How do I show that:

$$\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$$

This is actually problem B $4371$ given at this link. Looks like a very interesting problem.

My attempts: Well, I have been thinking about this for the whole day, and I have got some insights. I don't believe my insights will lead me to a $\text{complete}$ solution.

  • First, I wrote $\sin\frac{5\pi}{14}$ as $\sin\frac{9 \pi}{14}$ so that if I put $A = \frac{\pi}{14}$ so that the given equation becomes, $$\frac{1}{\sin^{2}{A}} + \frac{1}{\sin^{2}{3A}} + \frac{1}{\sin^{2}{9A}} =24$$ Then I tried working with this by taking $\text{lcm}$ and multiplying and doing something, which appeared futile.

  • Next, I actually didn't work it out, but I think we have to look for a equation which has roots as $\sin$ and then use $\text{sum of roots}$ formulas to get $24$. I think I haven't explained this clearly.

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    $\begingroup$ In your first bullet, you probably mean $\sin \frac{5\pi}{14} = \sin \frac{9\pi}{14}$. I would be more interested in the arithmetic progression of 1,3,5 than the geometric one because of the angle-sum identities. $\endgroup$ – Ross Millikan Jun 13 '11 at 19:18
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    $\begingroup$ Most people would write complete, not $\text{complete}$. :) $\endgroup$ – Mateen Ulhaq Jun 14 '11 at 1:19

The roots idea should work, but first convert to $\cos$ using the formula $1 - 2\sin^2 x = \cos 2x$.

You will need to get a polynomial of which $\cos (2k+1)\pi/7$ is a root (polynomial corresponding to $\cos 7\theta = -1$) and you are interested in finding out $\sum \frac{1}{1-r}$ over the roots $r$. By using the fact that $\cos 5\pi/7 = \cos 9\pi/7$ etc, you get your sum.

To complete it,

We have that the Chebyshev Polynomial $T_7(\cos x) = \cos 7x$

Thus the polynomial we seek is $\displaystyle Q(x) = T_7(x)+1 = 64x^7 - 112 x^5 + 56x^3 -7x +1$

Its roots are $\cos (2k+1) \pi /7$, $0 \le k \le 6$.

For any polynomial $P(x)$ with roots $r_1, r_2, \dots, r_n$ we have by differentiating $\log P(x)$ that

$$ \sum_{j=1}^{n} \frac{1}{x - r_j} = \frac{P'(x)}{P(x)}$$

Thus the value we seek is $\displaystyle \frac{Q'(1)}{Q(1)} - \frac{1}{2}$ (one of the roots is $\cos \pi = -1$) and this can easily be calculated to be $24$.

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  • $\begingroup$ I believe the roots of $T_7$ are $\cos(\frac{(2k+1)\pi}{14})$ $\endgroup$ – Thomas Andrews Jun 14 '11 at 3:31
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    $\begingroup$ Without knowing chebyshev polynomials, can't this be solved? $\endgroup$ – user9413 Jun 14 '11 at 4:56
  • $\begingroup$ @Thomas: That might be true, but we are looking at $Q(x) = T_7(x) + 1$. $\endgroup$ – Aryabhata Jun 14 '11 at 5:18
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    $\begingroup$ @Chandru: Yes you can. You can actually derive the polynomial yourself using $(\cos x + i \sin x)^7 = \cos 7x + i \sin 7x$. $\endgroup$ – Aryabhata Jun 14 '11 at 5:18
  • $\begingroup$ @Aryabhata: one help. I need to find value of $\cos\frac{2\pi}{13} + \cos\frac{6\pi}{13} + \cos\frac{8\pi}{13}$. In this link: isibang.ac.in/~sury/luckyoct10.pdf Prof.Sury finds using Gauss sums magar another method se mey verify karna chahta hoon. Aap suggest karo. $\endgroup$ – user9413 Feb 24 '12 at 19:48

Use $\sin(x) = \cos(\frac{\pi}2 - x)$, we can rewrite this as:

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

Let $a_k = \frac{1}{\cos \frac{k\pi}7}$.
Let $f(x) = (x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)(x-a_6)$.

