June 2024 Paper 3 Q8

OCR MEICurrent spec8 marksTrigonometry

8 In this question you must show detailed reasoning.

(a) Express \(\cos x + \sqrt{3}\sin x\) in the form \(R\sin(x + \alpha)\), where \(R \gt 0\) and \(0 \lt \alpha \lt \frac{1}{2}\pi\). Give the values of \(R\) and \(\alpha\) in exact form. [4]

(b) Hence solve the equation \(\cos x = \sqrt{3}(1 - \sin x)\) for values of \(x\) in the interval \(-\pi \leqslant x \leqslant \pi\). Give the roots of this equation in exact form. [4]

Mark scheme (a)

Scheme	Marks	AO
DR \(\left[\cos x + \sqrt{3}\sin x = R\sin(x + \alpha)\right]\) \(\Rightarrow R\cos\alpha = \sqrt{3}\), \(R\sin\alpha = 1\)	M1	1.1a
\(\tan\alpha = \dfrac{1}{\sqrt{3}}\)	M1	1.1 1.2
\([\alpha] = \dfrac{\pi}{6}\)	A1
\(\Rightarrow R^2 = 4,\ R = 2\)	B1	1.1
	[4]

Notes

M1: M0 if \(R\) missing but next M mark available.
Must be using \(\alpha\) and not \(x\).

M1: Allow other equivalent methods using their values of sin and cos e.g. drawing a right angled triangle or simply dividing the two equations. Allow equivalent surds

A1: Dependent on M2

B1: soi anywhere

Additional guidance

We are expecting to see the 2 equations \(R\cos\alpha = \sqrt{3}\) and \(R\sin\alpha = 1\) for the first M1. Must be in terms of \(\alpha\).

The B1 mark is for finding R=2. In addition to the method in the MS, they might find R first and use it to solve, say \(2\sin\alpha = 1\), leading to \(\alpha = \pi/6\) for the M1A1. The A1 is dependent on scoring M2.

Award M0 if the R is missing. They can still get next M1 but not the A1 (as an M0 has already been given – see section 6c of the marking instructions). M0M1A0B1 is likely to be quite common.

They can get M1 A0 if they draw a right-angled triangle and label it incorrectly giving \(\tan\alpha = \sqrt{3}\)

Mark scheme (b)

Scheme	Marks	AO
DR \(\cos x + \sqrt{3}\sin x = \sqrt{3} \Rightarrow 2\sin\left(x + \dfrac{\pi}{6}\right) = \sqrt{3}\)	M1	3.1a
\(\sin\left(x + \dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}\)	M1	1.1
\(x + \dfrac{\pi}{6} = \dfrac{\pi}{3}\) or \(\dfrac{2\pi}{3}\)	M1	1.1
\(x = \dfrac{\pi}{6}\), \(\dfrac{\pi}{2}\)	A1	1.1
	[4]

Notes

M1: Use of their result from (a)

M1: Value for a trig function following on from first step

M1: At least one value for their \(x + \dfrac{\pi}{6}\).
Condone use of inequality symbols if recovered.
Condone working consistently in degrees up to this point.

A1: Both roots correct in radians only.
If extra roots then A0
Candidates who square both sides and solve from there, should check their solutions and discard any erroneous ones. This gains a maximum of M3A0. If no check then M2A0 max.

Additional guidance

For the first M1 they use their result from part (a) to get an equation equal to \(\sqrt{3}\). An incorrect value of \(\alpha\) from Q8a means they can also access the next two M1s as per the MS.

The second M1 is for getting their trig expression = a value (probably \(\sqrt{3}/2\)).

The third M1 is for solving their equation to get at least one solution.

The A1 is given for the two correct values in radians and no others.

Note that candidates who square both sides and solve from there will get extraneous solutions so any of those must be discarded. There is a maximum of M3A0 if they do discard erroneous solutions but only M2A0 if they do not.