Problem source

Let $y=\cos\phi+\cos2\phi$, where $\phi=\dfrac{2\pi}{5}.$ Verify by direct substitution that $y$ satisfies the quadratic equation $2y^{2}=3y+2$ and deduce that the value of $y$ is $-\frac{1}{2}.$
	Let $\theta=\dfrac{2\pi}{17}.$ Show that 
	\[
	\sum_{k=0}^{16}\cos k\theta=0.
	\]
	If $z=\cos\theta+\cos2\theta+\cos4\theta+\cos8\theta,$ show that the value of $z$ is $-(1-\sqrt{17})/4$.

Solution source

Note that $\cos 4 \phi = \cos \phi, \cos 3 \phi = \cos 2 \phi$

\begin{align*}
&& LHS & = 2y^2 \\
&&&=  2 \left ( \cos \phi + \cos 2 \phi \right)^2 \\
&&&= 2 \cos ^2 \phi + 2 \cos^2 2 \phi + 4 \cos \phi \cos 2 \phi \\
&&&= \cos 2 \phi+1+ \cos4 \phi+1+2 \left ( \cos \phi + \cos 3 \phi \right) \\
&&&= \cos 2 \phi + 2 + \cos \phi + 2 \cos \phi + 2 \cos 2 \phi \\
&&&= 3(\cos \phi + \cos 2 \phi) + 2 \\
&&&= 3 y  + 2 \\
&&&= RHS
\end{align*}

Therefore $y$ satisfies $2y^2 = 3y+2$, which we can solve:

\begin{align*}
&& 0 &= 2y^2-3y-2 \\
&&&= (2y+1)(y-2) \\
\Rightarrow && y &= -\frac12,2
\end{align*}

Since $\cos \phi \neq 1$, $y \neq 2$, therefore $y = -\frac12$.

\begin{align*}
&& \sum_{k=0}^{16} \cos k \theta &=  \sum_{k=0}^{17} \textrm{Re} \left ( e^{ k \theta i} \right ) \\
&&&=  \textrm{Re} \left ( \sum_{k=0}^{16}e^{ k \theta i} \right ) \\
&&&= \textrm{Re} \left ( \frac{1-e^{17 \theta i}}{1-e^{i \theta}} \right ) \\
&&&= 0
\end{align*}

Suppose $z = \cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta$

\begin{align*}
z^2 &= \left (\cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta \right)^2 \\
&= \cos^2 \theta + \cos^2 2 \theta + \cos^2 4 \theta + \cos^2 8 \theta \\
& \quad \quad 2( \cos \theta \cos 2 \theta + \cos \theta \cos 4 \theta + \cos \theta \cos 8 \theta +  \\
& \quad \quad \quad \cos 2 \theta \cos 4 \theta + \cos 2 \theta \cos 8 \theta + \cos 4 \theta \cos 8 \theta) \\
&= \frac12 \left (\cos 2 \theta + 1+ \cos 4 \theta + 1 + \cos 8 \theta + 1 + \cos 16 \theta + 1 \right ) + \\
&\quad \quad ( \cos \theta + \cos 3 \theta + \cos 3 \theta + \cos 5 \theta + \cos 7 \theta + \cos 9 \theta +   \\
& \quad \quad \quad  \cos 2 \theta + \cos 6 \theta + \cos 6 \theta + \cos 10 \theta +\cos 4 \theta + \cos 12 \theta ) \\
&= \frac12 z + 2 + \\
& \quad \quad ( \cos 3 \theta + \cos 6 \theta - \cos 8 \theta - \cos 11 \theta  \\
& \quad \quad \quad - \cos 13 \theta - \cos 14 \theta - \cos 15 \theta - \cos 16 \theta  - 1) \\
&= \frac12 z + 1 - z \\
&= -\frac12 z +1 
\end{align*}

Therefore $z$ satisfies $z^2=-\frac12 z+1 \Rightarrow z = \frac{-\frac12  \pm \sqrt{\frac14+4}}{2} = \frac{-1 \pm \sqrt{17}}{4}$

Therefore $z = \frac{\sqrt{17}-1}{4}$ since $z > 0$