Show that: i) $ \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \quad x \neq (n - \frac{1}{2})\pi, \quad 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 $.

A-Level Edexcel Maths Pure Trigonometric Proof: Show that: i) \( \frac{\cos 2x

Show that: i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \quad x \neq (n - \frac{1}{2})\pi, \quad n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 5

Question 7

$Show-that:--i)--\(-\frac{\cos-2x}{\cos-x-+-\sin-x}-=-\cos-x---\sin-x,-\quad-x-\neq-(n---\frac{1}{2})\pi,-\quad-n-\in-\mathbb{Z}-Edexcel-A-Level Maths Pure-Question 7-2006-Paper 5.png$

Show that: i) $ \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \quad x \neq (n - \frac{1}{2})\pi, \quad n \in \mathbb{Z}. $ ii) \( \frac{1}{2} (\cos 2x -... show full transcript

Worked Solution & Example Answer:Show that: i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \quad x \neq (n - \frac{1}{2})\pi, \quad n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 5

Step 1

i) $ \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x $

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Answer

To prove the identity, we start with the left-hand side:

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

Now, use the identity (\cos^2 x - \sin^2 x = (\cos x + \sin x)(\cos x - \sin x)):

[ \frac{(\cos x + \sin x)(\cos x - \sin x)}{\cos x + \sin x} = \cos x - \sin x. ]

Thus, we have shown that ( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x ).

Step 2

ii) $ \frac{1}{2} (\cos 2x - \sin 2x) = \cos^2 x - \cos x \sin x - \frac{1}{2} $

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Answer

Starting from 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) = \frac{1}{2}((\cos^2 x - \sin^2 x) - 2\sin x \cos x) ]

Now apply the identities for (\cos 2x) and (\sin 2x) to simplify further:

[ \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x). ]

Thus, we can equate to find the right-hand side:

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

Step 3

b) $ \cos \left( \frac{\cos 2\theta}{\cos \theta + \sin \theta} \right) = \frac{1}{2} $

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Answer

From the derived result in part (ii), we substitute into the equation:

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

Step 4

c) Solve, for $ 0 < \theta < 2\pi, \sin 2\theta = \cos 2\theta $

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Answer

To solve (\sin 2\theta = \cos 2\theta), we can divide both sides by ( \cos 2\theta ):

[ \tan 2\theta = 1. ]

The solutions for (2\theta) in the range (0 < 2\theta < 4\pi) are:

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

Thus, solving for (\theta) gives:

[ \theta = \frac{\pi}{8} + \frac{n\pi}{2}\text{, for } \theta = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}. ]

This falls within the specified range.

Show that: i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \quad x \neq (n - \frac{1}{2})\pi, \quad n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 5

Worked Solution & Example Answer:Show that: i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \quad x \neq (n - \frac{1}{2})\pi, \quad n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 5

i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x \)

ii) \( \frac{1}{2} (\cos 2x - \sin 2x) = \cos^2 x - \cos x \sin x - \frac{1}{2} \)

b) \( \cos \left( \frac{\cos 2\theta}{\cos \theta + \sin \theta} \right) = \frac{1}{2} \)

c) Solve, for \( 0 < \theta < 2\pi, \sin 2\theta = \cos 2\theta \)

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