June 2024 Paper 3 Q7
7 Prove that \(\sin 8\theta \tan 4\theta + \cos 8\theta = 1\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(2\sin 4\theta\cos 4\theta\tan 4\theta + \cos 8\theta\) | M1 | 3.1a |
| \(2\sin 4\theta\cos 4\theta\dfrac{\sin 4\theta}{\cos 4\theta} + \cos 8\theta\) | M1 | 3.1a |
| \(2\sin^2 4\theta + 1 - 2\sin^2 4\theta = 1\) | A1 | 2.1 |
| [3] |
Notes
M1: e.g. Use of \(\sin 2A = 2\sin A\cos A\)
Use of means more than just quoting the formula - usually it will be part of an argument
Small angle approximations score M0
M1: e.g. Use of \(\tan A = \dfrac{\sin A}{\cos A}\)
The 2 double angle formulas along with the tan identity may be applied in any order. If proof is sound award the 3 marks for equivalent steps.
A1: e.g. Use of any correct identity for \(\cos 2A\) and convincing completion to given result
Working it through as an equation with the ‘ = 1 ‘ is not ‘convincing’
Alternative incorrect approach.
If \(\sin 8\theta = 4\sin 2\theta\), and \(\sin 2\theta\) is then expanded then SC1 can be awarded. A similar approach applies to \(\cos 8\theta\). However the SC1 can only be awarded once.
Additional guidance
The first M1 can be awarded for using \(\sin 8\theta = 2\sin 4\theta\cos 4\theta\) or for using \(\cos 8\theta = 1 - 2\sin^2 4\theta\) (or equivalent form) or for using \(\tan 4\theta = \sin 4\theta / \cos 4\theta\). The mark is not available for just writing the formula down.
The second M1 is for using another one.
The final A1 is for using the third one and completing the solution to show it equals 1. (We saw an incorrect factorisation of \(\sin 8\theta = 4\sin 2\theta = 4 \times 2\sin\theta\cos\theta\) which qualifies for SC1. A similar approach can be used for \(\cos 8\theta\) but the SC1 can only be awarded once).
Some candidates are using small angle approximations which does lead to an answer of 1, but is the incorrect method, so scores M0M0A0.