4
$\begingroup$

Calculate the value of $\sec^4\frac{\pi}{9}+\sec^4\frac{2\pi}{9}+\sec^4\frac{3\pi}{9}+\sec^4\frac{4\pi}{9}$.

I tried a lot of ways to calculate the sum, like converting to $\cos$, applying the triple angle formula, or taking $2$, $3$, or even all $4$ at a time and simplifying. But I couldn't get anywhere.

$\endgroup$
2
  • $\begingroup$ Rewrite as $$\sec^4\frac{\pi}{9}+\sec^4\frac{2\pi}{9}+\sec^4\frac{3\pi}{9}+\sec^4\frac{4\pi}{9}=\\(1+\tan^2\frac{\pi}{9})^2+ (1+\tan^2\frac{2\pi}{9})^2+(1+\tan^2\frac{4\pi}{9})^2+\sec^4\frac{3\pi}{9}$$ expan them and use $$\boxed{\tan ^ { 2 } 20 ^ { \circ } + \tan ^ { 2 } 4 0 ^ { \circ } + \tan ^ { 2 } 80 ^ { \circ } = 33.} $$ look this page math.stackexchange.com/questions/175736/… $\endgroup$
    – Khosrotash
    Commented Apr 18 at 18:26
  • 1
    $\begingroup$ plugging it into wolfram.alpha you get its equal to 1120 though $\endgroup$ Commented Apr 18 at 18:53

4 Answers 4

8
$\begingroup$

Wildly enough, you can show that the eigenvalues of the $4\times 4$ min-matrix (where $a_{ij}=\min(i,j)$), $$ {M}=\left( \begin{matrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \\ \end{matrix} \right), $$ are exactly $\frac{1}{4}\sec^2(k\pi / 9)$ for $k=1,2,3,4$. (A related formula holds for larger min-matrices.) The sum of the eigenvalues is just the trace; the sum of the squares of the eigenvalues is the trace of $M^2$ (easily found to be $70$). So you have $$ \sum_{k=1}^{4}\sec^4\left(\frac{k\pi}{9}\right)=16 \text{ Tr}(M^2)=16\cdot 70 = 1120. $$ Of course, this generalizes to larger denominators than $9$, and to higher even powers of the secant.

$\endgroup$
6
  • 1
    $\begingroup$ How did you come to this matrix? $\endgroup$
    – user
    Commented Apr 18 at 19:34
  • 1
    $\begingroup$ Well, it's the inverse of a particular tridiagonal matrix (the one with $2$s and a final $1$ on the diagonal, and $-1$ for all the adjacent elements); and that matrix has eigenvalues that can be expressed in terms of $\cos^2$. $\endgroup$
    – mjqxxxx
    Commented Apr 18 at 20:48
  • 1
    $\begingroup$ I would not say this is a trivial fact. $\endgroup$
    – user
    Commented Apr 20 at 21:29
  • 1
    $\begingroup$ Therefore I would suggest to start the answer with derivation of the matrix. $\endgroup$
    – user
    Commented Apr 21 at 8:35
  • 1
    $\begingroup$ A generalisation of this matrix was also seen in this post. $\endgroup$
    – Sahaj
    Commented May 2 at 12:19
3
$\begingroup$

@mjqxxxx 's solution seems fine fine but here is an alternative simpler approach i found.

$$\sec^4\frac{3\pi}9 \implies \sec^4\frac{\pi}3 = 16$$

for $$\theta=\frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9} \implies \cos3\theta=\frac 12$$

$$\cos3\theta=4\cos^3\theta-3\cos\theta\qquad \cos\theta=t$$

$$\implies 4t^3-3t=\frac 12 \implies 8t^3-6t=1\qquad\dots(1)$$ has roots for $$t=\cos\frac{\pi}9,\;\cos\frac{5\pi}9,\;\cos\frac{7\pi}9$$

