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


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*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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*$\text{Thirdly, is there a trick proving such type of identities using Gauss sums ?}$ One post related to this is: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$ I don't know how this will help as I haven't studied anything yet regarding Gauss sums.
  

 A: 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 $\frac{n^2-1}2$.
A: Another method would involve use of complex numbers.
** added **
OK, elaboration.

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).
A: 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.
A: 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
[
Proof:
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 $2\cos\frac{5\pi}7=2\cos\frac{9\pi}7$
]
A: 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 $24$.
