C3 June 2010 Q1

EdexcelOld spec5 marksProof Trigonometry

(a) Show that\[\frac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta\] (2)

(b) Hence find, for \(-180^\circ \leqslant \theta \lt 180^\circ\), all the solutions of\[\frac{2\sin 2\theta}{1 + \cos 2\theta} = 1\]Give your answers to 1 decimal place. (3)

Mark scheme (a)

Scheme	Marks
\(\dfrac{2\sin\theta\cos\theta}{1 + 2\cos^2\theta - 1}\)	M1
\(\dfrac{\cancel{2}\sin\theta\cancel{\cos\theta}}{\cancel{2}\cos\theta\cancel{\cos\theta}} = \tan\theta\) (as required) AG	A1 cso
	(2)

Notes

(a) M1: Uses both a correct identity for \(\sin 2\theta\) and a correct identity for \(\cos 2\theta\). Also allow a candidate writing \(1 + \cos 2\theta = 2\cos^2\theta\) on the denominator. Also note that angles must be consistent in when candidates apply these identities.

A1: Correct proof. No errors seen.

Mark scheme (b)

Scheme	Marks
\(2\tan\theta = 1 \Rightarrow \tan\theta = \dfrac{1}{2}\)	M1
\(\theta_1 = \text{awrt } 26.6^\circ\)	A1
\(\theta_2 = \text{awrt } -153.4^\circ\)	A1ft
	(3)
	(5 marks)

Notes

(b) 1^st M1 for either \(2\tan\theta = 1\) or \(\tan\theta = \dfrac{1}{2}\), seen or implied.

A1: awrt 26.6

A1ft: awrt \(-153.4^\circ\) or \(\theta_2 = -180^\circ + \theta_1\)

Special Case: For candidate solving, \(\tan\theta = k\), where \(k \ne \tfrac{1}{2}\), to give \(\theta_1\) and \(\theta_2 = -180^\circ + \theta_1\), then award M0A0B1 in part (b).

Special Case: Note that those candidates who writes \(\tan\theta = 1\), and gives ONLY two answers of \(45^\circ\) and \(-135^\circ\) that are inside the range will be awarded SC M0A0B1.