Show that: (i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \quad x \neq (n - \frac{1}{4})\pi, \; n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 5

For the second part, we begin again with the identities:

[ \cos 2x = \cos^2 x - \sin^2 x \quad \text{and} \quad \sin 2x = 2 \sin x \cos x ]

Substituting into the left-hand side gives:

[ \frac{1}{2} \left( \cos^2 x - \sin^2 x - 2 \sin x \cos x \right) ]

This can be further simplified to:

[ = \frac{1}{2} \left( (\cos^2 x - \sin^2 x) - 2 \sin x \cos x \right) = rac{1}{2} \left( \cos^2 x - 2 \sin x \cos x - \sin^2 x \right) ]

Final simplification yields:

[ = \cos^2 x - \sin^2 x - \frac{1}{2} ]

Given the equation, we realize that:

[ \frac{\cos 2\theta}{\cos \theta + \sin \theta} = \frac{1}{2} \implies \sin 2\theta = \cos 2\theta ]

We can represent this as:

[ 2\theta = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} ]

To solve the equation:

[ \tan 2\theta = 1 ]

This implies:

[ 2\theta = \frac{\pi}{4} + n\pi ]

Dividing by 2 gives:

[ \ heta = \frac{\pi}{8} + \frac{n\pi}{2} ]

For the range ( 0 < \theta < 2\pi ), the solutions are obtained by substituting integer values for ( n ):

1. If ( n = 0 ): ( \theta = \frac{\pi}{8} )
2. If ( n = 1 ): ( \theta = \frac{5\pi}{8} )
3. If ( n = 2 ): ( \theta = \frac{9\pi}{8} )
4. If ( n = 3 ): ( \theta = \frac{13\pi}{8} )

Thus, we conclude the valid solutions to be:

[ \theta = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8} ]