What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to write $\textrm{cosec}^2\left(\frac{4\pi}7\right)$ as $\textrm{cosec}^2\left(\frac{3\pi}7\right)$. Then converted in $\sin$... But in vain.. Is there any other approach?
When in doubt, convert trigonometric functions to complex exponentials.
If $w = e^{i\pi/7}$, $\csc^2(\pi/7) = \dfrac{-4}{(w-1/w)^2}$ and similarly for the others with $w$ replaced by $w^2$ and $w^4$. Simplifying, $$ \csc^2(\pi/7) + \csc^2(2\pi/7) + \csc^2(4\pi/7) - 8 \\= -4\,{\frac {2\,{w}^{16}+{w}^{14}+3\,{w}^{12}+3\,{w}^{10}+3\,{w}^{8}+3 \,{w}^{6}+3\,{w}^{4}+{w}^{2}+2}{\left({w}^{8}-1 \right) ^{2}}} $$ and the numerator is divisible by $w^6+w^5+w^4+w^3+w^2+w+1 = 0$
EDIT: This cries out for generalization. We also have $$\csc^2(\pi/3)+ \csc^2(2\pi/3) = 8/3$$ $$\csc^2(\pi/15) + \csc^2(2\pi/15) + \csc^4(4\pi/15) + \csc^2(8\pi/15) = 32$$ but unfortunately $$\csc^2(\pi/31) + \csc^2(2\pi/31) + \csc^2(4\pi/31) + \csc^2(8\pi/31) + \csc^2(16\pi/31)$$ is irrational (it seems to be a root of $z^3-160 z^2+3904 z-23552 = 0$)
EDIT: Instead we have $$ \sum_{j=1}^{15} \csc^2(j \pi/31) = 160$$
Actually it seems $$ \sum_{j=1}^n \csc^2(j \pi/(2n+1)) = \dfrac{2n(n+1)}{3}$$ for all positive integers $n$. In the case $n=3$, since $\csc(4\pi/7) = \csc(3\pi/7)$, $$\csc^2(\pi/7) + \csc^2(2\pi/7) + \csc^2(4\pi/7) = \csc^2(\pi/7) + \csc^2(2\pi/7) + \csc^2(3\pi/7) = 8$$ In the case $n=7$, there are actually four "basic" equations involving $\csc^2(j \pi/15)$: $$\eqalign{\csc^2(5\pi/15) &= 4/3\cr \csc^2(3\pi/15) + \csc^2(6\pi/15) &= 4\cr 10 \csc^2(\pi/15) + \csc^2(2\pi/15)+\csc^2(7\pi/15) &= 36\cr \csc^2(\pi/15) - 10 \csc^2(3 \pi/15) + \csc^2(4\pi/15) &= -4\cr}$$ and $\csc^2(\pi/15) + \csc^2(2\pi/15) + \csc^4(4\pi/15) + \csc^2(8\pi/15) = 32$ is the sum of the last two (using the fact that $\csc(8 \pi/15) = \csc(7 \pi/15)$).
If $7x=\pi,4x=\pi-3x$
$\implies \sin4x=\sin(\pi-3x)=\sin3x$
$\implies 2\sin2x\cos2x=3\sin x-4\sin^3x$
$\implies 4\sin x\cos x\cos2x=3\sin x-4\sin^3x$
If $\sin x\ne0,$ we have $4\cos x\cos2x=3-4\sin^2x\implies 4\cos x(1-2\sin^2x)=3-4\sin^2x$
On squaring & rearrangement, $64(\sin^2x)^3-112(\sin^2x)^2+56\sin^2x-7=0$
which is a cubic equation in $\sin^2x$ with roots being $\sin^2\frac{r\pi}7$
where $r=(1$ or $6),(2$ or $5)$ and $(3$ or $4)$ as $\sin\frac{(7-r)\pi}7=\sin (\pi-\frac{r\pi}7)=\sin\frac{r\pi}7$
Using Vieta's Formula we have,
$\displaystyle\sin^2\frac{\pi}7\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7=\frac7{64}$ and $\displaystyle\sin^2\frac{\pi}7\sin^2\frac{2\pi}7+\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7+\sin^2\frac{4\pi}7\sin^2\frac{\pi}7=\frac{56}{64}$
We need to find
$\displaystyle\frac1{\sin^2\frac{\pi}7}+\frac1{\sin^2\frac{2\pi}7}+\frac1{\sin^2\frac{4\pi}7}=\displaystyle\frac{\sin^2\frac{\pi}7\sin^2\frac{2\pi}7+\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7+\sin^2\frac{4\pi}7\sin^2\frac{\pi}7}{\sin^2\frac{\pi}7\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7}=\frac{\frac{56}{64}}{\frac7{64}}=\frac{56}7=8$
