June 2025 Paper 2 Q4
4
| Scheme | Marks | AO |
|---|---|---|
| LHS \(\equiv \cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta\) | M1 | 2.1 |
| \([\cos^2\theta + \sin^2\theta \equiv 1\) hence\(]\) \((\cos\theta + \sin\theta)^2 \equiv 1 + \sin 2\theta\) AG | A1 | 2.4 |
| [2] |
Notes
Allow “=” throughout Q4
M1: M1 for expanding correctly
A1: Must see final statement www
Alternative method
| Scheme | Marks | AO |
|---|---|---|
| \(1 + \sin 2\theta \equiv \cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta\) | M1 | |
| \(1 + \sin 2\theta \equiv (\cos\theta + \sin\theta)^2\) AG | A1 |
A1: Must see final statement www
| Scheme | Marks | AO |
|---|---|---|
| LHS \(\equiv \left[\dfrac{1}{\cos 2\theta} +\right] \dfrac{\sin 2\theta}{\cos 2\theta}\) | M1 | 3.1a |
| \(\equiv \dfrac{1 + \sin 2\theta}{\cos 2\theta}\) or \(\dfrac{1 + \sin 2\theta}{\cos^2\theta - \sin^2\theta}\) | A1 | 1.1 |
| \(\equiv \dfrac{(\cos\theta + \sin\theta)^2}{\cos^2\theta - \sin^2\theta}\) | A1 | 2.1 |
| \(\equiv \dfrac{(\cos\theta + \sin\theta)^2}{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}\) \(\equiv \dfrac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\) AG | A1 | 2.2a |
| [4] |
Notes
M1: For writing \(\tan 2\theta\) as \(\dfrac{\sin 2\theta}{\cos 2\theta}\) or \(\dfrac{2\tan\theta}{1 - \tan^2\theta}\). May be seen embedded or implied by next A1 (Allow this mark if \(\sec 2\theta\) still used)
A1: May see \(1 + 2\cos\theta\sin\theta\) for the numerator here
A1: Allow A1 for correct numerator or correct denominator seen in terms of \(\theta\), not \(2\theta\). Allow this mark even if subsequently \(\surd\) incorrectly.
A1: correctly obtained www, ie must see previous line
Alternative method
| Scheme | Marks | AO |
|---|---|---|
| RHS \(\equiv \dfrac{(\cos\theta + \sin\theta)^2}{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}\) | M1 | |
| \(\equiv \dfrac{1 + \sin 2\theta}{\cos^2\theta - \sin^2\theta}\) | A1 | |
| \(\equiv \dfrac{1 + \sin 2\theta}{\cos 2\theta}\) | A1 | |
| \(\equiv \sec 2\theta + \tan 2\theta\) AG | A1 |
M1: or \(\dfrac{(\cos\theta + \sin\theta)^2}{\cos^2\theta - \sin^2\theta}\) seen
A1: correct numerator seen
A1: correct denominator seen in terms of \(2\theta\), not \(\theta\)
A1: correctly obtained www, ie must see previous line
Alternative method using \(t\)-formulae
| Scheme | Marks | AO |
|---|---|---|
| LHS \(\equiv \dfrac{1 + t^2}{1 - t^2} + \dfrac{2t}{1 - t^2} \quad (t = \tan\theta)\) | M1 | |
| \(\equiv \left[\dfrac{(1 + t)^2}{1 - t^2}\right] \equiv \dfrac{(1 + t)^2}{(1 - t)(1 + t)}\) | A1 | |
| \(\equiv \dfrac{1 + t}{1 - t}\) | A1 | |
| \(\equiv \dfrac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\) AG | A1 |
A1: www