Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach : 
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(8\pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(\pi + \pi/7) $
$= 2\sin(3\pi)/7\times\cos(\pi/7)-\sin(\pi/7) $
I am not getting any clue how to proceed further or whether it is correct or not. Please help thanks..
 A: The idea in this answer is: if you can introduce a trigonometric sum with arguments in arithmetic progression, then you have a nice formula that does all the job (see here for the statement and two proofs).
Let
$$S=\sin \frac{2\pi}{7}+\sin \frac{4\pi}{7}+\sin \frac{8\pi}{7}$$
Then
$$S^2=\sin^2 \frac{2\pi}{7}+\sin^2 \frac{4\pi}{7}+\sin^2 \frac{8\pi}{7}\\
+2\sin \frac{2\pi}{7}\sin \frac{4\pi}{7}+2\sin \frac{2\pi}{7}\sin \frac{8\pi}{7}+2\sin \frac{4\pi}{7}\sin \frac{8\pi}{7}$$
$$S^2=\frac{1-\cos \frac{4\pi}{7}}{2}+\frac{1-\cos \frac{8\pi}{7}}{2}+\frac{1-\cos \frac{16\pi}{7}}{2}+\cos \frac{2\pi}{7}-\cos \frac{6\pi}{7}\\
+\cos \frac{6\pi}{7}-\cos \frac{10\pi}{7}+\cos \frac{4\pi}{7}-\cos \frac{12\pi}{7}$$
$$S^2=\frac32+\cos \frac{2\pi}{7}+\frac12\cos \frac{4\pi}{7}-\frac12\cos \frac{8\pi}{7}-\cos \frac{10\pi}{7}-\cos \frac{12\pi}{7}-\frac12\cos \frac{16\pi}{7}$$
$$S^2=\frac32+\cos \frac{2\pi}{7}-\frac12\cos \frac{3\pi}{7}+\frac12\cos \frac{\pi}{7}+\cos \frac{3\pi}{7}-\cos \frac{2\pi}{7}-\frac12\cos \frac{2\pi}{7}$$
$$S^2=\frac32+\frac12\cos \frac{\pi}{7}-\frac12\cos \frac{2\pi}{7}+\frac12\cos \frac{3\pi}{7}$$
$$S^2=\frac32-\frac12\left(\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}\right)$$
And using the following formula from this answer, with $n=3, a=b=\frac{2\pi}{7}$:
$$\sum_{k=0}^{n-1} \cos (a+kb) = \frac{\sin \frac{nb}{2}}{\sin \frac{b}{2}} \cos \left( a+ (n-1)\frac{b}{2}\right)$$
$$\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}=\frac{\sin \frac{3\pi}{7}\cos \frac{4\pi}{7}}{\sin \frac{\pi}{7}}=\frac{\sin\frac{7\pi}{7}-\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac12$$
Hence
$$S^2=\frac32+\frac14=\frac74$$
And we know that $S>0$ because the only negative term is $\sin\frac{8\pi}{7}=-\sin\frac{\pi}{7}$, and $0<\sin\frac{\pi}{7}<\sin\frac{2\pi}{7}$. So we have finally
$$\sin \frac{2\pi}{7}+\sin \frac{4\pi}{7}+\sin \frac{8\pi}{7}=\frac{\sqrt7}{2}$$
A: Here's my approach, it only use basic trigonometry identities (i.e, it requires no complex numbers), so it's quite lengthy. I would be glad if you guys can help me shorten it a bit.


*

*Use Product to Sum Formula to prove that:


$\displaystyle \begin{align*}& \ \cos\left(\frac{2\pi}{7} \right) \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{2\pi}{7} \right)\cos\left(\frac{8\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right)\cos\left(\frac{8\pi}{7} \right)\\
= & \ \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{6\pi}{7} \right) \\
= & \ \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{8\pi}{7} \right)\end{align*}$


*

*Let $\displaystyle A = \cos\left(\frac{2\pi}{7} \right) + \cos\left(\frac{4\pi}{7} \right) + \cos\left(\frac{8\pi}{7} \right)$. Prove that $A < 0$, then square $A$ and using (1) along with Power Reduction Formula to prove that $A$ is a solution to the equation: $\displaystyle X^2 - \frac{5}{2}X - \frac{3}{2} = 0$. Hence $\displaystyle A = -\frac{1}{2}$.

*Again use the Product to Sum Formula to prove that $\displaystyle \sin\left(\frac{2\pi}{7} \right) \sin\left(\frac{4\pi}{7} \right) + \sin\left(\frac{2\pi}{7} \right)\sin\left(\frac{8\pi}{7} \right) + \sin\left(\frac{4\pi}{7} \right)\sin\left(\frac{8\pi}{7} \right) = 0$.

