June 2023 Paper 1 Q7

OCR ACurrent spec9 marksTrigonometry

(a) Use the result \(\cos(A + B) = \cos A\cos B - \sin A\sin B\) to show that \[\cos(A - B) = \cos A\cos B + \sin A\sin B.\] [2]

The function \(\mathrm{f}(\theta)\) is defined as \(\cos(\theta + 30^\circ)\cos(\theta - 30^\circ)\), where \(\theta\) is in degrees.

(b) Show that \(\mathrm{f}(\theta) = \cos^2\theta - \dfrac{1}{4}\). [3]

(c)

(i) Determine the following.

The maximum value of \(\mathrm{f}(\theta)\)
The smallest positive value of \(\theta\) for which this maximum value occurs

[2]

(ii) Determine the following.

The minimum value of \(\mathrm{f}(\theta)\)
The smallest positive value of \(\theta\) for which this minimum value occurs

[2]

Mark scheme (a)

Scheme	Marks	AO
\(\cos(A - B) = \cos A\cos(-B) - \sin A\sin(-B)\)	M1	2.1
\(\cos(-B) = \cos B,\ \sin(-B) = -\sin B\), \(\cos(A - B) = \cos A\cos B - \sin A(-\sin B)\) \(\cos(A - B) = \cos A\cos B + \sin A\sin B\) A.G.	A1	2.4
	[2]

Notes

M1: Replace \(B\) with \(-B\) in given identity

A1: State \(\cos(-B) = \cos B\) and \(\sin(-B) = -\sin B\), and conclude with correct identity
Condone \(-\sin A\sin(-B)\) becoming \(\sin A\sin B\) with no intermediate step
\(\cos(-B) = \cos B,\ \sin(-B) = -\sin B\) must be stated, but no justification needed

Mark scheme (b)

Scheme	Marks	AO
\(\left(\frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta\right)\left(\frac{\sqrt{3}}{2}\cos\theta + \frac{1}{2}\sin\theta\right)\)	B1	2.1
\(\frac{3}{4}\cos^2\theta - \frac{1}{4}\sin^2\theta\)	M1	2.1
\(\frac{3}{4}\cos^2\theta - \frac{1}{4}\left(1 - \cos^2\theta\right)\) \(\cos^2\theta - \frac{1}{4}\) A.G.	A1	2.1
	[3]

Notes

B1: Use correct identities, with exact trig values, to obtain a correct expression
Allow BOD for ambiguous positioning of + and – signs in a product, but penalise explicit errors if a single identity is seen in isolation
If expansion done before exact trig values used, then the expression must still be correct at the point that the B1 is awarded

M1: Expand brackets
May be recognised as difference of two squares so no need to see \(\frac{\sqrt{3}}{4}\cos\theta\sin\theta - \frac{\sqrt{3}}{4}\cos\theta\sin\theta\)
To obtain answer of form \(a\cos^2\theta - b\sin^2\theta\ (a \gt 0,\ b \gt 0)\), with possibly \(c\cos\theta\sin\theta - c\cos\theta\sin\theta\) also present

A1: Use Pythagorean identity and simplify to given answer
www eg if middle terms shown for expansion, then these must be correct

Mark scheme (c)

Scheme	Marks	AO
(i) max value is \(\frac{3}{4}\)	B1	1.1
when \(\theta\) is \(180^\circ\)	B1	1.1
	[2]
(ii) min value is \(-\frac{1}{4}\)	B1	1.1
when \(\theta\) is \(90^\circ\)	B1	1.1
	[2]

Notes

(c)(i)

B1: Correct max value

B1: Correct angle
B0 if any extra angles given
Must be ‘positive’ so B0 for \(0^\circ\)
Must be in degrees

Marks are independent

(c)(ii)

B1: Correct min value

B1: Correct angle
B0 if any extra angles given
Must be in degrees
SC If angles in both parts are correct, but in radians, then penalise only once (mark as B0 in (i) and B1 in (ii))

Marks are independent