TITLE: Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$
QUESTION [19 upvotes]: 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.

REPLY [3 votes]: $$S=\csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}$$


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