Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$

Question

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?

Answer 1

When in doubt, convert trig 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 $\csc(8 \pi/15) = \csc(7 \pi/15)$).

Answer 2

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, $\displaystyle \frac1{\sin^2\frac{\pi}7}+\frac1{\sin^2\frac{2\pi}7}+\frac1{\sin^2\frac{4\pi}7}=\frac{56}7=8$

Answer 3

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

$\displaystyle\frac1{\sin^2\frac{\pi}7}+\frac1{\sin^2\frac{2\pi}7}+\frac1{\sin^2\frac{4\pi}7}=\displaystyle\frac{\sin^2\frac{\pi}7\sin^2\frac{2\pi}7+\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7+\sin^2\frac{4\pi}7\sin^2\frac{\pi}7}{\sin^2\frac{\pi}7\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7}=\frac{\frac{56}{64}}{\frac7{64}}=\frac{56}7=8$

Answer 4

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 $7x=\pi$