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$ $$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2},$$ $$\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=$$ $$=\frac{1}{2}\left(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\right)=$$ $$=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$ and $$\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{1}{8}.$$ Id est, $$\frac{1}{\sin^2\frac{\pi}{7}}+\frac{1}{\sin^2\frac{2\pi}{7}}+\frac{1}{\sin^2\frac{4\pi}{7}}=$$ $$=2\left(\frac{1}{1-\cos\frac{2\pi}{7}}+\frac{1}{1-\cos\frac{4\pi}{7}}+\frac{1}{1-\cos\frac{6\pi}{7}}\right)=$$ $$=\frac{2\left(\left(1-\cos\frac{2\pi}{7}\right)\left(1-\cos\frac{4\pi}{7}\right)+\left(1-\cos\frac{2\pi}{7}\right)\left(1-\cos\frac{6\pi}{7}\right)+\left(1-\cos\frac{4\pi}{7}\right)\left(1-\cos\frac{6\pi}{7}\right)\right)}{\left(1-\cos\frac{2\pi}{7}\right)\left(1-\cos\frac{4\pi}{7}\right)\left(1-\cos\frac{6\pi}{7}\right)}=$$ $$=\frac{2\left(3-2\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)\right)}{1-\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)-\frac{1}{8}}=8.$$
From this and this, $\sin7x=7t-56t^3+112t^5-64t^7$ where $t=\sin x$
Now, if $\sin7x=0, 7x=n\pi, x=\frac{n\pi}7$ where $n=0,1,2,3,4,5,6$
Clearly, $\sin\frac{r\pi}7$ are the roots of $7-56t^2+112t^4-64t^6=0$ where $r=1,2,3,4,5,6$
As $\sin\frac{(7-r)\pi}7=\sin (\pi-\frac{r\pi}7)=\sin\frac{r\pi}7,$
$\sin^2\frac{r\pi}7$ are the roots of $7-56s+112s^2-64s^3=0$ where $r=1,2,4$
Putting $y=\frac1s,$ $\displaystyle7-\frac{56}y+\frac{112}{y^2}-\frac{64}{y^3}=0$
$\displaystyle\implies 7y^3-56y^2+112y-64=0$
Now, using Vieta's Formula, $\displaystyle \frac1{\sin^2\frac{\pi}7}+\frac1{\sin^2\frac{2\pi}7}+\frac1{\sin^2\frac{4\pi}7}=\frac{56}7=8$
Let $7x=\pi\implies 4x=\pi-3x$
$$\frac1{\sin2x}+\frac1{\sin4x}=\frac{\sin4x+\sin2x}{\sin4x\sin2x}$$
$$=\frac{2\sin3x\cos x}{\sin(\pi-3x)\sin2x}(\text{ using } \sin2A+\sin2B=2\sin(A+B)\cos(A-B))$$
$$=\frac{2\sin3x\cos x}{\sin(3x)2\sin x\cos x}(\text{ using } \sin2C=2\sin C\cos C \text{ and }\sin(\pi-y)=\sin y)$$ $$=\frac1{\sin x}$$
$$\implies \frac1{\sin x}-\frac1{\sin2x}-\frac1{\sin4x}=0$$
Squaring we get,
$$\frac1{\sin^2x}+\frac1{\sin^22x}+\frac1{\sin^24x}=2\left(\frac1{\sin x\sin4x}+\frac1{\sin x\sin2x}-\frac1{\sin2x\sin4x}\right)=2\frac{(\sin2x+\sin4x-\sin x)}{\sin x\sin2x\sin4x}$$
Now, $$\sin2x+\sin4x-\sin x=2\sin3x\cos x-\sin(\pi-6x)\text{ as }x=\pi-6x$$
$$\implies \sin2x+\sin4x-\sin x=2\sin3x\cos x-\sin6x$$
$$=2\sin3x\cos x-2\sin3x\cos3x=2\sin3x(\cos x-\cos3x)=2\sin3x(2\sin2x\sin x)$$ using $\cos2C-\cos2D=\sin(D-C)\sin(C+D)$
$$\implies \frac{\sin2x+\sin4x-\sin x}{\sin x\sin2x\sin4x}=4$$ as $\sin x\sin2x\sin4x\ne0$ if $7x=\pi$
I provide one more solution, where we don't use sines and cosines.
First, some preparation.