Now, using that $a_k = - a_{7-k}$, this can be written as:

$$f(x) = (x^2-a_1^2)(x^2-a_2^2)(x^2-a_3^2)$$

Now, our problem is to find the sum $a_1^2 + a_2^2 + a_3^2$, which is just the negative of the coefficient of $x^4$ in the polynomial $f(x)$.

Let $U_6(x)$ be the Chebyshev polynomial of the second kind - that is:

$$U_6(\cos \theta) = \frac{\sin 7\theta }{\sin \theta}$$

It is a polynomial of degree $6$ with roots equal to $\cos(\frac{k\pi}7)$, for $k=1,...,6$.

So the polynomials $f(x)$ and $x^6U_6(1/x)$ have the same roots, so:

$$f(x) = C x^6 U_6(\frac{1}x)$$

for some constant $C$.

But $U_6(x) = 64x^6-80x^4+24x^2-1$, so $x^6 U_6(\frac{1}x) = -x^6 + 24 x^4 - 80x^2 + 64$. Since the coefficient of $x^6$ is $-1$, and it is $1$ in $f(x)$, $C=-1.$ So:

$$f(x) = x^6 - 24x^4 +80x^2 - 64$$

In particular, the sum you are looking for is $24$.

In general, if $n$ is odd, then the sum:

$$\sum_{k=1}^{\frac{n-1}2} \frac{1}{\cos^2 \frac{k\pi}{n}}$$

is the absolute value of the coefficient of $x^2$ in the polynomial $U_{n-1}(x)$, which turns out to have closed form $\frac{n^2-1}2$.

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  • $\begingroup$ Thanks a lot. I could have never thought of Chebyshev's polynomials even after @aryabhata had given a hint. $\endgroup$ – user9413 Jun 14 '11 at 4:55

Another method would involve use of complex numbers.

** added **

OK, elaboration.

Maple used for writing...

Let $w = \exp(i \pi/14)$ so that $w^7 = i$. In (1) I factored $w^7-i$ and in (2) obtained the relation satisfied by $w$. (3) is what we want to compute. (4) is the relations of the trig functions to $w$. In (5) we wrote the thing to compute in terms of $w$. In (6) we took the denominator, and reduced it using the relation satisfied by $w$. In (7) the same thing for the numerator. So (8) is our answer, which is simplified in (9).

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  • 8
    $\begingroup$ Could you elaborate? $\endgroup$ – ttt Jun 14 '11 at 5:15

We can derive $\cos7x=64c^7-112c^5+56c^3-7c$ where $c=\cos x$(Proof Below)

If $\cos7x=0, 7x=\frac{(2r+1)\pi}2,x=\frac{(2r+1)\pi}{14}$ where $r=0,1,2,3,4,5,6$

So, the roots of $64c^7-112c^5+56c^3-7c=0$ are $\cos\frac{(2r+1)\pi}{14}$ where $r=0,1,2,3,4,5,6$

So, the roots of $64c^6-112c^4+56c^2-7=0$ are $\cos\frac{(2r+1)\pi}{14}$ where $r=0,1,2,4,5,6$ as $\cos x=c=0$ corresponds to $r=3$

So, the roots of $64d^3-112d^2+56d-7=0$ are $\cos^2\frac{(2r+1)\pi}{14}$ where $r=0,1,2$ or $r=4,5,6$ as $\cos \frac{(7-k)\pi}7=\cos(\pi-\frac{k\pi}7)=-\cos\frac{k\pi}7$

If $$y=\frac1{\sin^2\frac{(2r+1)\pi}{14}}, y=\frac1{1-d}\implies d=\frac{y-1}y$$

So, the equation whose roots are $\frac1{\sin^2\frac{(2r+1)\pi}{14}}$ where $r=0,1,2$ is $$64\left(\frac{y-1}y\right)^3-112\left(\frac{y-1}y\right)^2+56\left(\frac{y-1}y\right)-7=0$$