Squaring $(1)$: $$64t^6+36t^2-96t^4=1\qquad\dots(2)$$ has roots for $$t=\cos\frac{\pi}9,\;\cos\frac{\pi}9,\;\cos\frac{5\pi}9,\;\cos\frac{5\pi}9,\;\cos\frac{7\pi}9,\;\cos\frac{7\pi}9$$

substituting: $t=\frac 1{\sqrt x}$ in $(2)$

$$\implies x^3-36x^2+96x-64=0$$ has roots for $$x=\sec^2\frac{\pi}9,\;\sec^2\frac{5\pi}9,\;\sec^2\frac{7\pi}9$$

Now,

$$\sec^2\frac{\pi}9+\sec^2\frac{5\pi}9+\sec^2\frac{7\pi}9=36$$ $$\sec^2\frac{\pi}9\sec^2\frac{5\pi}9+\sec^2\frac{\pi}9\sec^2\frac{7\pi}9+\sec^2\frac{5\pi}9\sec^2\frac{7\pi}9=96$$

$$\sec^4\frac{\pi}9+\sec^4\frac{5\pi}9+\sec^4\frac{7\pi}9=\left(\sec^2\frac{\pi}9+\sec^2\frac{5\pi}9+\sec^2\frac{7\pi}9\right)^2 - 2\left(\sec^2\frac{\pi}9\sec^2\frac{5\pi}9+\sec^2\frac{\pi}9\sec^2\frac{7\pi}9+\sec^2\frac{5\pi}9\sec^2\frac{7\pi}9\right)$$

$$\implies \sec^4\frac{\pi}9+\sec^4\frac{5\pi}9+\sec^4\frac{7\pi}9 = 1104$$

So,

$$\sec^4\frac{\pi}9+\sec^4\frac{3\pi}9+\sec^4\frac{5\pi}9+\sec^4\frac{7\pi}9=1104+16=1120$$

$\endgroup$
0
2
$\begingroup$

Consider the expansion of $\tan9\theta$. Using De Moivre's Theorem, this is $$\tan9\theta=\frac{9\tan\theta-84\tan^3\theta+126\tan^5\theta-36\tan^7\theta+\tan^9\theta}{1-36\tan^2\theta+126\tan^4\theta-84\tan^6\theta+9\tan^8\theta}$$Using the roots of $\tan9\theta=0$, we get that $$\begin{align}\tan^9\theta-36\tan^7\theta+126\tan^5\theta-84\tan^3\theta+9\tan\theta&=\prod_{k=0}^8\left(\tan\theta-\tan\left(\frac{k\pi}9\right)\right)\\&=\tan\theta\prod_{k=1}^4\left((\tan\theta-\tan\left(\frac{k\pi}9\right)\right)\prod_{k=5}^8\left((\tan\theta-\tan\left(\frac{k\pi}9\right)\right)\\&=\tan\theta\prod_{k=1}^4\left((\tan\theta-\tan\left(\frac{k\pi}9\right)\right)\sum_{k=1}^4\left((\tan\theta+\tan\left(\frac{k\pi}9\right)\right)\\&=\tan\theta\prod_{k=1}^4\left((\tan^2\theta-\tan^2\left(\frac{k\pi}9\right)\right)\end{align}$$This means that $\tan^2\left(\frac\pi9\right)$, $\tan^2\left(\frac{2\pi}9\right)$, $\tan^2\left(\frac{3\pi}9\right)$ and $\tan^2\left(\frac{4\pi}9\right)$ are the roots of $t^4-36t^3+126t^2-84t+9=0$.