We all know that $$ \tan (x \pm y) = \frac {\tan (x) \pm \tan (y)} {1 \mp \tan (x)\tan (y)}. $$ Then $$ \cot (x - y) = \frac {\cot (x) \cot (y) + 1} { \cot (y) - \cot (x)}, $$ or $$ \cot (x) \cot (y) = \cot(x-y) (\cot(y) -\cot(x)) - 1 $$ Especially, $$ \cot (2x) = \frac {\cot^2 (x) -1} {2 \cot (x)}, $$ which yields $$ \cot^2(x) - 1 = 2\cot(x)\cot(2x). $$ Also, $$ \csc^2(x) = \frac {\sin^2(x) + \cos^2(x)} {\sin ^2(x)} = 1 + \cot^2(x). $$ Now let's start. Let $x = \pi /7$. $$ P := \csc^2(x) + \csc^2(2x) + \csc^2(4x) = 3 + \cot^2(x) + \cot^2(2x) + \cot^2(3x). $$ By the formula above, $$ P = 6 + 2(\cot(x) \cot(2x) + \cot(2x) \cot (4x) + \cot (4x) \cot (8x)) =: 6 +2Q. $$ Now change the angle: since $\cot(\pi \pm y) = \mp\cot (y)$, we have $$ Q = \cot(x)\cot(2x) - \cot(2x) \cot (3x) - \cot(3x) \cot (x). $$ Now, \begin{align*} Q &= 1+ \cot(x)(\cot(x) - \cot(2x)) - \cot(x) (\cot(2x) -\cot(3x)) - \cot(2x)(\cot(x) - \cot (3x))\\ &= 1+ \cot(x) (\cot(x) -3\cot(2x) + \cot(3x)) + \cot(2x) \cot(3x)\\ &= \cot(x) (\cot(x) -3\cot(2x) + \cot(3x)) + (\cot(2x)-\cot(3x))\cot(x) \\ &= \cot(x)(\cot(x) -2\cot(2x))\\ &= \cot^2(x) - 2\cot(2x) \cot(x)\\ &=1. \end{align*} Therefore $P = 6+2Q=8$.
Consider the polynomial $$P(z):=z^6+z^5+z^4+z^3+z^2+z+1\,.$$ Let $t:=z+\dfrac{1}{z}$. Therefore, $P(z)=z^3\,Q(t)$, where $$Q(t):=t^3+t^2-2t-1\,.$$
Let $\omega:=\exp(\text{i}\theta)$, where $\theta:=\dfrac{2\pi}{7}$. Then, $\omega$, $\omega^2$, $\omega^3$, $\omega^4$, $\omega^5$, and $\omega^6$ are all the roots of $P(z)$. That is, $t_1:=2\cos(\theta)$, $t_2:=2\cos(2\theta)$, and $t_3:=2\cos(3\theta)$ are all the roots of $Q(t)$. Note that, for an arbitrary $\phi$, $$\text{csc}^2(\phi)=\frac{1}{1-\cos^2(\phi)}=\frac{4}{4-\big(2\cos(\phi)\big)^2}\,.$$ That is, $$S:=\text{csc}^2(\theta)+\text{csc}^2(2\theta)+\text{csc}^2(4\theta)=\text{csc}^2(\theta)+\text{csc}^2(2\theta)+\text{csc}^2(3\theta)=\sum_{j=1}^3\,\frac{4}{4-t_j^2}\,.$$
Since $Q(t_j)=0$ for each $j$, we get that $$t_j^3+t_j^2-2t_j-1=0\text{ or }(t_j^2-4)(t_j+1)+2t_j+3=0\,.$$ That is, $$t_j+1=\frac{2t_j+3}{4-t_j^2}=\frac{2(t_j-2)+7}{4-t_j^2}=\frac{7}{4-t_j^2}-\frac{2}{t_j+2}\,.$$ Furthermore, $$0=t_j^3+t_j^2-2t_j-1=(t_j+2)(t_j^2-t_j)-1\text{ or }\frac{1}{t_j+2}=t_j^2-t_j\,.$$ Hence, $$\frac{4}{4-t_j^2}=\frac{4}{7}\,\left(2(t_j^2-t_j)+t_j+1\right)=\frac{4}{7}\,\left(2t_j^2-t_j+1\right)\,.$$
Finally, $$S=\sum_{j=1}^3\,\frac{4}{4-t_j^2}=\frac{4}{7}\,\sum_{j=1}^3\,\left(2t_j^2-t_j+1\right)\,.$$ Since $\sum\limits_{j=1}^3\,t_j=-1$ and $\sum\limits_{j=1}^3\,t_j^2=(-1)^2-2(-2)=5$ by Vieta's Formulas, we end up with $$S=\frac{4}{7}\,\big(2\cdot 5-(-1)+3\big)=8\,.$$
$$p(x)=\frac{(x+i)^7-(x-i)^7}{2i}=7x^6-35x^4+21x^2-1$$ has the roots $\cot(\tfrac{k\pi}7)$, $1\leq k\leq6.$ For this polynomial, $e_1=0$, $e_2=\frac{-35}7=-5$ and by Newton-Girard identity $p_2=e_1^2-2e_2=10.$ Hence the wanted sum is $S=\frac{6+p_2}2=8.$
