C3 June 2012 Q5

EdexcelOld spec9 marksTrigonometry

(a) Express \(4\operatorname{cosec}^2 2\theta - \operatorname{cosec}^2\theta\) in terms of \(\sin\theta\) and \(\cos\theta\). (2)

(b) Hence show that \[4\operatorname{cosec}^2 2\theta - \operatorname{cosec}^2\theta = \sec^2\theta\] (4)

(c) Hence or otherwise solve, for \(0 \lt \theta \lt \pi\), \[4\operatorname{cosec}^2 2\theta - \operatorname{cosec}^2\theta = 4\] giving your answers in terms of \(\pi\). (3)

Mark scheme (a)

Scheme	Marks
\(4\operatorname{cosec}^2 2\theta - \operatorname{cosec}^2\theta = \dfrac{4}{\sin^2 2\theta} - \dfrac{1}{\sin^2\theta}\)
\(= \dfrac{4}{(2\sin\theta\cos\theta)^2} - \dfrac{1}{\sin^2\theta}\)	B1 B1
	(2)

Notes

Note (a) and (b) can be scored together

B1 One term correct. Eg. writes \(4\operatorname{cosec}^2 2\theta\) as \(\dfrac{4}{(2\sin\theta\cos\theta)^2}\) or \(\operatorname{cosec}^2\theta\) as \(\dfrac{1}{\sin^2\theta}\). Accept terms like \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta = 1 + \dfrac{\cos^2\theta}{\sin^2\theta}\). The question merely asks for an expression in \(\sin\theta\) and \(\cos\theta\)

B1 A fully correct expression in \(\sin\theta\) and \(\cos\theta\). Eg. \(\dfrac{4}{(2\sin\theta\cos\theta)^2} - \dfrac{1}{\sin^2\theta}\) Accept equivalents
Allow a different variable say \(x\)’s instead of \(\theta\)’s but do not allow mixed units.

Mark scheme (b)

Scheme	Marks
\(\dfrac{4}{(2\sin\theta\cos\theta)^2} - \dfrac{1}{\sin^2\theta} = \dfrac{\cancel{4}}{\cancel{4}\sin^2\theta\cos^2\theta} - \dfrac{1}{\sin^2\theta}\)
\(= \dfrac{1}{\sin^2\theta\cos^2\theta} - \dfrac{\cos^2\theta}{\sin^2\theta\cos^2\theta}\)	M1
Using \(1 - \cos^2\theta = \sin^2\theta\) \(= \dfrac{\cancel{\sin^2\theta}}{\cancel{\sin^2\theta}\cos^2\theta}\)	M1
\(= \dfrac{1}{\cos^2\theta} = \sec^2\theta\)	M1A1*
	(4)

Notes

M1 Attempts to combine their expression in \(\sin\theta\) and \(\cos\theta\) using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted

M1 Attempts to form a ‘single’ term on the numerator by using the identity \(1 - \cos^2\theta = \sin^2\theta\)

M1 Cancels correctly by \(\sin^2\theta\) terms and replaces \(\dfrac{1}{\cos^2\theta}\) with \(\sec^2\theta\)

A1* Cso. This is a given answer. All aspects must be correct

IF IN ANY DOUBT SEND TO REVIEW OR CONSULT YOUR TEAM LEADER

Mark scheme (c)

Scheme	Marks
\(\sec^2\theta = 4 \Rightarrow \sec\theta = \pm 2 \Rightarrow \cos\theta = \pm\dfrac{1}{2}\)	M1
\(\theta = \dfrac{\pi}{3}, \dfrac{2\pi}{3}\)	A1,A1
	(3)
	(9 marks)

Notes

M1 For \(\sec^2\theta = 4\) leading to a solution of \(\cos\theta\) by taking the root and inverting in either order.
Similarly accept \(\tan^2\theta = 3\), \(\sin^2\theta = \dfrac{3}{4}\) leading to solutions of \(\tan\theta, \sin\theta\). Also accept \(\cos 2\theta = -\dfrac{1}{2}\)

A1 Obtains one correct answer usually \(\theta = \dfrac{\pi}{3}\) Do not accept decimal answers or degrees

A1 Obtains both correct answers. \(\theta = \dfrac{\pi}{3}, \dfrac{2\pi}{3}\) Do not award if there are extra solutions inside the range.
Ignore solutions outside the range.