C3 June 2006 Q8

(a) Given that \(\cos A = \dfrac{3}{4}\), where \(270^\circ \lt A \lt 360^\circ\), find the exact value of \(\sin 2A\). (5)

(b)

(i) Show that \(\cos\left(2x + \dfrac{\pi}{3}\right) + \cos\left(2x - \dfrac{\pi}{3}\right) \equiv \cos 2x\). (3)

Given that\[y = 3\sin^2 x + \cos\left(2x + \frac{\pi}{3}\right) + \cos\left(2x - \frac{\pi}{3}\right),\]

(ii) show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sin 2x\). (4)

Mark scheme (a)

Scheme	Marks
Method for finding \(\sin A\)	M1
\(\sin A = -\dfrac{\sqrt{7}}{4}\)	A1 A1
Use of \(\sin 2A \equiv 2\sin A\cos A\)	M1
\(\sin 2A = -\dfrac{3\sqrt{7}}{8}\) or equivalent exact	A1ft
	(5)

First A1 for \(\dfrac{\sqrt{7}}{4}\), exact.
Second A1 for sign (even if dec. answer given)

A1ft: \(\pm\) f.t. Requires exact value, dependent on 2nd M

Mark scheme (b)

Scheme	Marks
(i) \(\cos\left(2x + \dfrac{\pi}{3}\right) + \cos\left(2x - \dfrac{\pi}{3}\right) \equiv \cos 2x\cos\dfrac{\pi}{3} - \sin 2x\sin\dfrac{\pi}{3} + \cos 2x\cos\dfrac{\pi}{3} + \sin 2x\sin\dfrac{\pi}{3}\)	M1
\(\equiv 2\cos 2x\cos\dfrac{\pi}{3}\) [This can be just written down (using factor formulae) for M1A1]	A1
\(\equiv \cos 2x\) AG	A1*
	(3)
(ii) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\sin x\cos x - 2\sin 2x\) or \(6\sin x\cos x - 2\sin\left(2x + \dfrac{\pi}{3}\right) - 2\sin\left(2x - \dfrac{\pi}{3}\right)\)	B1 B1
\(= 3\sin 2x - 2\sin 2x\)	M1
\(= \sin 2x\) AG	A1*
	(4)
	(12 marks)

(i) M1A1 earned, if \(\equiv 2\cos 2x\cos\dfrac{\pi}{3}\) just written down, using factor theorem
Final A1* requires some working after first result.

(ii) First B1 for \(6\sin x\cos x\); second B1 for remaining term(s)