2025 DSS PRELIMS P2 Q6

The diagram below shows an isosceles triangle \(ABC\), where \(AB = AC = 2\) cm, and a square \(BCDE\). \(M\) is the midpoint of \(BC\). Angle \(BAM = \theta\) and can vary.

1. Show that the total area of \(ABEDC\) is \(16\sin^2\theta + 4\sin\theta\cos\theta\) cm\(^2\).[3]
2. Show that \(16\sin^2\theta + 4\sin\theta\cos\theta = 2\sin 2\theta - 8\cos 2\theta + 8\).[2]
3. Express \(2\sin 2\theta - 8\cos 2\theta\) in the form \(R\sin(2\theta - \alpha)\), where \(R > 0\) and \(0^\circ < \alpha < 90^\circ\).[3]