C3 June 2016 Q3

EdexcelOld spec10 marksTrigonometry

(a) Express \(2\cos\theta - \sin\theta\) in the form \(R\cos(\theta + \alpha)\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha < 90^\circ\). Give the exact value of \(R\) and give the value of \(\alpha\) to 2 decimal places. (3)

(b) Hence solve, for \(0 \leqslant \theta < 360^\circ\),\[\frac{2}{2\cos\theta - \sin\theta - 1} = 15\]Give your answers to one decimal place. (5)

(c) Use your solutions to parts (a) and (b) to deduce the smallest positive value of \(\theta\) for which\[\frac{2}{2\cos\theta + \sin\theta - 1} = 15\]Give your answer to one decimal place. (2)

Mark scheme (a)

Scheme	Marks
\(R = \sqrt{5}\)	B1
\(\tan\alpha = \dfrac{1}{2} \Rightarrow \alpha = 26.57^\circ\)	M1A1
	(3)

Notes

B1: \(R = \sqrt{5}\). Condone \(R = \pm\sqrt{5}\) Ignore decimals

M1: \(\tan\alpha = \pm\dfrac{1}{2}, \tan\alpha = \pm\dfrac{2}{1} \Rightarrow \alpha = \ldots\)
If their value of \(R\) is used to find the value of \(\alpha\) only accept \(\cos\alpha = \pm\dfrac{2}{R}\) OR \(\sin\alpha = \pm\dfrac{1}{R} \Rightarrow \alpha = \ldots\)

A1: \(\alpha = \text{awrt } 26.57^\circ\)

Mark scheme (b)

Scheme	Marks
\(\dfrac{2}{2\cos\theta - \sin\theta - 1} = 15 \Rightarrow \dfrac{2}{\sqrt{5}\cos(\theta + 26.6^\circ) - 1} = 15\)
\(\Rightarrow \cos(\theta + 26.6^\circ) = \dfrac{17}{15\sqrt{5}} = (\textit{awrt } 0.507)\)	M1A1
\(\theta + 26.57^\circ = 59.54^\circ\) \(\Rightarrow \theta = \textit{awrt } 33.0^\circ\) or \(\textit{awrt } 273.9^\circ\)	A1
\(\theta + 26.6^\circ = 360^\circ - \text{their '}59.5^\circ\text{'}\)	dM1
\(\Rightarrow \theta = \textit{awrt } 273.9^\circ\) and \(\textit{awrt } 33.0^\circ\)	A1
	(5)

Notes

M1: Attempts to use part (a) \(\Rightarrow \cos(\theta \pm \text{their } 26.6^\circ) = K\), \(|K| \leqslant 1\)

A1: \(\cos(\theta \pm \text{their } 26.6^\circ) = \dfrac{17}{15\sqrt{5}} = (\text{awrt } 0.507)\). Can be implied by \((\theta \pm \text{their } 26.6^\circ) = \text{awrt } 59.5^\circ / 59.6^\circ\)

A1: One solution correct, \(\theta = \textit{awrt } 33.0^\circ\) or \(\theta = \textit{awrt } 273.9^\circ\) Do not accept 33 for 33.0.

dM1: Obtains a second solution in the range. It is dependent upon having scored the previous M.
Usually for \(\theta \pm \text{their } 26.6^\circ = 360^\circ - \text{their } 59.5^\circ \Rightarrow \theta = \ldots\)

A1: Both solutions \(\theta = \textit{awrt } 33.0^\circ\) and \(\textit{awrt } 273.9^\circ\). Do not accept 33 for 33.0.
Extra solutions inside the range withhold this A1. Ignore solutions outside the range \(0 \leqslant \theta < 360^\circ\)

Mark scheme (c)

Scheme	Marks
\(\theta - \text{their } 26.57^\circ = \text{their } 59.54^\circ \Rightarrow \theta = \ldots\)	M1
\(\theta = \text{awrt } 86.1^\circ\)	A1
	(2)
	(10 marks)

Notes

M1: \(\theta - \text{their } 26.57^\circ = \text{their } 59.54^\circ \Rightarrow \theta = \ldots\)
Alternatively \(-\theta + \text{their } 26.6^\circ = -\text{their } 59.5^\circ \Rightarrow \theta = \ldots\)
If the candidate has an incorrect sign for \(\alpha\), for example they used \(\cos(\theta - 26.57^\circ)\) in part (b) it would be scored for \(\theta + \text{their } 26.57^\circ = \text{their } 59.54^\circ \Rightarrow \theta = \ldots\)

A1: awrt \(86.1^\circ\) ONLY. Allow both marks following a correct (a) and (b)
They can restart the question to achieve this result. Do not award if 86.1 was their smallest answer in (b). This occurs when they have \(\cos(\theta - 26.57^\circ)\) instead of \(\cos(\theta + 26.57^\circ)\) in (b)

Answers in radians: Withhold only one A mark, the first time a solution in radians appears
FYI (a) \(\alpha = 0.46\) (b) \(\theta_1 = \text{awrt } 0.58\) and \(\theta_2 = \text{awrt } 4.78\) (c) \(\theta_3 = \text{awrt } 1.50\). Require 2 dp accuracy