If $u=t+1$, then $t=u-1$, so the equation becomes $$\begin{align}t^4-36t^3+126t^2-84t+9&=(u-1)^4-36(u-1)^3+126(u-1)^2-84(u-1)+9\\&=u^4-40u^3+240u^2-448u+256\end{align}$$which has roots $\sec^2\left(\frac{\pi}9\right)$, $\sec^2\left(\frac{2\pi}9\right)$, $\sec^2\left(\frac{3\pi}9\right)$ and $\sec^2\left(\frac{4\pi}9\right)$. This means that $$\sec^4\left(\frac{\pi}9\right)+\sec^4\left(\frac{2\pi}9\right)+\sec^4\left(\frac{3\pi}9\right)+\sec^4\left(\frac{4\pi}9\right)=40^2-2(240)=\boxed{1120}$$

$\endgroup$
1
$\begingroup$

I shall provide a generalized answer. I start with the following preliminary claim:

Claim: Consider the set $S := \{s_1, s_2,\dots\}$. Then by $e_n$ denote the sum of elements of $S$ taken $n$ at a time. To be precise, $e_n$ is the $n$th elementary symmetric polynomial of the elements in $S$. Then we have $$\sum_{s\in S} s^2 = e_1^2 - 2e_2$$ and $$\sum_{s \in S} s^4 = e_1^4 - 4e_1^2e_2 + 2e_2^2 + 4e_1e_3-4e_4$$

Proof: Left as an exercise. This can be proven easily using Newton's identity.

Theorem: For a fixed odd natural $n$ we have $$\sum_{r=1}^{n-1} \sec^4\left(\frac{r\pi}n\right) =\frac{1}3(n^2+3)(n+1)(n-1)$$

Proof: Recall the expansion $$\sin(nx) = \cos^n (x)\cdot \left(\binom{n}{1} \tan x - \binom{n}3 \tan^3 x + \dots + (-1)^{\frac{n-1}2}\binom{n}{n} \tan^{n} x\right)$$ for odd natural $n$ which comes from Binomial theorem & De-Moivre's theorem. Take $p(t, n) = \binom{n}{1} t - \binom{n}3 t^3 + \dots + (-1)^{\frac{n-1}2}\binom{n}{n} t^n$ where $p(\tan x, n)$ is the polynomial in $\tan x$ in the above expansion. Since $\cos x$ and $\sin nx$ never vanish simultaneously in this context, it follows that the roots of $\sin(nx)$ are also the roots of $p(\tan x,n)$. So $p(t,n)$ has for its roots all the $n$ elements of $S$, where $$S := \left\{\tan 0, \tan\left(\frac{\pi}n\right), \tan\left(\frac{2\pi}n\right), \dots, \tan\left(\frac{(n-1)\pi}n\right)\right\}$$ Define $e_n$ to be the n-th elementary symmetric polynomial for the elements of $S$. Note that we can evaluate $e_n$ from the coefficients of the polynomial $p(t,n)$ via Vieta's relations. Also note that $\sec^4 x = 1 + 2\tan^2x + \tan^4x$. With all this it follows $$\sum_{r=1}^{n-1} \sec^4\left(\frac{r\pi}n\right) = \sum_{s \in S} (1 + 2s^2 + s^4) - \underbrace{1}_{\text{because of }\tan 0} = n - 1 + 2(e_1^2 - 2e_2) + (e_1^4 - 4e_1^2e_2 + 2e_2^2 + 4e_1e_3 - 4e_4) = 2e_2^2 - 4e_2 - 4e_4 + n-1 = 2\binom{n}{n-2}^2 + 4\binom{n}{n-2} - 4\binom{n}{n-4} + n - 1 = \frac{1}3(n^2+3)(n+1)(n-1)$$

From the theorem, it follows that $$\sum_{r=1}^{9} \sec^4\left(\frac{r\pi}9\right) = \frac13\cdot 84 \cdot 10 \cdot 8 = 2240$$ but recall that $\cos(x) = -\cos(\pi - x)$ so that $\sec^4(x) = \sec^4(\pi - x)$. Then we get $\sum_{r=1}^{9} \sec^4\left(\frac{r\pi}9\right) = 2\sum_{r=1}^{4} \sec^4\left(\frac{r\pi}9\right)$ so it follows at once that $$\boxed{\sum_{r=1}^{4} \sec^4\left(\frac{r\pi}9\right) = 1120}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .