Problem source

If $u$ and $v$ are the two roots of $z^{2}+az+b=0,$ show that $a=-u-v$
and $b=uv.$

Let $\alpha=\cos(2\pi/7)+\mathrm{i}\sin(2\pi/7).$ Show
that $\alpha$ is a root of $z^{6}-1=0$ and express the roots in
terms of $\alpha.$ The number $\alpha+\alpha^{2}+\alpha^{4}$ is
a root of a quadratic equation 
\[
z^{2}+Az+B=0
\]
where $A$ and $B$ are real. By guessing the other root, or otherwise,
find the numerical values of $A$ and $B$. 

Show that 
\[
\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=-\frac{1}{2},
\]
and evaluate 
\[
\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7},
\]
making it clear how you determine the sign of your answer.

Solution source

\begin{align*}
0 &= z^2+az+b  \\
&= (z-u)(z-v) \\
&= z^2-(u+v)z+uv
\end{align*}

Therefore by comparing coefficients, $a = -u-v$ and $b = uv$.

Suppose $\alpha = \cos(2\pi/7) + i \sin (2\pi/7)$, then by De Moivre, $\alpha^7 = \cos(2\pi) + i \sin (2\pi) = 1$, ie $\alpha^7-1 = 0$.

Notice that $(\alpha+\alpha^2 + \alpha^4) + (\alpha^3+\alpha^5+\alpha^6) = -1$ and 
\begin{align*}
P &= (\alpha+\alpha^2 + \alpha^4)(\alpha^3+\alpha^5+\alpha^6)  \\
&= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^{7}+\alpha^{9}+\alpha^{10} \\
&= 3 + \alpha+ \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 \\
&= 2
\end{align*}

Therefore it is a root of $x^2+x+2 = 0 \Rightarrow x = \frac{-1 \pm i\sqrt{7}}{2}$

Therefore $\cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + 
\cos \frac{8\pi}{7} = \textrm{Re}(\alpha + \alpha^2 + \alpha^4) = -\frac12$

And $\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7} = \textrm{Im}(\alpha + \alpha^2 + \alpha^4) = \pm\frac{\sqrt{7}}2$ since it is positive it is $\frac{\sqrt{7}}{2}$