C3 June 2017 Q9

Notes

M1: Uses a correct double angle identity involving \(\sin 2x\) Accept \(\sin(x + x) = \sin x\cos x + \cos x\sin x\)

M1: Uses \(\tan x = \dfrac{\sin x}{\cos x}\) with \(\sin 2x = 2\sin x\cos x\) and attempts to combine the two terms using a common denominator. This can be awarded on two separate terms with a common denominator.
Alternatively uses \(\sin x = \tan x\cos x\) and attempts to combine two terms using factorisation of \(\tan x\)

dM1: Both M's must have been scored. Uses a correct double angle identity involving \(\cos 2x\).

A1*: A fully correct solution with no errors or omissions. All notation must be correct and variables must be consistent
Withhold this mark if for instance they write \(\tan x = \dfrac{\sin}{\cos}\)
If the candidate \(\times\cos x\) on line 1 and/or \(\div\sin x\) they cannot score any more than one mark unless they are working with both sides of the equation or it is fully explained.

Alternatives to parts (a) and (b): (a) Alt 1

a) Alt 1 Starting from the rhs

M1: Uses a correct double angle identity for \(\cos 2x\). Accept any correct version including \(\cos(x + x) = \cos x\cos x - \sin x\sin x\)

M1: Uses \(\tan x = \dfrac{\sin x}{\cos x}\) with \(\cos 2x = 2\cos^2 x - 1\) and attempts to multiply out the bracket

dM1: Both M's must have been scored. It is for using \(2\sin x\cos x = \sin 2x\)

A1*: A fully correct solution with no errors or omissions. All notation must be correct and variables must be consistent.
See Main scheme for how to deal with candidates who \(\div\tan x\)

(a) Alt 2

a) Alt 2 Candidates who use both sides

M1: Uses a correct double angle identity involving \(\sin 2x\) or \(\cos 2x\). Can be scored from either side
Accept \(\sin(x + x) = \sin x\cos x + \cos x\sin x\) or \(\cos(x + x) = \cos x\cos x - \sin x\sin x\)

M1: Uses \(\tan x = \dfrac{\sin x}{\cos x}\) with \(\cos 2x = 2\cos^2 x - 1\) and cancels the \(\tan x\) term from both sides

dM1: Uses a correct double angle identity involving \(\sin 2x\) Both previous M's must have been scored

A1*: A fully correct solution with no errors or omissions AND statement "hence true", "a tick", "QED".
All notation must be correct and variables must be consistent

Notes

M1: The \(\tan x\) must be cancelled or factorised out to produce \(\cos 2x - 3\sin x = 0\) or \(\dfrac{\cos 2x}{\sin x} = 3\) oe Condone slips

M1: Uses \(\cos 2x = 1 - 2\sin^2 x\) to form a 3TQ=0 in \(\sin x\) The = 0 may be implied by later work

M1: Uses the formula/completion of square or GC with invsin to produce at least one value for \(x\)
It may be implied by one correct value.
This mark can be scored from factorisation of their 3TQ in \(\sin x\) but only if their quadratic factorises.

A1: Two of \(x = 0, 180^\circ, \text{awrt } 16.3^\circ, \text{awrt } 163.7^\circ\) or in radians two of awrt 0.28, 2.86, 0 and \(\pi\) or 3.14
This mark can be awarded as a SC for those students who just produce \(0, 180^\circ\) ( or 0 and \(\pi\)) from \(\tan x = 0\) or \(\sin x = 0\).

A1: All four values in degrees \(x = 0, 180^\circ, \text{awrt } 16.3^\circ, \text{awrt } 163.7^\circ\) and no extra's inside the range \(0 \leqslant x < 360^\circ\).
Condone \(0 = 0.0\) and \(180^\circ = 180.0^\circ\) Ignore any roots outside range.

It is possible to solve part (b) without using the given identity. There are various ways of doing this, one of which is shown below.