On simplification, $y^3(64-112+56-7)+y^2\{64(-3)-112(-2)+56(-1)\}+y()+()=0$

So using Vieta's Formulas, $$\sum_{r=0}^2\frac1{\sin^2\frac{(2r+1)\pi}{14}}=\frac{24}1$$



(1) Using $\cos C+\cos D=2\cos\left(\frac{C-D}2\right)\cos\left(\frac{C+D}2\right)$

$$\cos7x+\cos x=2\cos3x\cos4x=2\cos3x(2\cos^22x-1)\text{ using }\cos2y=2\cos^2y-1$$

$$\text{ So, }\cos7x=-c+2(4c^3-3c)\{2(2c^2-1)^2-1\} \text{ using } \cos3y=4\cos^3y-3\cos y \text{ where } c=\cos x$$ $$\cos 7x=-c+(8c^3-6c)(8c^4-8c^2+1)=64c^7-112c^5+56c^3-7c$$

(2) Alternatively using de Moivre's formula,

$$\cos 7x+i\sin7x=(\cos x+i\sin x)^7$$

Expanding and equating the real parts $\cos7x=c^7-\binom72c^5s^2+\binom72c^3s^4-\binom76cs^6$ where $c=\cos x,s=\sin x$

So, $\cos7x=c^7-\binom72c^5(1-c^2)+\binom72c^3(1-c^2)^2-\binom76c(1-c^2)^3$


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This may be a $3$ year old question, but I would like to add to the list an answer that relies on the sum of $\tan^2$ identity.

Let $$\begin{align}S &= \frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}}\\ &=\frac{1}{\cos^{2}\frac{3\pi}{7}} + \frac{1}{\sin^{2}\frac{2\pi}{7}} + \frac{1}{\sin^{2}\frac{\pi}{7}}\\ &= \sec^2{\frac{3\pi}{7}} + \sec^2{\frac{2\pi}{7}} + \sec^2{\frac{\pi}{7}}\\ &= \tan^2\frac{\pi}{7} + \tan^2\frac{2\pi}{7} + \tan^2\frac{3\pi}{7} + 3\\ &= \sum_{k = 1}^3 \tan^2\frac{k\pi}{7} + 3\end{align}$$

From the sum of $\tan^2$ identity discussed over here, we have

$$ \sum_{k=1}^n\tan^2\frac{k\pi}{2n+1} = 2n^2+n,\quad n\in\mathbb{N}^+. $$

In our case, set $n = 3$. Then,

$$\begin{align}S &= 2(3)^2 + 3 + 3 \\&=24 \end{align}$$

as desired.

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  • $\begingroup$ If I'm not mistaken, this generalizes to give $\sum_{k=1}^n \csc^2 (k\pi/(2n+1)) = 2n(n+1)$, with $n = 3$ being the original question. $\endgroup$ – Michael Lugo Sep 18 '14 at 14:33

We can prove (below), the roots of $$z^3-z^2-2z+1=0 \ \ \ \ (1)$$ are $2\cos\frac{(2r+1)\pi}7$ where $r=0,1,2$

So,if we set $\displaystyle t=\frac1{\sin^2{\frac{(2r+1)\pi}{14}}}$ where $r=0,1,2$

$\implies 2\cos\frac{(2r+1)\pi}7=2\left(1-2\sin^2{\frac{(2r+1)\pi}{14}}\right)=2\left(1-\frac2t\right)=\frac{2(t-2)}t$ which will satisfy the equation $(1)$

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

On simplification we have $$8(t-2)^3-4t(t-2)^2-4t^2(t-2)+t^3=0$$

$$\text{or, }t^3(8-4-4+1)-t^2(8\cdot3\cdot2-4\cdot4-8)+()t+()=0$$

$$\text{or, }t^3-24t^2+()t+()=0$$

Now, use Vieta's Formulas



Let $7x=\pi$ and $y=\cos x+i\sin x$

Using De Moivre's formula, $y^7=(\cos x+i\sin x)^7=\cos \pi+\sin\pi=-1$

So, the roots of $y^7+1=0\ \ \ \ (1)$

are $\cos \theta+i\sin\theta$ where $\theta=\frac{(2r+1)\pi}7$ where $r=0,1,2,3,4,5,6$

