June 2024 Paper 2 Q6
6 It is given that
\[(2\sin\theta + 3\cos\theta)^2 + (6\sin\theta - \cos\theta)^2 = 30\]and that \(\theta\) is obtuse.
Find the exact value of \(\sin\theta\).
Fully justify your answer. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Expands brackets, at least one completely correct | M1 | 1.1a |
| Expands both brackets correctly | A1 | 1.1b |
| Uses \(\sin^2\theta + \cos^2\theta = 1\) correctly to eliminate \(\cos^2\theta\) or \(\sin^2\theta\) from their equation PI by \(30\sin^2\theta + 10 = 30\) | M1 | 3.1a |
| Obtains an equation of the form \(\sin^2\theta = k\) or \(\cos^2\theta = k\) where \(0 \lt k \lt 1\) PI by \(\sin\theta = \sqrt{k}\) or \(\sin\theta = -\sqrt{k}\) | M1 | 1.1a |
| Obtains \(\sin\theta = \pm\sqrt{\dfrac{2}{3}}\) | A1 | 1.1b |
| Completes reasoned argument to obtain \(\sin\theta = \sqrt{\dfrac{2}{3}}\) and explains why \(\sin\theta = \sqrt{\dfrac{2}{3}}\) Must come from \(\sin\theta = \pm\sqrt{\dfrac{2}{3}}\) FT their \(\sin^2\theta = k\) where \(0 \lt k \lt 1\) | R1F | 2.4 |
| (6 marks) |
Typical solution
\[4\sin^2\theta + 12\sin\theta\cos\theta + 9\cos^2\theta\]\[+ 36\sin^2\theta - 12\sin\theta\cos\theta + \cos^2\theta = 30\]\[40\sin^2\theta + 10\cos^2\theta = 30\]\[40\sin^2\theta + 10\left(1 - \sin^2\theta\right) = 30\]\[30\sin^2\theta + 10 = 30\]\[\sin^2\theta = \frac{2}{3}\]\[\sin\theta = \pm\frac{\sqrt{6}}{3}\]since \(\theta\) is obtuse \(\sin\theta = \dfrac{\sqrt{6}}{3}\)