C3 June 2012 Q8

EdexcelOld spec12 marksTrigonometry

8. \[\mathrm{f}(x) = 7\cos 2x - 24\sin 2x\]

Given that \(\mathrm{f}(x) = R\cos(2x + \alpha)\), where \(R \gt 0\) and \(0 \lt \alpha \lt 90^\circ\),

(a) find the value of \(R\) and the value of \(\alpha\). (3)

(b) Hence solve the equation \[7\cos 2x - 24\sin 2x = 12.5\] for \(0 \leqslant x \lt 180^\circ\), giving your answers to 1 decimal place. (5)

(c) Express \(14\cos^2 x - 48\sin x\cos x\) in the form \(a\cos 2x + b\sin 2x + c\), where \(a\), \(b\), and \(c\) are constants to be found. (2)

(d) Hence, using your answers to parts (a) and (c), deduce the maximum value of \[14\cos^2 x - 48\sin x\cos x\] (2)

Mark scheme (a)

Scheme	Marks
\(R = 25\)	B1
\(\tan\alpha = \dfrac{24}{7} \Rightarrow \alpha = (\text{awrt})73.7^\circ\)	M1A1
	(3)

Notes

B1 Accept 25, awrt 25.0, \(\sqrt{625}\). Condone \(\pm 25\)

M1 For \(\tan\alpha = \pm\dfrac{24}{7}\) \(\tan\alpha = \pm\dfrac{7}{24}\) \(\sin\alpha = \pm\dfrac{24}{\text{their } R}\), \(\cos\alpha = \pm\dfrac{7}{\text{their } R}\)

A1 \(\alpha = (\text{awrt})73.7^\circ\). The answer 1.287 (radians) is A0

Mark scheme (b)

Scheme	Marks
\(\cos(2x + \text{their } \alpha) = \dfrac{12.5}{\text{their } R}\)	M1
\(2x + \text{their } '\alpha' = 60^\circ\)	A1
\(2x + \text{their } '\alpha' = \text{their } 300^\circ \text{ or their } 420^\circ \Rightarrow x = ..\)	M1
\(x = \text{awrt } 113.1^\circ, 173.1^\circ\)	A1A1
	(5)

Notes

M1 For using part (a) and dividing by their R to reach \(\cos(2x + \text{their } \alpha) = \dfrac{12.5}{\text{their } R}\)

A1 Achieving \(2x + \text{their } \alpha = 60^{(0)}\). This can be implied by 113.1⁽⁰⁾/113.2⁽⁰⁾ or 173.1⁽⁰⁾/173.2⁽⁰⁾ or - 6.8⁽⁰⁾/ -6.85⁽⁰⁾/-6.9⁽⁰⁾

M1 Finding a secondary value of x from their principal value. A correct answer will imply this mark
Look for \(\dfrac{360 \pm \text{'their' principal value} \pm \text{'their' } \alpha}{2}\)

A1 \(x = \text{awrt } 113.1^\circ/113.2^\circ\) OR \(173.1^\circ/173.2^\circ\).

A1 \(x = \text{awrt } 113.1^\circ\) AND \(173.1^\circ\). Ignore solutions outside of range. Penalise this mark for extra solutions inside the range

Radian solutions- they will lose the first time it occurs (usually in a with 1.287 radians) Part b will then be marked as follows

M1 For using part (a) and dividing by their R to reach \(\cos(2x + \text{their } \alpha) = \dfrac{12.5}{\text{their } R}\)

A1 The correct principal value of \(\dfrac{\pi}{3}\) or awrt 1.05 radians. Accept 60⁽⁰⁾
This can be implied by awrt – 0.12 radians or awrt or 1.97 radians or awrt 3.02 radians

M1 Finding a secondary value of x from their principal value. A correct answer will imply this mark
Look for \(\dfrac{2\pi \pm \text{'their' principal value} \pm \text{'their' } \alpha}{2}\) Do not allow mixed units.

A1 \(x = \text{awrt } 1.97\) OR 3.02.

A1 \(x = \text{awrt } 1.97\) AND 3.02. Ignore solutions outside of range. Penalise this mark for extra solutions inside the range

Mark scheme (c)

Scheme	Marks
Attempts to use \(\cos 2x = 2\cos^2 x - 1\) AND \(\sin 2x = 2\sin x\cos x\) in the expression	M1
\(14\cos^2 x - 48\sin x\cos x = 7(\cos 2x + 1) - 24\sin 2x\) \(= 7\cos 2x - 24\sin 2x + 7\)	A1
	(2)

Notes

M1 Attempts to use \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x\cos x\) in expression.
Allow slips in sign on the \(\cos 2x\) term. So accept \(2\cos^2 x = \pm\cos 2x \pm 1\)

A1 Cao \(= 7\cos 2x - 24\sin 2x + 7\). The order of terms is not important. Also accept a=7, b=-24, c=7

Mark scheme (d)

Scheme	Marks
\(14\cos^2 x - 48\sin x\cos x = R\cos(2x + \alpha) + 7\)
Maximum value = ’\(R\)’+’\(c\)’	M1
= 32 cao	A1
	(2)
	(12 marks)

Notes

M1 This mark is scored for adding their R to their c

A1 cao 32