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.


*

*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.

  
*
  
*$\text{Thirdly, is there a trick proving such type of identities using Gauss sums ?}$ 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.
  


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 $2\cos\frac{5\pi}7=2\cos\frac{9\pi}7$
]
