Problem Statement
Let $\omega = e^{i \frac{2\pi}{5}}$ be a primitive fifth root of unity.
**(a)** [2 marks] Show that $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$.
**(b)** [4 marks] Deduce that $\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2}$.
**(c)** [3 marks] Hence, show that $\cos\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5} - 1}{4}$.
Verified Solution & Marking Scheme
Part (a) [M1]
Sum of a geometric progression: $1 + \omega + \omega^2 + \omega^3 + \omega^4 = \frac{1 - \omega^5}{1 - \omega}$
Part (a) [A1]
Since $\omega^5 = (e^{i 2\pi/5})^5 = e^{i 2\pi} = 1$, the numerator is $1 - 1 = 0$. Since $\omega \neq 1$, sum is $0$.
Part (b) [M1]
Take real parts: $\text{Re}(1 + \omega + \omega^2 + \omega^3 + \omega^4) = 1 + \cos(2\pi/5) + \cos(4\pi/5) + \cos(6\pi/5) + \cos(8\pi/5) = 0$
Part (b) [M1]
Note that $\cos(8\pi/5) = \cos(2\pi - 2\pi/5) = \cos(2\pi/5)$ and $\cos(6\pi/5) = \cos(2\pi - 4\pi/5) = \cos(4\pi/5)$
Part (b) [A1]
$1 + 2\cos(2\pi/5) + 2\cos(4\pi/5) = 0$
Part (b) [A1]
$\cos(2\pi/5) + \cos(4\pi/5) = -\frac{1}{2}$ (AG)
Part (c) [M1]
Use double angle identity: $\cos(4\pi/5) = 2\cos^2(2\pi/5) - 1$. Let $u = \cos(2\pi/5)$.
Part (c) [M1]
$u + 2u^2 - 1 = -\frac{1}{2} \implies 4u^2 + 2u - 1 = 0$
Part (c) [A1]
Using quadratic formula: $u = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4}$. Since $0 < \frac{2\pi}{5} < \frac{\pi}{2}$, $\cos(2\pi/5) > 0$. Thus $\cos(2\pi/5) = \frac{\sqrt{5} - 1}{4}$.