Show that: (i) \(\frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \ x \neq (n - \frac{1}{2})\pi, \ n \in \mathbb{Z}.\) (ii) \(\frac{1}{2} (\cos 2x - \sin 2x) = \cos^2 x - \cos x \sin x - \frac{1}{2}.\) (b) Hence, or otherwise, show that the equation \(\cos \left( \frac{\cos 2\theta}{\cos \theta + \sin \theta} \right) = \frac{1}{2}\) can be written as \(\sin 2\theta = \cos 2\theta.\) (c) Solve, for \(0 < \theta < 2\pi\), \(\sin 2\theta = \cos 2\theta\) giving your answers in terms of \(\pi\). - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 5

Starting with the left-hand side:

$\frac{1}{2} (\cos 2x - \sin 2x) = \frac{1}{2} (\cos^2 x - \sin^2 x - 2\sin x \cos x).$
Now we can rewrite it, focusing on identities:

$= \frac{1}{2} ((\cos^2 x - \sin^2 x) - 2\sin x \cos x).$
This indicates:

$= \cos^2 x - \sin^2 x - \sin 2x.$
So consider:

$\cos^2 x - \sin x \to \cos^2 x - (\sin^2 x + \cos x \sin x) - \frac{1}{2}.$

Using the result from part (ii), we can view it as:

If (\sin 2\theta = \cos 2\theta), we use the identity that:

Here, we know:

$\frac{\cos 2\theta}{\cos \theta + \sin \theta} = \frac{1}{2}$ gives rise to a derived condition of (\sin^2 2\theta + \cos^2 2\theta = 1.)
From rearranging, (\cos 2\theta = \frac{1}{2}) yields the desired form.

Given that (\sin 2\theta = \cos 2\theta), we rewrite it as:

$\tan 2\theta = 1.$
This means:

$2\theta = \frac{\pi}{4} + n\pi, \ n \in \mathbb{Z}.$
Thus solving for (\theta):

1. For (n = 0):

$\theta = \frac{\pi}{8}.$

2. For (n = 1):

$\theta = \frac{5\pi}{8}.$

3. For (n = 2):

$\theta = \frac{9\pi}{8}.$
4. For (n = 3):

$\theta = \frac{13\pi}{8}.$

This gives all solutions satisfying the condition ((0 < \theta < 2\pi)).