$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$ let $$f(x)=\frac{1}{1+2\cos x}$$
prove that :
$$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$$
My Try :
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{4\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{6\pi}{7})}$$
$$L=\frac{1}{1+2\cos (\frac{6\pi}{7})}+\frac{1}{1+2\cos (\frac{4\pi}{7})}+\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
what now ?
 A: Use factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$ 
or  if $7x=2m\pi$ where  $m$ is any integer
$\sin4x=\sin(2m\pi-3x)=-\sin3x$
$\implies4\sin x\cos x\cos2x=\sin x(4\sin^2x-3)$
$\implies4\sin x\cos x(2\cos^2x-1)=\sin x\{4(1-\cos^2x)-1\}$
So, the roots of $4\cos x(2\cos^2x-1)=4(1-\cos^2x)-1\iff8\cos^3x+4\cos^2x-4\cos x-3=0\ \ \ \ (1)$
are $7x=2m\pi$ where  $m\equiv\pm1,\pm2,\pm3\pmod7$
Now if $u=\dfrac1{1+2\cos x}\iff \cos x=?$
Replace the values of $\cos x$ in $(1)$
A: Let $2\cos\frac{2\pi}{7}=a$, $2\cos\frac{4\pi}{7}=b$ and $2\cos\frac{6\pi}{7}=c$.
Hence,
$$a+b+c=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{\sin\frac{\pi}{7}}=-1;$$
$$ab+ac+bc=4\left(\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}\right)=$$
$$=2\left(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\right)=-2$$ and
$$abc=8\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{\sin\frac{2\pi}{7}}=1.$$
Id est, $$f\left(\frac{2\pi}{7}\right)+f\left(\frac{4\pi}{7}\right)+f\left(\frac{6\pi}{7}\right)=\sum_{cyc}\frac{1}{1+a}=\frac{\sum\limits_{cyc}(a+1)(b+1)}{\prod\limits_{cyc}(1+a)}=$$
$$=\frac{ab+ac+bc+2(a+b+c)+3}{1+a+b+c+ab+ac+bc+abc}=\frac{-2-2+3}{1-1-2+1}=1.$$
Done!
A: The minimal polynomial of $2\cos\frac{2\pi}{7}$ over $\mathbb{Q}$ is given by $x^3+x^2-2x-1$, and the conjugated roots are $2\cos\frac{4\pi}{7}$ and $2\cos\frac{6\pi}{7}$. It follows that $1+2\cos\frac{2\pi}{7}$, $1+2\cos\frac{4\pi}{7}$ and $1+2\cos\frac{6\pi}{7}$ are roots of
$$ p(x)= x^3-2x^2-x+1 $$
and by Vieta's theorem
$$\sum_{k=1}^{3}\frac{1}{1+2\cos\frac{2\pi k}{7}} = -\frac{[x^1]\,p(x)}{[x^0]\,p(x)}=\color{red}{1} $$
as wanted.
A: $f(2x)=\dfrac1{1+2\cos2x}=\dfrac1{1+2(1-2\sin^2x)}=\dfrac{\sin x}{\sin3x}$ for $\sin x\ne0$
$x=\dfrac\pi7\implies f\left(\dfrac{2\pi}7\right)=\dfrac{\sin\dfrac{\pi}7}{\sin\dfrac{3\pi}7}=2\cos\dfrac{3\pi}7$  as $\sin\dfrac{\pi}7=\sin\left(\pi-
\dfrac\pi7\right)=2\sin\dfrac{3\pi}7\cos\dfrac{3\pi}7$
$x=\dfrac{2\pi}7\implies f\left(\dfrac{4\pi}7\right)=\dfrac{\sin\dfrac{2\pi}7}{\sin\dfrac{6\pi}7}=2\cos\dfrac{\pi}7$ as $\sin\dfrac{6\pi}7=\sin\left(\pi-\dfrac{6\pi}7\right)$
$x=\dfrac{3\pi}7\implies f\left(\dfrac{6\pi}7\right)=\dfrac{\sin\dfrac{3\pi}7}{\sin\dfrac{9\pi}7}=\dfrac{-\sin\left(\pi+\dfrac{3\pi}7\right)}{-\sin\left(2\pi-\dfrac{5\pi}7\right)}=2\cos\dfrac{5\pi}7$
So, we need $S=2\cos\dfrac{\pi}7+2\cos\dfrac{3\pi}7+2\cos\dfrac{5\pi}7$
Like $\sum \cos$ when angles are in arithmetic progression,
$\implies\sin\dfrac\pi7\cdot S=2\cos\dfrac{\pi}7\cdot\sin\dfrac\pi7+2\cos\dfrac{3\pi}7\cdot\sin\dfrac\pi7+2\cos\dfrac{5\pi}7\cdot\sin\dfrac\pi7$
Using Werner Formulas,
$\sin\dfrac\pi7\cdot S=\sin\dfrac{6\pi}7=\cdots=\sin\dfrac\pi7$
