C3 January 2006 Q7

EdexcelOld spec12 marksProof Trigonometry

(a) Show that

(i) \(\dfrac{\cos 2x}{\cos x + \sin x} \equiv \cos x - \sin x, \quad x \neq (n - \tfrac{1}{4})\pi,\ n \in \mathbb{Z}\), (2)

(ii) \(\tfrac{1}{2}(\cos 2x - \sin 2x) \equiv \cos^2 x - \cos x\sin x - \tfrac{1}{2}\). (3)

(b) Hence, or otherwise, show that the equation\[\cos\theta\left(\frac{\cos 2\theta}{\cos\theta + \sin\theta}\right) = \frac{1}{2}\]can be written as\[\sin 2\theta = \cos 2\theta.\] (3)

(c) Solve, for \(0 \leqslant \theta \lt 2\pi\),\[\sin 2\theta = \cos 2\theta,\]giving your answers in terms of \(\pi\). (4)

Mark scheme (a)

Scheme	Marks
(i) Use of \(\cos 2x = \cos^2 x - \sin^2 x\) in an attempt to prove the identity.	M1
\(\dfrac{\cos 2x}{\cos x + \sin x} = \dfrac{\cos^2 x - \sin^2 x}{\cos x + \sin x} = \dfrac{(\cos x - \sin x)(\cos x + \sin x)}{\cos x + \sin x} = \cos x - \sin x\) * cso	A1
	(2)
(ii) Use of \(\cos 2x = 2\cos^2 x - 1\) in an attempt to prove the identity.	M1
Use of \(\sin 2x = 2\sin x\cos x\) in an attempt to prove the identity.	M1
\(\dfrac{1}{2}(\cos 2x - \sin 2x) = \dfrac{1}{2}(2\cos^2 x - 1 - 2\sin x\cos x) = \cos^2 x - \cos x\sin x - \dfrac{1}{2}\) * cso	A1
	(3)

Mark scheme (b)

Scheme	Marks
\(\cos\theta(\cos\theta - \sin\theta) = \dfrac{1}{2}\) Using (a)(i)	M1
\(\cos^2\theta - \cos\theta\sin\theta - \dfrac{1}{2} = 0\)
\(\dfrac{1}{2}(\cos 2\theta - \sin 2\theta) = 0\) Using (a)(ii)	M1
\(\cos 2\theta = \sin 2\theta\) *	A1
	(3)

Mark scheme (c)

Scheme	Marks
\(\tan 2\theta = 1\)	M1
\(2\theta = \dfrac{\pi}{4}, \left(\dfrac{5\pi}{4}, \dfrac{9\pi}{4}, \dfrac{13\pi}{4}\right)\) any one correct value of \(2\theta\)	A1
\(\theta = \dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{8}, \dfrac{13\pi}{8}\) Obtaining at least 2 solutions in range	M1
The 4 correct solutions	A1
	(4)
	(12 marks)

Notes

If decimals \((0.393, 1.963, 3.534, 5.105)\) or degrees \((22.5^\circ, 112.5^\circ, 202.5^\circ, 292.5^\circ)\) are given, but all 4 solutions are found, penalise one A mark only.
Ignore solutions out of range.