C3 January 2010 Q3

EdexcelOld spec9 marksTrigonometry

(a) Express \(5\cos x - 3\sin x\) in the form \(R\cos(x + \alpha)\), where \(R \gt 0\) and \(0 \lt \alpha \lt \tfrac{1}{2}\pi\). (4)

(b) Hence, or otherwise, solve the equation\[5\cos x - 3\sin x = 4\]for \(0 \leqslant x \lt 2\pi\), giving your answers to 2 decimal places. (5)

Mark scheme (a)

Scheme	Marks
\(5\cos x - 3\sin x = R\cos(x + \alpha),\ \ R \gt 0,\ 0 \lt x \lt \tfrac{\pi}{2}\)
\(5\cos x - 3\sin x = R\cos x\cos\alpha - R\sin x\sin\alpha\)
Equate \(\cos x\): \(5 = R\cos\alpha\) Equate \(\sin x\): \(3 = R\sin\alpha\)
\(R = \sqrt{5^2 + 3^2};= \sqrt{34}\ \{= 5.83095\ldots\}\)	M1; A1
\(\tan\alpha = \tfrac{3}{5} \Rightarrow \alpha = 0.5404195003\ldots^{c}\)	M1 A1
Hence, \(5\cos x - 3\sin x = \sqrt{34}\cos(x + 0.5404)\)
	(4)

Notes

M1: \(R^2 = 5^2 + 3^2\)

A1: \(\sqrt{34}\) or awrt 5.8

M1: \(\tan\alpha = \pm\tfrac{3}{5}\) or \(\tan\alpha = \pm\tfrac{5}{3}\) or \(\sin\alpha = \pm\frac{3}{\text{their } R}\) or \(\cos\alpha = \pm\frac{5}{\text{their } R}\)

A1: \(\alpha = \text{awrt } 0.54\) or \(\alpha = \text{awrt } 0.17\pi\) or \(\alpha = \dfrac{\pi}{\text{awrt } 5.8}\)

Mark scheme (b)

Scheme	Marks
\(5\cos x - 3\sin x = 4\) \(\sqrt{34}\cos(x + 0.5404) = 4\)
\(\cos(x + 0.5404) = \dfrac{4}{\sqrt{34}}\ \{= 0.68599\ldots\}\)	M1
\((x + 0.5404) = 0.814826916\ldots^{c}\)	M1
\(x = 0.2744\ldots^{c}\)	A1
\((x + 0.5404) = 2\pi - 0.814826916\ldots^{c}\ \{= 5.468358\ldots^{c}\}\)	ddM1
\(x = 4.9279\ldots^{c}\)	A1
Hence, \(x = \{0.27, 4.93\}\)
	(5)
	(9 marks)

Notes

M1: \(\cos(x \pm \text{their } \alpha) = \dfrac{4}{\text{their } R}\)

M1: For applying \(\cos^{-1}\left(\dfrac{4}{\text{their } R}\right)\)

A1: awrt \(0.27^{c}\)

ddM1: \(2\pi - \text{their } 0.8148\)

A1: awrt \(4.93^{c}\)

Part (b): If there are any EXTRA solutions inside the range \(0 \leqslant x \lt 2\pi\), then withhold the final accuracy mark if the candidate would otherwise score all 5 marks. Also ignore EXTRA solutions outside the range \(0 \leqslant x \lt 2\pi\).