Leaving the factor $y+1$ which corresponds to $r=3,$

we get $y^6-y^5+y^4-y^3+y^2-y+1=0$

Dividing either sides by $y^3,$ $$y^3+\frac1{y^3}-\left(y^2+\frac1{y^2}\right)+y+\frac1y-1=0$$

$$\implies \left(y+\frac1y\right)^3-3\left(y+\frac1y\right)-\{\left(y+\frac1y\right)^2-\}+y+\frac1y-1=0$$

$$\implies z^3-z^2-2z+1=0\ \ \ \ (2)$$ where $z=y+\frac1y=2\cos\theta$

Now, since $\cos(2\pi-A)=\cos A,\cos\left(2\pi-\frac{(2r+1)\pi}7\right)=\cos\left(\frac{(13-2r)\pi}7\right)$ where $r=0,1,2$

So, the roots of equation $(2)$ are $2\cos\frac\pi7=2\cos\frac{13\pi}7, 2\cos\frac{3\pi}7=2\cos\frac{11\pi}7$ and $2\cos\frac{5\pi}7=2\cos\frac{9\pi}7$


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For people unknown to complex-number: Let $s:=\sin,c:=\cos$ and the subscript as in $s_k$ denote $\sin\left(\frac{k\pi}{14}\right)$ and similiarly for other trigonometric ratios. $$\csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}=\frac{(s_3s_5)^2+(s_5s_1)^2+(s_1s_3)^2}{(s_1s_3s_5)^2}\\ =\frac{(c_2-c_8)^2+(c_4-c_6)^2+(c_2-c_4)^2}{4s_1^2s_3^2s_5^2} $$ Now, numerator is: $$\begin{align} (c_2-c_8)^2+(c_4-c_6)^2+(c_2-c_4)^2 &=2c_2^2+2c_4^2+c_6^2+c_8^2-2(c_2c_4+c_4c_6+c_2c_8)\\ &=1+c_4+1+c_8+1+\frac12(c_{12}+c_{16})-2(c_6+c_2+c_{10})\\ &=3+c_4+c_8-c_2-2(c_2+c_6+c_{10})\\ &=3-c_2+c_4-c_6-2(c_2-c_4+c_6)\\ &=3(1-c_2+c_4-c_6)\\ &=3(c_0+c_4+c_8+c_{12})\\ &=\frac3{2s_2}2s_2(c_0+c_4+c_8+c_{12})\\ &=\frac3{2s_2}(2s_2c_0+2s_2c_4+2s_2c_8+2s_2c_{12})\\ &=\frac3{2s_2}(s_2+s_2+s_6-s_2+s_{10}-s_6+s_{14}-s_{10})\\ &=\frac3{2s_2}(s_2)\\ &=\frac32 \end{align}$$ One thing to note is $\displaystyle c_0+c_4+c_8+c_{12}=\frac12$.

Now denominator is (do similarly for this):

$$\begin{align} 4s_1^2s_3^2s_5^2 &=4\frac1{16}(2(2s_1s_3)s_5)^2\\ &=\frac1{4}(2(c_2-c_4)s_5)^2\\ &=\frac1{4}(s_7+s_3-s_9-s_1)^2\\ &=\frac1{4}(1-s_1+s_3-s_5)^2\\ &=\frac1{4}(c_0+c_8+c_4+c_{12})^2\\ &=\frac1{4}\left(\frac12\right)^2\\ &=\frac{1}{16} \end{align}$$

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  • $\begingroup$ +1 You have a nice proof, but I don't think people who have not studied complex numbers would be given this type of problem. $\endgroup$ – Toby Mak Oct 3 '19 at 10:51

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