OCR MEI Paper 3 2024 June — Question 7 3 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2024
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeProve algebraic trigonometric identity
DifficultyChallenging +1.2 This requires proving a non-standard identity involving double/multiple angles (8θ and 4θ). Students must recognize to use double angle formulas strategically, likely expressing sin 8θ and cos 8θ in terms of 4θ, then simplify using tan 4θ = sin 4θ/cos 4θ. While systematic, it requires more algebraic manipulation and insight than routine identity proofs, placing it moderately above average difficulty.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities

7 Prove that \(\sin 8 \theta \tan 4 \theta + \cos 8 \theta = 1\).

Question 7:
AnswerMarks Guidance
Using \(\sin 8\theta = 2\sin 4\theta\cos 4\theta\) OR \(\cos 8\theta = 1 - 2\sin^2 4\theta\) or equivalent OR \(\tan 4\theta = \sin 4\theta / \cos 4\theta\)M1 Not awarded for just writing formula down
Using another identityM1
Using third identity and completing to show equals 1A1 SC1 for incorrect factorisation of \(\sin 8\theta = 4\sin 2\theta = 4 \times 2\sin\theta\cos\theta\) (SC1 awarded once only)
*Note: Small angle approximations lead to answer of 1 but score M0M0A0.*
Question 8a:
AnswerMarks Guidance
\(R\cos\alpha = \sqrt{3}\) and \(R\sin\alpha = 1\)M1 Must be in terms of \(\alpha\)
\(R = 2\)B1
\(\alpha = \frac{\pi}{6}\)M1A1 A1 dependent on scoring M2; award M0 if R is missing
Question 8b:
AnswerMarks Guidance
Using *their* result from (a) to get equation equal to \(\sqrt{3}\)M1 Incorrect \(\alpha\) from Q8a can still access next two M1s
*Their* trig expression \(=\) a value (probably \(\frac{\sqrt{3}}{2}\))M1
Solving *their* equation for at least one solutionM1
Two correct values in radians, no othersA1
*Note: Squaring both sides may produce extraneous solutions which must be discarded. Maximum M3A0 if not discarded, M2A0 if not discarded.*
Question 9a:
AnswerMarks
Finding either \(f(1)\) or \(f(2)\)B1
Finding other value AND stating sign changeB1
Question 9b:
AnswerMarks Guidance
Any value between 1.125 and 1.25B1 Halving the range and using 1.0625 scores B0
Question 9c:
AnswerMarks Guidance
Valid iteration values e.g. \(f(1.15) = -0.09\), \(f(1.16) = -0.025\), \(f(1.163) = -0.05\), \(f(1.164) = 0.0015\); final answer 1.2 (1dp)B1 Must give a value for \(f\); 1.2 with no other working scores B0
Question 9di:
AnswerMarks Guidance
Change in signB1 Change in sign is all that is required
Question 9dii:
AnswerMarks Guidance
Various valid commentsB1 ISW after correct answer seen
## Question 7:
| Using $\sin 8\theta = 2\sin 4\theta\cos 4\theta$ OR $\cos 8\theta = 1 - 2\sin^2 4\theta$ or equivalent OR $\tan 4\theta = \sin 4\theta / \cos 4\theta$ | M1 | Not awarded for just writing formula down |
| Using another identity | M1 | |
| Using third identity and completing to show equals 1 | A1 | SC1 for incorrect factorisation of $\sin 8\theta = 4\sin 2\theta = 4 \times 2\sin\theta\cos\theta$ (SC1 awarded once only) |

*Note: Small angle approximations lead to answer of 1 but score M0M0A0.*

## Question 8a:
| $R\cos\alpha = \sqrt{3}$ and $R\sin\alpha = 1$ | M1 | Must be in terms of $\alpha$ |
| $R = 2$ | B1 | |
| $\alpha = \frac{\pi}{6}$ | M1A1 | A1 dependent on scoring M2; award M0 if R is missing |

## Question 8b:
| Using *their* result from (a) to get equation equal to $\sqrt{3}$ | M1 | Incorrect $\alpha$ from Q8a can still access next two M1s |
| *Their* trig expression $=$ a value (probably $\frac{\sqrt{3}}{2}$) | M1 | |
| Solving *their* equation for at least one solution | M1 | |
| Two correct values in radians, no others | A1 | |

*Note: Squaring both sides may produce extraneous solutions which must be discarded. Maximum M3A0 if not discarded, M2A0 if not discarded.*

## Question 9a:
| Finding either $f(1)$ or $f(2)$ | B1 | |
| Finding other value AND stating sign change | B1 | |

## Question 9b:
| Any value between 1.125 and 1.25 | B1 | Halving the range and using 1.0625 scores B0 |

## Question 9c:
| Valid iteration values e.g. $f(1.15) = -0.09$, $f(1.16) = -0.025$, $f(1.163) = -0.05$, $f(1.164) = 0.0015$; final answer 1.2 (1dp) | B1 | Must give a value for $f$; 1.2 with no other working scores B0 |

## Question 9di:
| Change in sign | B1 | Change in sign is all that is required |

## Question 9dii:
| Various valid comments | B1 | ISW after correct answer seen |
7 Prove that $\sin 8 \theta \tan 4 \theta + \cos 8 \theta = 1$.

\hfill \mbox{\textit{OCR MEI Paper 3 2024 Q7 [3]}}