*Let $\displaystyle B = \sin\left(\frac{2\pi}{7} \right) + \sin\left(\frac{4\pi}{7} \right) + \sin\left(\frac{8\pi}{7} \right)$. Prove that $B > 0$. Using the result in (3), we have $\displaystyle B^2 = \sin^2\left(\frac{2\pi}{7} \right) + \sin^2\left(\frac{4\pi}{7} \right) + \sin^2\left(\frac{8\pi}{7} \right)$, using Power Reduction Formula, and (2), you'll get $\displaystyle B^2 = \frac{3}{2} - \frac{1}{2} A = \frac{7}{4}$. From the fact that $B > 0$, we'll have $\displaystyle B = \frac{\sqrt{7}}{2}$.
The above are just kind of hints, and you'll have to apply some trig formulae (identities) to get the result.
A: Setting $7\theta=\pi,S=\sin2\theta+\sin4\theta+\sin8\theta=(\sin2\theta+\sin8\theta)+\sin4\theta$
Using Prosthaphaeresis Formula & Double angle formula,
$S=2\sin5\theta\cos3\theta+2\sin2\theta\cos2\theta$
As $5\theta=\pi-2\theta,\sin5\theta=\sin2\theta$ 
and $3\theta=\pi-4\theta,\cos3\theta=-\cos4\theta$
$S=-2\sin2\theta\cos4\theta+2\sin2\theta\cos2\theta=2\sin2\theta(\cos2\theta-\cos4\theta)$
Again using Prosthaphaeresis Formula,  $S=2\sin2\theta(2\sin3\theta\sin\theta)$ which is clearly $>0$
But, $\sin2\theta=\sin(\pi-2\theta)=\sin5\theta;\sin\theta=\sin6\theta;\sin3\theta=\sin4\theta$
$\implies \dfrac S4=+\sqrt{\prod_{r=1}^6\sin r\theta}\ \ \ \ (1)$
We can derive (See below) $\sin7x=7\sin x+\cdots-64\sin^7x$
If $\sin7x=0,7x=n\pi$ where $n$ is any integer, $x=\dfrac{n\pi}7,0\le n\le6$
So, $\sin\dfrac{n\pi}7,0\le n\le6$ are the roots of $7\sin x+\cdots-64\sin^7x=0$
So, $\sin\dfrac{n\pi}7,1\le n\le6$ are the roots of $7+\cdots-64\sin^6x=0\iff64\sin^6x+\cdots-7=0$
Using Vieta's formula, $\prod_{r=1}^6\sin r\theta=\dfrac7{64}$
Apply this in $(1)$

Derivation $\#1:$
Using de Moivre's Theorem, $\cos7y+i\sin7y=(\cos y+i\sin y)^7$
$=\cdots +i\left(\sin^7y -\binom72\sin^5y\cos^2y+\binom74\sin^3y\cos^4y-\binom76\sin y\cos^6y\right)$
Writing $\cos^2y=1-\sin^2y$
$\cos7y+i\sin7y=\cdots +i\left(7\sin y-64\sin^7y\right)$
Derivation $\#2:$
Using Prosthaphaeresis Formula, $\sin7x+\sin x=2\sin4x\cos3x=2(2\sin2x\cos2x)(4\cos^3x-3\cos x)=4(2\sin x\cos x)(1-2\sin^2x)(4\cos^3x-3\cos x)$
$=8(\sin x-2\sin^3x)\cos^2x(4\cos^2x-1)$
$=8(\sin x-2\sin^3x)(1-\sin^2x)\{4(1-\sin^2x)-1\}$
$=8\sin x+\cdots-64\sin^7x$
$\implies\sin7x=7\sin x+\cdots-64\sin^7x$
A: Let $\Phi_7(x)$ the minimal polynomial of $\xi=\exp\left(\frac{2\pi i}{7}\right)$. Its Galois group over $\mathbb{Q}$ is cyclic and generated by $\xi\mapsto\xi^3$, since $3$ is a generator of $\mathbb{Z}/(7\mathbb{Z})^*$. The quadratic residues $\!\!\pmod{7}$ are $1,2,4$ and the quadratic non-residues are $3,5,6$, hence:
$$ \alpha = \xi+\xi^2+\xi^4,\qquad \beta=\xi^3+\xi^5+\xi^6 $$
are two conjugated algebraic numbers with degree $\frac{6}{3}=2$. In particular, both $\alpha+\beta$ and $\alpha\beta$ are rational numbers:
$$ \alpha+\beta=\sum_{k=1}^{6}\xi^k=-1,\qquad \alpha\beta=3+\sum_{k=1}^{6}\xi^k=2 $$
and the minimal polynomial of $\alpha$ and $\beta$ is given by $x^2+x+2$.
It is straightforward to check that $\text{Im}(\alpha)>0$ and $\text{Im}(\beta)<0$, hence by computing the discriminant of  $x^2+x+2$ we have:
$$ \alpha-\beta = \sqrt{-7} $$
and the claim follows by De Moivre's formula, since, for instance, 
$$ \xi-\xi^6 = \exp\left(\frac{2\pi i}{7}\right)-\exp\left(-\frac{2\pi i}{7}\right)=2i\sin\left(\frac{2\pi}{7}\right).$$
A: You can evaluate this by using complex methods.
Let $\alpha=e^{2\pi i/7}$.  Then
$$\sin\frac{2\pi}{7}=\frac{\alpha-\alpha^{-1}}{2i}\ ,\quad
  \sin\frac{4\pi}{7}=\frac{\alpha^2-\alpha^{-2}}{2i}\ ,\quad
  \sin\frac{8\pi}{7}=\frac{\alpha^4-\alpha^{-4}}{2i}\ .$$
Now let $S$ be the sum of these three numbers.  Write $S$ in terms of powers of $\alpha$ and calculate $S^2$.  Initially it's a mess, but you can use the relations
$$\alpha^7=1\quad\hbox{and}\quad \alpha^6+\alpha^5+\cdots+\alpha=-1$$
to show that it simplifies, amazingly, to
$$S^2=\frac{7}{4}\ .$$
It's not hard to see that $S$ is positive, so
$$S=\frac{\sqrt7}{2}\ .$$
Comment.  You can also write the sum as
$$S=\frac{1}{2i}\sum_{k=1}^6 \Bigl(\frac{k}{7}\Bigr)\alpha^k\ ,$$
where $(\frac{k}{7})$ is a Legendre symbol, and this connects the sum with some very interesting and often difficult mathematics.
