$$\frac{a^n+b^n}{a+b}=\sum_{r=0}^{n-1} a^{(n-1)-r}b^r,$$

and the fact that for $n$th roots of unity $\xi$,

$$\sum_{k=0}^{n-1} \xi^k =\begin{cases} n & \xi=1 \\ 0 & \xi\ne1 \end{cases} $$

we may derive

$$\sum_{k=0}^{n-1}\frac{1}{\cos(\frac{2\pi k}{n})^m}=\sum_k \left(\frac{2}{\zeta^k+\zeta^{-k}}\right)^m=\sum_k \left(\frac{\zeta^{nk}+\zeta^{-nk}}{\zeta^k+\zeta^{-k}}\right)^m $$

$$=\sum_k\left(\sum_{r=0}^{n-1}(-1)^r \zeta^{-(2r+1)k}\right)^m=\sum_k \sum_{\substack{r_1,\cdots,r_m \\ \sum r_i=r}}(-1)^r\zeta^{-(2r+m)k}$$

$$ =\sum_{\substack{r_1,\cdots,r_m \\ \sum r_i=r}} (-1)^r \sum_k (\zeta^{-2r-m})^k=n(A-B).$$

Therefore in conclusion we have

Theorem.

$$\sum_{k=0}^{n-1} \frac{1}{\cos(\frac{2\pi k}{n})^m}=n(A-B)$$

where $A$ and $B$ count the solutions to $r_1+\cdots+r_m\equiv -m/2$ mod $n$ with $\sum_i r_i$ even and odd respectively (and $0\le r_1,\cdots,r_m<n$).

As a special case, if $m=2$ we see that $A=n$ and $B=0$, yielding the corollary

$$\sum_{k=0}^{n-1}\frac{1}{\cos(\frac{2\pi k}{n})^2}=n^2.$$

Answer 4

Incomplete solution

The idea behind this solution is using Vieta's cubic formulas to rewrite the equality in a simpler, more manageable way. Letting $\cos^2\frac{\pi}{7}=\alpha$, $\cos^2\frac{2\pi}{7}=\beta$, and $\cos^2\frac{3\pi}{7}=\gamma$, we have this:

$$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=24$$

$$\frac{\alpha\beta+\alpha\gamma+\beta\gamma}{\alpha\beta\gamma}=24$$

By letting $\alpha,\beta,$ and $\gamma$ be roots to a cubic, we use Vieta's formula's to see that $-\frac{c}{d}=24$

---

Now, consider the three denominators as roots to a cubic polynomial. Let's make a third degree equation using them:

$$(x-\cos^2\frac{\pi}{7})(x-\cos^2\frac{2\pi}{7})(x-\cos\frac{3\pi}{7})$$

Let's use substitutions to avoid some nasty simplified expression. Let $u=\cos\frac{\pi}{7}$. Using some angle addition trig identities, we transform our equation into this:

$$(x-u^2)(x-4u^2+4u-1)(x-16u^6+24u^4-9u^2)$$

Multiply them out:

$$x^3+(-16u^6+24u^4-14u^2+4u-1)x^2+(80u^8-64u^7-104u^6+96u^5+25u^4-40u^3+10u^2)x+(-64u^{10}+64u^9+80u^8-96u^7-12u^6+36u^5-9u^4)$$

Consider the form $ax^3+bx^2+cx+d$. Per Vietas' formulas, $-c/d=24$.

$$\frac{80u^6-64u^5-104u^4+96u^3+25u^2-40u+10}{64u^8-64u^7-80u^6+96u^5+12u^4-36u^3+9u^2}=24$$

I will continue to work on proving this last part, but I figured at least simplifying your problem to this might be of some asssistance, and someone else may be able to validate the last equation before I do. Basically, if you prove the last equality you prove the question posted.