Let $\alpha = \cos\theta + i \sin\theta$, where $0 < \theta < 2\pi$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2017 - Paper 1

The series can be expressed as follows:

$C = 1 + \alpha + \alpha^2 + \cdots + \alpha^n = \frac{1 - \alpha^{n+1}}{1 - \alpha}.$

From previous steps, we know that:

$1 + 2(\cos \theta + \cos 2\theta + \cdots + \cos n\theta) = \frac{(1 - \alpha^{n+1})}{(1 - \alpha)(1 - \alpha^n)}.$

Using the result derived in part (ii), we can rewrite:

$1 + 2(\cos\theta + \cos 2\theta + \cdots + \cos n\theta) = C = \frac{\alpha^{n+1} - (\alpha^{n+1} - \alpha)}{(1 - \alpha)(1 - \alpha^n)}.$

By substituting for $\alpha$ and applying results from part (i), we can simplify this to find the required identity.

To demonstrate that this sum is independent of $n$, we note that as $n$ increases, the angles $\frac{k\pi}{n}$ uniformly distribute over the interval. Thus, the relationship from parts (i) and (ii) shows that this sum converges to a constant value as $n$ varies. Employing trigonometric identities and the symmetry in cosine values yields:

$\cos \frac{k \pi}{n}$ which confirms independence from $n$.

The eccentricity $e$ of a hyperbola is given by:

$e = \frac{c}{a}$ where $c = \sqrt{a^2 + b^2}$. Given that $e = 2$, we have $c = 2a$. Knowing that the distance from one of the foci to one of the vertices is given as 1, we can set up the equation:

$c - a = 1.$ Substituting $c = 2a$ leads to:

$2a - a = 1,$ which simplifies to:

a = 1. Thus, $c = 2$.