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Show that that (i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \: x \neq (n - \frac{1}{2})\pi, \: n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 5
Question 1
Show that 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 - \... show full transcript
Worked Solution & Example Answer:Show that that (i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x, \: x \neq (n - \frac{1}{2})\pi, \: n \in \mathbb{Z} - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 5
Step 1
(i) \( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x \)
Answer
To show that ( \frac{\cos 2x}{\cos x + \sin x} = \cos x - \sin x ), we start by using the double angle formula for cosine:
[ \cos 2x = \cos^2 x - \sin^2 x ]
Now substituting this into the left-hand side, we have:
[ \frac{\cos^2 x - \sin^2 x}{\cos x + \sin x} ]
We can factor the numerator as follows:
[ \cos^2 x - \sin^2 x = (\cos x + \sin x)(\cos x - \sin x) ]
So, we rewrite it as:
[ \frac{(\cos x + \sin x)(\cos x - \sin x)}{\cos x + \sin x} ]
Provided that ( \cos x + \sin x \neq 0 ), we simplify to:
[ \cos x - \sin x ]
Step 2
(ii) \( \frac{1}{2}(\cos 2x - \sin 2x) = \cos^2 x - \cos x \sin x - \frac{1}{2} \)
Answer
Starting with the left-hand side, we can again use the double angle identities:
[ \frac{1}{2}(\cos 2x - \sin 2x) = \frac{1}{2}(\cos^2 x - \sin^2 x - 2\sin x \cos x) ]
This simplifies to:
[ \frac{1}{2}(\cos^2 x - \sin^2 x - 2\sin x \cos x) = \frac{1}{2}\left(\cos^2 x + \sin^2 x - 1 - 2\sin x \cos x \right) ]
Since ( \cos^2 x + \sin^2 x = 1 ), we substitute to get:
[ \frac{1}{2}(1 - 1 - 2\sin x \cos x) = -\sin x \cos x ]
Rearranging gives us:
[ \cos^2 x - \cos x \sin x - \frac{1}{2} = 0 ]
Step 3
b) \( \cos \left( \frac{\cos 2\theta}{\cos \theta + \sin \theta} \right) = \frac{1}{2} \)
Answer
To express the equation in terms of ( \sin 2\theta ) and ( \cos 2\theta ), we first use the known identity for cosines:
If ( \cos A = \frac{1}{2} ), then ( A = \frac{\pi}{3} + 2k\pi ) or ( A = -\frac{\pi}{3} + 2k\pi ). Thus,
This means we can equate:
[ \frac{\cos 2\theta}{\cos \theta + \sin \theta} = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad \frac{\cos 2\theta}{\cos \theta + \sin \theta} = -\frac{\pi}{3} + 2k\pi ]
From these, we can derive:\n[ \sin 2\theta = \cos 2\theta ]
Step 4
c) Solve, for \( 0 < \theta < 2\pi \), \( \sin 2\theta = \cos 2\theta \)
Answer
From ( \sin 2\theta = \cos 2\theta ), we know that:
[ \tan 2\theta = 1 ]
This leads to:
[ 2\theta = \frac{\pi}{4} + n\pi ]
From here, we can solve for ( \theta ):
[ \theta = \frac{\pi}{8} + \frac{n\pi}{2} ]
Thus, for ( n = 0 ) and ( n = 1 ), we find:
[ \theta = \frac{\pi}{8} \quad \text{and} \quad \theta = \frac{5\pi}{8} ]
Hence, the solutions in the range are:
[ \theta = \frac{\pi}{8}, \frac{5\pi}{8} ]