1 problem found
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: 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 \(z > 0\)